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- Introduction
- Linear Programming
- Norm
- Inner Product
- Minima and Maxima
- Convex Set
- Affine Set
- Convex Hull
- Caratheodory Theorem
- Weierstrass Theorem
- Closest Point Theorem
- Fundamental Separation Theorem
- Convex Cones
- Polar Cone
- Conic Combination
- Polyhedral Set
- Extreme point of a convex set
- Direction
- Convex & Concave Function
- Jensen's Inequality
- Differentiable Convex Function
- Sufficient & Necessary Conditions for Global Optima
- Quasiconvex & Quasiconcave functions
- Differentiable Quasiconvex Function
- Strictly Quasiconvex Function
- Strongly Quasiconvex Function
- Pseudoconvex Function
- Convex Programming Problem
- Fritz-John Conditions
- Karush-Kuhn-Tucker Optimality Necessary Conditions
- Algorithms for Convex Problems
Convex Optimization - Quick Guide
Convex Optimization - Introduction
This course is useful for the students who want to solve non-linear optimization problems that arise in various engineering and scientific applications. This course starts with basic theory of linear programming and will introduce the concepts of convex sets and functions and related terminologies to explain various theorems that are required to solve the non linear programming problems. This course will introduce various algorithms that are used to solve such problems. These type of problems arise in various applications including machine learning, optimization problems in electrical engineering, etc. It requires the students to have prior knowledge of high school maths concepts and calculus.
In this course, the students will learn to solve the optimization problems like minf(x) subject to some constraints.
These problems are easily solvable if the function f(x) is a linear function and if the constraints are linear. Then it is called a linear programming problem LPP. But if the constraints are non-linear, then it is difficult to solve the above problem. Unless we can plot the functions in a graph, then try to analyse the optimization can be one way, but we can't plot a function if it's beyond three dimensions. Hence there comes the techniques of non-linear programming or convex programming to solve such problems. In these tutorial, we will focus on learning such techniques and in the end, a few algorithms to solve such problems. first we will bring the notion of convex sets which is the base of the convex programming problems. Then with the introduction of convex functions, we will some important theorems to solve these problems and some algorithms based on these theorems.
Terminologies
The space Rn − It is an n-dimensional vector with real numbers, defined as follows − Rn={(x1,x2,...,xn)τ:x1,x2,....,xn∈R}
The space RmXn − It is a set of all real values matrices of order mXn.
Convex Optimization - Linear Programming
Methodology
Linear Programming also called Linear Optimization, is a technique which is used to solve mathematical problems in which the relationships are linear in nature. the basic nature of Linear Programming is to maximize or minimize an objective function with subject to some constraints. The objective function is a linear function which is obtained from the mathematical model of the problem. The constraints are the conditions which are imposed on the model and are also linear.
- From the given question, find the objective function.
- find the constraints.
- Draw the constraints on a graph.
- find the feasible region, which is formed by the intersection of all the constraints.
- find the vertices of the feasible region.
- find the value of the objective function at these vertices.
- The vertice which either maximizes or minimizes the objective function accordingtothequestion is the answer.
Examples
Step 1 − Maximize 5x+3y subject to
x+y≤2,
3x+y≤3,
x≥0andy≥0
Solution −
The first step is to find the feasible region on a graph.

Clearly from the graph, the vertices of the feasible region are
(0,0)(0,2)(1,0)(12,32)
Let f(x,y)=5x+3y
Putting these values in the objective function, we get −
f(0,0)=0
f(0,2)=6
f(1,0)=5
f(12,32)=7
Therefore, the function maximizes at (12,32)
Step 2 − A watch company produces a digital and a mechanical watch. Long-term projections indicate an expected demand of at least 100 digital and 80 mechanical watches each day. Because of limitations on production capacity, no more than 200 digital and 170 mechanical watches can be made daily. To satisfy a shipping contract, a total of at least 200 watches much be shipped each day.
If each digital watch sold results in a $2 loss, but each mechanical watch produces a $5 profit, how many of each type should be made daily to maximize net profits?
Solution −
Let x be the number of digital watches produced
y be the number of mechanical watches produced
According to the question, at least 100 digital watches are to be made daily and maximaum 200 digital watches can be made.
⇒100≤x≤200
Similarly, at least 80 mechanical watches are to be made daily and maximum 170 mechanical watches can be made.
⇒80≤y≤170
Since at least 200 watches are to be produced each day.
⇒x+y≤200
Since each digital watch sold results in a $2 loss, but each mechanical watch produces a $5 profit,
Total profit can be calculated as
Profit=−2x+5y
And we have to maximize the profit, Therefore, the question can be formulated as −
Maximize −2x+5y subject to
100≤x≤200
80≤y≤170
x+y≤200
Plotting the above equations in a graph, we get,

The vertices of the feasible region are
(100,170)(200,170)(200,180)(120,80)and(100,100)
The maximum value of the objective function is obtained at (100,170) Thus, to maximize the net profits, 100 units of digital watches and 170 units of mechanical watches should be produced.
Convex Optimization - Norm
A norm is a function that gives a strictly positive value to a vector or a variable.
Norm is a function f:Rn→R
The basic characteristics of a norm are −
Let X be a vector such that X∈Rn
‖x‖≥0
‖x‖=0⇔x=0∀x∈X
‖αx‖=|α|‖x‖∀x∈Xandαisascalar
‖x+y‖≤‖x‖+‖y‖∀x,y∈X
‖x−y‖≥‖‖x‖−‖y‖‖
By definition, norm is calculated as follows −
‖x‖1=n∑i=1|xi|
‖x‖2=(n∑i=1|xi|2)12
‖x‖p=(n∑i=1|xi|p)1p,1≤p≤∞
Norm is a continuous function.
Proof
By definition, if xn→x in X⇒f(xn)→f(x) then f(x) is a constant function.
Let f(x)=‖x‖
Therefore, |f(xn)−f(x)|=|‖xn‖−‖x‖|≤||xn−x||
Since xn→x thus, ‖xn−x‖→0
Therefore |f(xn)−f(x)|≤0⇒|f(xn)−f(x)|=0⇒f(xn)→f(x)
Hence, norm is a continuous function.
Convex Optimization - Inner Product
Inner product is a function which gives a scalar to a pair of vectors.
Inner Product − f:Rn×Rn→κ where κ is a scalar.
The basic characteristics of inner product are as follows −
Let X∈Rn
⟨x,x⟩≥0,∀x∈X
⟨x,x⟩=0⇔x=0,∀x∈X
⟨αx,y⟩=α⟨x,y⟩,∀α∈κand∀x,y∈X
⟨x+y,z⟩=⟨x,z⟩+⟨y,z⟩,∀x,y,z∈X
⟨¯y,x⟩=(x,y),∀x,y∈X
Note −
Relationship between norm and inner product: ‖x‖=√(x,x)
∀x,y∈Rn,⟨x,y⟩=x1y1+x2y2+...+xnyn
Examples
1. find the inner product of x=(1,2,1)andy=(3,−1,3)
Solution
⟨x,y⟩=x1y1+x2y2+x3y3
⟨x,y⟩=(1×3)+(2×−1)+(1×3)
⟨x,y⟩=3+(−2)+3
⟨x,y⟩=4
2. If x=(4,9,1),y=(−3,5,1) and z=(2,4,1), find (x+y,z)
Solution
As we know, ⟨x+y,z⟩=⟨x,z⟩+⟨y,z⟩
⟨x+y,z⟩=(x1z1+x2z2+x3z3)+(y1z1+y2z2+y3z3)
⟨x+y,z⟩={(4×2)+(9×4)+(1×1)}+
{(−3×2)+(5×4)+(1×1)}
⟨x+y,z⟩=(8+36+1)+(−6+20+1)
⟨x+y,z⟩=45+15
⟨x+y,z⟩=60
Convex Optimization - Minima and Maxima
Local Minima or Minimize
ˉx∈S is said to be local minima of a function f if f(ˉx)≤f(x),∀x∈Nε(ˉx) where Nε(ˉx) means neighbourhood of ˉx, i.e., Nε(ˉx) means $\left \| x-\bar{x} \right \|
Local Maxima or Maximizer
ˉx∈S is said to be local maxima of a function f if f(ˉx)≥f(x),∀x∈Nε(ˉx) where Nε(ˉx) means neighbourhood of ˉx, i.e., Nε(ˉx) means $\left \| x-\bar{x} \right \|
Global minima
ˉx∈S is said to be global minima of a function f if f(ˉx)≤f(x),∀x∈S
Global maxima
ˉx∈S is said to be global maxima of a function f if f(ˉx)≥f(x),∀x∈S
Examples
Step 1 − find the local minima and maxima of f(ˉx)=|x2−4|
Solution −

From the graph of the above function, it is clear that the local minima occurs at x=±2 and local maxima at x=0
Step 2 − find the global minima af the function f(x)=|4x3−3x2+7|
Solution −

From the graph of the above function, it is clear that the global minima occurs at x=−1.
Convex Optimization - Convex Set
Let S⊆Rn A set S is said to be convex if the line segment joining any two points of the set S also belongs to the S, i.e., if x1,x2∈S, then λx1+(1−λ)x2∈S where λ∈(0,1).
Note −
- The union of two convex sets may or may not be convex.
- The intersection of two convex sets is always convex.
Proof
Let S1 and S2 be two convex set.
Let S3=S1∩S2
Let x1,x2∈S3
Since S3=S1∩S2 thus x1,x2∈S1and x1,x2∈S2
Since Si is convex set, ∀ i∈1,2,
Thus λx1+(1−λ)x2∈Si where λ∈(0,1)
Therfore, λx1+(1−λ)x2∈S1∩S2
⇒λx1+(1−λ)x2∈S3
Hence, S3 is a convex set.
Weighted average of the form k∑i=1λixi,where k∑i=1λi=1 and λi≥0,∀i∈[1,k] is called conic combination of x1,x2,....xk.
Weighted average of the form k∑i=1λixi, where k∑i=1λi=1 is called affine combination of x1,x2,....xk.
Weighted average of the form k∑i=1λixi is called linear combination of x1,x2,....xk.
Examples
Step 1 − Prove that the set S={x∈Rn:Cx≤α} is a convex set.
Solution
Let x1 and x2∈S
⇒Cx1≤α and andCx2≤α
To show:y=(λx1+(1−λ)x2)∈S∀λ∈(0,1)
Cy=C(λx1+(1−λ)x2)=λCx1+(1−λ)Cx2
⇒Cy≤λα+(1−λ)α
⇒Cy≤α
⇒y∈S
Therefore, S is a convex set.
Step 2 − Prove that the set S={(x1,x2)∈R2:x21≤8x2} is a convex set.
Solution
Let x,y∈S
Let x=(x1,x2) and y=(y1,y2)
⇒x21≤8x2 and y21≤8y2
To show − λx+(1−λ)y∈S⇒λ(x1,x2)+(1−λ)(y1,y2)∈S⇒[λx1+(1−λ)y2]∈S)]
Now,[λx1+(1−λ)y1]2=λ2x21+(1−λ)2y21+2λ(1−λ)x1y1
But 2x1y1≤x21+y21
Therefore,
[λx1+(1−λ)y1]2≤λ2x21+(1−λ)2y21+2λ(1−λ)(x21+y21)
⇒[λx1+(1−λ)y1]2≤λx21+(1−λ)y21
⇒[λx1+(1−λ)y1]2≤8λx2+8(1−λ)y2
⇒[λx1+(1−λ)y1]2≤8[λx2+(1−λ)y2]
⇒λx+(1−λ)y∈S
Step 3 − Show that a set S∈Rn is convex if and only if for each integer k, every convex combination of any k points of S is in S.
Solution
Let S be a convex set. then, to show;
c1x1+c2x2+.....+ckxk∈S,k∑1ci=1,ci≥0,∀i∈1,2,....,k
Proof by induction
For k=1,x1∈S,c1=1⇒c1x1∈S
For k=2,x1,x2∈S,c1+c2=1 and Since S is a convex set
⇒c1x1+c2x2∈S.
Let the convex combination of m points of S is in S i.e.,
c1x1+c2x2+...+cmxm∈S,m∑1ci=1,ci≥0,∀i∈1,2,...,m
Now, Let x1,x2....,xm,xm+1∈S
Let x=μ1x1+μ2x2+...+μmxm+μm+1xm+1
Let x=(μ1+μ2+...+μm)μ1x1+μ2x2+μmxmμ1+μ2+.........+μm+μm+1xm+1
Let y=μ1x1+μ2x2+...+μmxmμ1+μ2+.........+μm
⇒x=(μ1+μ2+...+μm)y+μm+1xm+1
Now y∈S because the sum of the coeicients is 1.
⇒x∈S since S is a convex set and y,xm+1∈S
Hence proved by induction.
Convex Optimization - affine Set
A set A is said to be an affine set if for any two distinct points, the line passing through these points lie in the set A.
Note −
S is an affine set if and only if it contains every affine combination of its points.
-
Empty and singleton sets are both affine and convex set.
For example, solution of a linear equation is an affine set.
Proof
Let S be the solution of a linear equation.
By definition, S={x∈Rn:Ax=b}
Let x1,x2∈S⇒Ax1=b and Ax2=b
To prove : A[θx1+(1−θ)x2]=b,∀θ∈(0,1)
A[θx1+(1−θ)x2]=θAx1+(1−θ)Ax2=θb+(1−θ)b=b
Thus S is an affine set.
Theorem
If C is an affine set and x0∈C, then the set V=C−x0={x−x0:x∈C} is a subspace of C.
Proof
Let x1,x2∈V
To show: αx1+βx2∈V for some α,β
Now, x1+x0∈C and x2+x0∈C by definition of V
Now, αx1+βx2+x0=α(x1+x0)+β(x2+x0)+(1−α−β)x0
But α(x1+x0)+β(x2+x0)+(1−α−β)x0∈C because C is an affine set.
Therefore, αx1+βx2∈V
Hence proved.
Convex Optimization - Hull
The convex hull of a set of points in S is the boundary of the smallest convex region that contain all the points of S inside it or on its boundary.
OR
Let S⊆Rn The convex hull of S, denoted Co(S) by is the collection of all convex combination of S, i.e., x∈Co(S) if and only if x∈n∑i=1λixi, where n∑1λi=1 and λi≥0∀xi∈S
Remark − Conves hull of a set of points in S in the plane defines a convex polygon and the points of S on the boundary of the polygon defines the vertices of the polygon.
Theorem Co(S)={x:x=n∑i=1λixi,xi∈S,n∑i=1λi=1,λi≥0} Show that a convex hull is a convex set.
Proof
Let x1,x2∈Co(S), then x1=n∑i=1λixi and x2=n∑i=1λγxi where n∑i=1λi=1,λi≥0 and n∑i=1γi=1,γi≥0
For θ∈(0,1),θx1+(1−θ)x2=θn∑i=1λixi+(1−θ)n∑i=1γixi
θx1+(1−θ)x2=n∑i=1λiθxi+n∑i=1γi(1−θ)xi
θx1+(1−θ)x2=n∑i=1[λiθ+γi(1−θ)]xi
Considering the coefficients,
n∑i=1[λiθ+γi(1−θ)]=θn∑i=1λi+(1−θ)n∑i=1γi=θ+(1−θ)=1
Hence, θx1+(1−θ)x2∈Co(S)
Thus, a convex hull is a convex set.
Caratheodory Theorem
Let S be an arbitrary set in Rn.If x∈Co(S), then x∈Co(x1,x2,....,xn,xn+1).
Proof
Since x∈Co(S), then x is representated by a convex combination of a finite number of points in S, i.e.,
x=k∑j=1λjxj,k∑j=1λj=1,λj≥0 and xj∈S,∀j∈(1,k)
If k≤n+1, the result obtained is obviously true.
If k≥n+1, then (x2−x1)(x3−x1),.....,(xk−x1) are linearly dependent.
⇒∃μj∈R,2≤j≤k notallzero such that k∑j=2μj(xj−x1)=0
Define μ1=−k∑j=2μj, then k∑j=1μjxj=0,k∑j=1μj=0
where not all μ′js are equal to zero. Since k∑j=1μj=0, at least one of the μj>0,1≤j≤k
Then, x=k∑1λjxj+0
x=k∑1λjxj−αk∑1μjxj
x=k∑1(λj−αμj)xj
Choose α such that α=min{λjμj,μj≥0}=λjμj, for some i=1,2,...,k
If μj≤0,λj−αμj≥0
If μj>0,thenλjμj≥λiμi=α⇒λj−αμj≥0,j=1,2,...k
In particular, λi−αμi=0, by definition of α
x=k∑j=1(λj−αμj)xj,where
λj−αμj≥0 and k∑j=1(λj−αμj)=1 and λi−αμi=0
Thus, x can be representated as a convex combination of at most k−1 points.
This reduction process can be repeated until x is representated as a convex combination of n+1 elements.
Convex Optimization - Weierstrass Theorem
Let S be a non empty, closed and bounded set alsocalledcompactset in Rn and let f:S→R be a continuous function on S, then the problem min {f(x):x∈S} attains its minimum.
Proof
Since S is non-empty and bounded, there exists a lower bound.
α=Inf{f(x):x∈S}
Now let Sj={x∈S:α≤f(x)≤α+δj}∀j=1,2,... and δ∈(0,1)
By the definition of infimium, Sj is non-empty, for each j.
Choose some xj∈Sj to get a sequence {xj} for j=1,2,...
Since S is bounded, the sequence is also bounded and there is a convergent subsequence {yj}, which converges to ˆx. Hence ˆx is a limit point and S is closed, therefore, ˆx∈S. Since f is continuous, f(yi)→f(ˆx).
Since α≤f(yi)≤α+δk,α=lim
Thus, \hat{x} is the minimizing solution.
Remarks
There are two important necessary conditions for Weierstrass Theorem to hold. These are as follows −
-
Step 1 − The set S should be a bounded set.
Consider the function f\left x \right =x$.
It is an unbounded set and it does have a minima at any point in its domain.
Thus, for minima to obtain, S should be bounded.
-
Step 2 − The set S should be closed.
Consider the function f\left ( x \right )=\frac{1}{x} in the domain \left 0,1 \right .
This function is not closed in the given domain and its minima also does not exist.
Hence, for minima to obtain, S should be closed.
Convex Optimization - Closest Point Theorem
Let S be a non-empty closed convex set in \mathbb{R}^n and let y\notin S, then \exists a point \bar{x}\in S with minimum distance from y, i.e.,\left \| y-\bar{x} \right \| \leq \left \| y-x \right \| \forall x \in S.
Furthermore, \bar{x} is a minimizing point if and only if \left ( y-\hat{x} \right )^{T}\left ( x-\hat{x} \right )\leq 0 or \left ( y-\hat{x}, x-\hat{x} \right )\leq 0
Proof
Existence of closest point
Since S\ne \phi,\exists a point \hat{x}\in S such that the minimum distance of S from y is less than or equal to \left \| y-\hat{x} \right \|.
Define \hat{S}=S \cap \left \{ x:\left \| y-x \right \|\leq \left \| y-\hat{x} \right \| \right \}
Since \hat{S} is closed and bounded, and since norm is a continuous function, then by Weierstrass theorem, there exists a minimum point \hat{x} \in S such that \left \| y-\hat{x} \right \|=Inf\left \{ \left \| y-x \right \|,x \in S \right \}
Uniqueness
Suppose \bar{x} \in S such that \left \| y-\hat{x} \right \|=\left \| y-\hat{x} \right \|= \alpha
Since S is convex, \frac{\hat{x}+\bar{x}}{2} \in S
But, \left \| y-\frac{\hat{x}-\bar{x}}{2} \right \|\leq \frac{1}{2}\left \| y-\hat{x} \right \|+\frac{1}{2}\left \| y-\bar{x} \right \|=\alpha
It can't be strict inequality because \hat{x} is closest to y.
Therefore, \left \| y-\hat{x} \right \|=\mu \left \| y-\hat{x} \right \|, for some \mu
Now \left \| \mu \right \|=1. If \mu=-1, then \left ( y-\hat{x} \right )=-\left ( y-\hat{x} \right )\Rightarrow y=\frac{\hat{x}+\bar{x}}{2} \in S
But y \in S. Hence contradiction. Thus \mu=1 \Rightarrow \hat{x}=\bar{x}
Thus, minimizing point is unique.
For the second part of the proof, assume \left ( y-\hat{x} \right )^{\tau }\left ( x-\bar{x} \right )\leq 0 for all x\in S
Now,
\left \| y-x \right \|^{2}=\left \| y-\hat{x}+ \hat{x}-x\right \|^{2}=\left \| y-\hat{x} \right \|^{2}+\left \|\hat{x}-x \right \|^{2}+2\left (\hat{x}-x \right )^{\tau }\left ( y-\hat{x} \right )
\Rightarrow \left \| y-x \right \|^{2}\geq \left \| y-\hat{x} \right \|^{2} because \left \| \hat{x}-x \right \|^{2}\geq 0 and \left ( \hat{x}- x\right )^{T}\left ( y-\hat{x} \right )\geq 0
Thus, \hat{x} is minimizing point.
Conversely, assume \hat{x} is minimizimg point.
\Rightarrow \left \| y-x \right \|^{2}\geq \left \| y-\hat{x} \right \|^2 \forall x \in S
Since S is convex set.
\Rightarrow \lambda x+\left ( 1-\lambda \right )\hat{x}=\hat{x}+\lambda\left ( x-\hat{x} \right ) \in S for x \in S and \lambda \in \left ( 0,1 \right )
Now, \left \| y-\hat{x}-\lambda\left ( x-\hat{x} \right ) \right \|^{2}\geq \left \| y-\hat{x} \right \|^2
And
\left \| y-\hat{x}-\lambda\left ( x-\hat{x} \right ) \right \|^{2}=\left \| y-\hat{x} \right \|^{2}+\lambda^2\left \| x-\hat{x} \right \|^{2}-2\lambda\left ( y-\hat{x} \right )^{T}\left ( x-\hat{x} \right )
\Rightarrow \left \| y-\hat{x} \right \|^{2}+\lambda^{2}\left \| x-\hat{x} \right \|-2 \lambda\left ( y-\hat{x} \right )^{T}\left ( x-\hat{x} \right )\geq \left \| y-\hat{x} \right \|^{2}
\Rightarrow 2 \lambda\left ( y-\hat{x} \right )^{T}\left ( x-\hat{x} \right )\leq \lambda^2\left \| x-\hat{x} \right \|^2
\Rightarrow \left ( y-\hat{x} \right )^{T}\left ( x-\hat{x} \right )\leq 0
Hence Proved.
Fundamental Separation Theorem
Let S be a non-empty closed, convex set in \mathbb{R}^n and y \notin S. Then, there exists a non zero vector p and scalar \beta such that p^T y>\beta and $p^T x
Proof
Since S is non empty closed convex set and y \notin S thus by closest point theorem, there exists a unique minimizing point \hat{x} \in S such that
\left ( x-\hat{x} \right )^T\left ( y-\hat{x} \right )\leq 0 \forall x \in S
Let p=\left ( y-\hat{x} \right )\neq 0 and \beta=\hat{x}^T\left ( y-\hat{x} \right )=p^T\hat{x}.
Then \left ( x-\hat{x} \right )^T\left ( y-\hat{x} \right )\leq 0
\Rightarrow \left ( y-\hat{x} \right )^T\left ( x-\hat{x} \right )\leq 0
\Rightarrow \left ( y-\hat{x} \right )^Tx\leq \left ( y-\hat{x} \right )^T \hat{x}=\hat{x}^T\left ( y-\hat{x} \right ) i,e., p^Tx \leq \beta
Also, p^Ty-\beta=\left ( y-\hat{x} \right )^Ty-\hat{x}^T \left ( y-\hat{x} \right )
=\left ( y-\hat{x} \right )^T \left ( y-x \right )=\left \| y-\hat{x} \right \|^{2}>0
\Rightarrow p^Ty> \beta
This theorem results in separating hyperplanes. The hyperplanes based on the above theorem can be defined as follows −
Let S_1 and S_2 are be non-empty subsets of \mathbb{R} and H=\left \{ X:A^TX=b \right \} be a hyperplane.
The hyperplane H is said to separate S_1 and S_2 if A^TX \leq b \forall X \in S_1 and A_TX \geq b \forall X \in S_2
The hyperplane H is said to strictly separate S_1 and S_2 if A^TX b \forall X \in S_2
The hyperplane H is said to strongly separate S_1 and S_2 if A^TX \leq b \forall X \in S_1 and A_TX \geq b+ \varepsilon \forall X \in S_2, where \varepsilon is a positive scalar.
Convex Optimization - Cones
A non empty set C in \mathbb{R}^n is said to be cone with vertex 0 if x \in C\Rightarrow \lambda x \in C \forall \lambda \geq 0.
A set C is a convex cone if it convex as well as cone.
For example, y=\left | x \right | is not a convex cone because it is not convex.
But, y \geq \left | x \right | is a convex cone because it is convex as well as cone.
Note − A cone C is convex if and only if for any x,y \in C, x+y \in C.
Proof
Since C is cone, for x,y \in C \Rightarrow \lambda x \in C and \mu y \in C \:\forall \:\lambda, \mu \geq 0
C is convex if \lambda x + \left ( 1-\lambda \right )y \in C \: \forall \:\lambda \in \left ( 0, 1 \right )
Since C is cone, \lambda x \in C and \left ( 1-\lambda \right )y \in C \Leftrightarrow x,y \in C
Thus C is convex if x+y \in C
In general, if x_1,x_2 \in C, then, \lambda_1x_1+\lambda_2x_2 \in C, \forall \lambda_1,\lambda_2 \geq 0
Examples
The conic combination of infinite set of vectors in \mathbb{R}^n is a convex cone.
Any empty set is a convex cone.
Any linear function is a convex cone.
Since a hyperplane is linear, it is also a convex cone.
Closed half spaces are also convex cones.
Note − The intersection of two convex cones is a convex cone but their union may or may not be a convex cone.
Convex Optimization - Polar Cone
Let S be a non empty set in \mathbb{R}^n Then, the polar cone of S denoted by S^* is given by S^*=\left \{p \in \mathbb{R}^n, p^Tx \leq 0 \: \forall x \in S \right \}.
Remark
Polar cone is always convex even if S is not convex.
If S is empty set, S^*=\mathbb{R}^n.
Polarity may be seen as a generalisation of orthogonality.
Let C\subseteq \mathbb{R}^n then the orthogonal space of C, denoted by C^\perp =\left \{ y \in \mathbb{R}^n:\left \langle x,y \right \rangle=0 \forall x \in C \right \}.
Lemma
Let S,S_1 and S_2 be non empty sets in \mathbb{R}^n then the following statements are true −
S^* is a closed convex cone.
S \subseteq S^{**} where S^{**} is a polar cone of S^*.
S_1 \subseteq S_2 \Rightarrow S_{2}^{*} \subseteq S_{1}^{*}.
Proof
Step 1 − S^*=\left \{ p \in \mathbb{R}^n,p^Tx\leq 0 \: \forall \:x \in S \right \}
-
Let x_1,x_2 \in S^*\Rightarrow x_{1}^{T}x\leq 0 and x_{2}^{T}x \leq 0,\forall x \in S
For \lambda \in \left ( 0, 1 \right ),\left [ \lambda x_1+\left ( 1-\lambda \right )x_2 \right ]^Tx=\left [ \left ( \lambda x_1 \right )^T+ \left \{\left ( 1-\lambda \right )x_{2} \right \}^{T}\right ]x, \forall x \in S
=\left [ \lambda x_{1}^{T} +\left ( 1-\lambda \right )x_{2}^{T}\right ]x=\lambda x_{1}^{T}x+\left ( 1-\lambda \right )x_{2}^{T}\leq 0
Thus \lambda x_1+\left ( 1-\lambda \right )x_{2} \in S^*
Therefore S^* is a convex set.
-
For \lambda \geq 0,p^{T}x \leq 0, \forall \:x \in S
Therefore, \lambda p^T x \leq 0,
\Rightarrow \left ( \lambda p \right )^T x \leq 0
\Rightarrow \lambda p \in S^*
Thus, S^* is a cone.
-
To show S^* is closed, i.e., to show if p_n \rightarrow p as n \rightarrow \infty, then p \in S^*
\forall x \in S, p_{n}^{T}x-p^T x=\left ( p_n-p \right )^T x
As p_n \rightarrow p as n \rightarrow \infty \Rightarrow \left ( p_n \rightarrow p \right )\rightarrow 0
Therefore p_{n}^{T}x \rightarrow p^{T}x. But p_{n}^{T}x \leq 0, \: \forall x \in S
Thus, p^Tx \leq 0, \forall x \in S
\Rightarrow p \in S^*
Hence, S^* is closed.
Step 2 − S^{**}=\left \{ q \in \mathbb{R}^n:q^T p \leq 0, \forall p \in S^*\right \}
Let x \in S, then \forall p \in S^*, p^T x \leq 0 \Rightarrow x^Tp \leq 0 \Rightarrow x \in S^{**}
Thus, S \subseteq S^{**}
Step 3 − S_2^*=\left \{ p \in \mathbb{R}^n:p^Tx\leq 0, \forall x \in S_2 \right \}
Since S_1 \subseteq S_2 \Rightarrow \forall x \in S_2 \Rightarrow \forall x \in S_1
Therefore, if \hat{p} \in S_2^*, then \hat{p}^Tx \leq 0,\forall x \in S_2
\Rightarrow \hat{p}^Tx\leq 0, \forall x \in S_1
\Rightarrow \hat{p}^T \in S_1^*
\Rightarrow S_2^* \subseteq S_1^*
Theorem
Let C be a non empty closed convex cone, then C=C^**
Proof
C=C^{**} by previous lemma.
To prove : x \in C^{**} \subseteq C
Let x \in C^{**} and let x \notin C
Then by fundamental separation theorem, there exists a vector p \neq 0 and a scalar \alpha such that p^Ty \leq \alpha, \forall y \in C
Therefore, p^Tx > \alpha
But since \left ( y=0 \right ) \in C and p^Ty\leq \alpha, \forall y \in C \Rightarrow \alpha\geq 0 and p^Tx>0
If p \notin C^*, then there exists some \bar{y} \in C such that p^T \bar{y}>0 and p^T\left ( \lambda \bar{y} \right ) can be made arbitrarily large by taking \lambda sufficiently large.
This contradicts with the fact that p^Ty \leq \alpha, \forall y \in C
Therefore,p \in C^*
Since x \in C^*=\left \{ q:q^Tp\leq 0, \forall p \in C^* \right \}
Therefore, x^Tp \leq 0 \Rightarrow p^Tx \leq 0
But p^Tx> \alpha
Thus is contardiction.
Thus, x \in C
Hence C=C^{**}.
Convex Optimization - Conic Combination
A point of the form \alpha_1x_1+\alpha_2x_2+....+\alpha_nx_n with \alpha_1, \alpha_2,...,\alpha_n\geq 0 is called conic combination of x_1, x_2,...,x_n.
If x_i are in convex cone C, then every conic combination of x_i is also in C.
A set C is a convex cone if it contains all the conic combination of its elements.
Conic Hull
A conic hull is defined as a set of all conic combinations of a given set S and is denoted by coniS.
Thus, coni\left ( S \right )=\left \{ \displaystyle\sum\limits_{i=1}^k \lambda_ix_i:x_i \in S,\lambda_i\in \mathbb{R}, \lambda_i\geq 0,i=1,2,...\right \}
- The conic hull is a convex set.
- The origin always belong to the conic hull.
Convex Optimization - Polyhedral Set
A set in \mathbb{R}^n is said to be polyhedral if it is the intersection of a finite number of closed half spaces, i.e.,
S=\left \{ x \in \mathbb{R}^n:p_{i}^{T}x\leq \alpha_i, i=1,2,....,n \right \}
For example,
\left \{ x \in \mathbb{R}^n:AX=b \right \}
\left \{ x \in \mathbb{R}^n:AX\leq b \right \}
\left \{ x \in \mathbb{R}^n:AX\geq b \right \}
Polyhedral Cone
A set in \mathbb{R}^n is said to be polyhedral cone if it is the intersection of a finite number of half spaces that contain the origin, i.e., S=\left \{ x \in \mathbb{R}^n:p_{i}^{T}x\leq 0, i=1, 2,... \right \}
Polytope
A polytope is a polyhedral set which is bounded.
Remarks
- A polytope is a convex hull of a finite set of points.
- A polyhedral cone is generated by a finite set of vectors.
- A polyhedral set is a closed set.
- A polyhedral set is a convex set.
Extreme point of a convex set
Let S be a convex set in \mathbb{R}^n. A vector x \in S is said to be a extreme point of S if x= \lambda x_1+\left ( 1-\lambda \right )x_2 with x_1, x_2 \in S and \lambda \in\left ( 0, 1 \right )\Rightarrow x=x_1=x_2.
Example
Step 1 − S=\left \{ \left ( x_1,x_2 \right ) \in \mathbb{R}^2:x_{1}^{2}+x_{2}^{2}\leq 1 \right \}
Extreme point, E=\left \{ \left ( x_1, x_2 \right )\in \mathbb{R}^2:x_{1}^{2}+x_{2}^{2}= 1 \right \}
Step 2 − $S=\left \{ \left ( x_1,x_2 \right )\in \mathbb{R}^2:x_1+x_2
Extreme point, E=\left \{ \left ( 0, 0 \right), \left ( 2, 0 \right), \left ( 0, 1 \right), \left ( \frac{2}{3}, \frac{4}{3} \right) \right \}
Step 3 − S is the polytope made by the points \left \{ \left ( 0,0 \right ), \left ( 1,1 \right ), \left ( 1,3 \right ), \left ( -2,4 \right ),\left ( 0,2 \right ) \right \}
Extreme point, E=\left \{ \left ( 0,0 \right ), \left ( 1,1 \right ),\left ( 1,3 \right ),\left ( -2,4 \right ) \right \}
Remarks
Any point of the convex set S, can be represented as a convex combination of its extreme points.
It is only true for closed and bounded sets in \mathbb{R}^n.
It may not be true for unbounded sets.
k extreme points
A point in a convex set is called k extreme if and only if it is the interior point of a k-dimensional convex set within S, and it is not an interior point of a k+1- dimensional convex set within S. Basically, for a convex set S, k extreme points make k-dimensional open faces.
Convex Optimization - Direction
Let S be a closed convex set in \mathbb{R}^n. A non zero vector d \in \mathbb{R}^n is called a direction of S if for each x \in S,x+\lambda d \in S, \forall \lambda \geq 0.
Two directions d_1 and d_2 of S are called distinct if d \neq \alpha d_2 for \alpha>0.
A direction d of S is said to be extreme direction if it cannot be written as a positive linear combination of two distinct directions, i.e., if d=\lambda _1d_1+\lambda _2d_2 for \lambda _1, \lambda _2>0, then d_1= \alpha d_2 for some \alpha.
Any other direction can be expressed as a positive combination of extreme directions.
For a convex set S, the direction d such that x+\lambda d \in S for some x \in S and all \lambda \geq0 is called recessive for S.
Let E be the set of the points where a certain function f:S \rightarrow over a non-empty convex set S in \mathbb{R}^n attains its maximum, then E is called exposed face of S. The directions of exposed faces are called exposed directions.
A ray whose direction is an extreme direction is called an extreme ray.
Example
Consider the function f\left ( x \right )=y=\left |x \right |, where x \in \mathbb{R}^n. Let d be unit vector in \mathbb{R}^n
Then, d is the direction for the function f because for any \lambda \geq 0, x+\lambda d \in f\left ( x \right ).
Convex and Concave Function
Let f:S \rightarrow \mathbb{R}, where S is non empty convex set in \mathbb{R}^n, then f\left ( x \right ) is said to be convex on S if f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right ), \forall \lambda \in \left ( 0,1 \right ).
On the other hand, Let f:S\rightarrow \mathbb{R}, where S is non empty convex set in \mathbb{R}^n, then f\left ( x \right ) is said to be concave on S if f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\geq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right ), \forall \lambda \in \left ( 0, 1 \right ).
Let f:S \rightarrow \mathbb{R} where S is non empty convex set in \mathbb{R}^n, then f\left ( x\right ) is said to be strictly convex on S if $f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )
Let f:S \rightarrow \mathbb{R} where S is non empty convex set in \mathbb{R}^n, then f\left ( x\right ) is said to be strictly concave on S if f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )> \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right ), \forall \lambda \in \left ( 0, 1 \right ).
Examples
A linear function is both convex and concave.
f\left ( x \right )=\left | x \right | is a convex function.
f\left ( x \right )= \frac{1}{x} is a convex function.
Theorem
Let f_1,f_2,...,f_k:\mathbb{R}^n \rightarrow \mathbb{R} be convex functions. Consider the function f\left ( x \right )=\displaystyle\sum\limits_{j=1}^k \alpha_jf_j\left ( x \right ) where \alpha_j>0,j=1, 2, ...k, then f\left ( x \right )is a convex function.
Proof
Since f_1,f_2,...f_k are convex functions
Therefore, f_i\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f_i\left ( x_1 \right )+\left ( 1-\lambda \right )f_i\left ( x_2 \right ), \forall \lambda \in \left ( 0, 1 \right ) and i=1, 2,....,k
Consider the function f\left ( x \right ).
Therefore,
f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )
=\displaystyle\sum\limits_{j=1}^k \alpha_jf_j\left ( \lambda x_1 +1-\lambda \right )x_2\leq \displaystyle\sum\limits_{j=1}^k\alpha_j\lambda f_j\left ( x_1 \right )+\left ( 1-\lambda \right )f_j\left ( x_2 \right )
\Rightarrow f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda \left ( \displaystyle\sum\limits_{j=1}^k \alpha _jf_j\left ( x_1 \right ) \right )+\left ( \displaystyle\sum\limits_{j=1}^k \alpha _jf_j\left ( x_2 \right ) \right )
\Rightarrow f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f\left ( x_2 \right )\leq \left ( 1-\lambda \right )f\left ( x_2 \right )
Hence, f\left ( x\right ) is a convex function.
Theorem
Let f\left ( x\right ) be a convex function on a convex set S\subset \mathbb{R}^n then a local minima of f\left ( x\right ) on S is a global minima.
Proof
Let \hat{x} be a local minima for f\left ( x \right ) and \hat{x} is not global minima.
therefore, \exists \hat{x} \in S such that $f\left ( \bar{x} \right )
Since \hat{x} is a local minima, there exists neighbourhood N_\varepsilon \left ( \hat{x} \right ) such that f\left ( \hat{x} \right )\leq f\left ( x \right ), \forall x \in N_\varepsilon \left ( \hat{x} \right )\cap S
But f\left ( x \right ) is a convex function on S, therefore for \lambda \in \left ( 0, 1 \right )
we have \lambda \hat{x}+\left ( 1-\lambda \right )\bar{x}\leq \lambda f\left ( \hat{x} \right )+\left ( 1-\lambda \right )f\left ( \bar{x} \right )
$\Rightarrow \lambda \hat{x}+\left ( 1-\lambda \right )\bar{x}
$\Rightarrow \lambda \hat{x}+\left ( 1-\lambda \right )\bar{x}
But for some $\lambda
\lambda \hat{x}+\left ( 1-\lambda \right )\bar{x} \in N_\varepsilon \left ( \hat{x} \right )\cap S and $f\left ( \lambda \hat{x}+\left ( 1-\lambda \right )\bar{x} \right )
which is a contradiction.
Hence, \bar{x} is a global minima.
Epigraph
let S be a non-empty subset in \mathbb{R}^n and let f:S \rightarrow \mathbb{R} then the epigraph of f denoted by epif or E_f is a subset of \mathbb{R}^n+1 defined by E_f=\left \{ \left ( x,\alpha \right ):x \in \mathbb{R}^n, \alpha \in \mathbb{R}, f\left ( x \right )\leq \alpha \right \}
Hypograph
let S be a non-empty subset in \mathbb{R}^n and let f:S \rightarrow \mathbb{R}, then the hypograph of f denoted by hypf or H_f=\left \{ \left ( x, \alpha \right ):x \in \mathbb{R}^n, \alpha \in \mathbb{R}^n, \alpha \in \mathbb{R}, f\left ( x \right )\geq \alpha \right \}
Theorem
Let S be a non-empty convex set in \mathbb{R}^n and let f:S \rightarrow \mathbb{R}^n, then f is convex if and only if its epigraph E_f is a convex set.
Proof
Let f is a convex function.
To show E_f is a convex set.
Let \left ( x_1, \alpha_1 \right ),\left ( x_2, \alpha_2 \right ) \in E_f,\lambda \in\left ( 0, 1 \right )
To show \lambda \left ( x_1,\alpha_1 \right )+\left ( 1-\lambda \right )\left ( x_2, \alpha_2 \right ) \in E_f
\Rightarrow \left [ \lambda x_1+\left ( 1-\lambda \right )x_2, \lambda \alpha_1+\left ( 1-\lambda \right )\alpha_2 \right ]\in E_f
f\left ( x_1 \right )\leq \alpha _1, f\left ( x_2\right )\leq \alpha _2
Therefore, f\left (\lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f \left ( x_2 \right )
\Rightarrow f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda \alpha_1+\left ( 1-\lambda \right )\alpha_2
Converse
Let E_f is a convex set.
To show f is convex.
i.e., to show if x_1, x_2 \in S,\lambda \left ( 0, 1\right )
f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right )
Let x_1,x_2 \in S, \lambda \in \left ( 0, 1 \right ),f\left ( x_1 \right ), f\left ( x_2 \right ) \in \mathbb{R}
Since E_f is a convex set, \left ( \lambda x_1+\left ( 1-\lambda \right )x_2, \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )\right )f\left ( x_2 \right )\in E_f
Therefore, f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right )
Convex Optimization - Jensen's Inequality
Let S be a non-empty convex set in \mathbb{R}^n and f:S \rightarrow \mathbb{R}^n. Then f is convex if and only if for each integer k>0
x_1,x_2,...x_k \in S, \displaystyle\sum\limits_{i=1}^k \lambda_i=1, \lambda_i\geq 0, \forall i=1,2,s,k, we have f\left ( \displaystyle\sum\limits_{i=1}^k \lambda_ix_i \right )\leq \displaystyle\sum\limits_{i=1}^k \lambda _if\left ( x \right )
Proof
By induction on k.
k=1:x_1 \in S Therefore f\left ( \lambda_1 x_1\right ) \leq \lambda_i f\left (x_1\right ) because \lambda_i=1.
k=2:\lambda_1+\lambda_2=1 and x_1, x_2 \in S
Therefore, \lambda_1x_1+\lambda_2x_2 \in S
Hence by definition, f\left ( \lambda_1 x_1 +\lambda_2 x_2 \right )\leq \lambda _1f\left ( x_1 \right )+\lambda _2f\left ( x_2 \right )
Let the statement is true for $n
Therefore,
f\left ( \lambda_1 x_1+ \lambda_2 x_2+....+\lambda_k x_k\right )\leq \lambda_1 f\left (x_1 \right )+\lambda_2 f\left (x_2 \right )+...+\lambda_k f\left (x_k \right )
k=n+1: Let x_1, x_2,....x_n,x_{n+1} \in S and \displaystyle\sum\limits_{i=1}^{n+1}=1
Therefore \mu_1x_1+\mu_2x_2+.......+\mu_nx_n+\mu_{n+1} x_{n+1} \in S
thus,f\left (\mu_1x_1+\mu_2x_2+...+\mu_nx_n+\mu_{n+1} x_{n+1} \right )
=f\left ( \left ( \mu_1+\mu_2+...+\mu_n \right)\frac{\mu_1x_1+\mu_2x_2+...+\mu_nx_n}{\mu_1+\mu_2+\mu_3}+\mu_{n+1}x_{n+1} \right)
=f\left ( \mu_y+\mu_{n+1}x_{n+1} \right ) where \mu=\mu_1+\mu_2+...+\mu_n and
y=\frac{\mu_1x_1+\mu_2x_2+...+\mu_nx_n}{\mu_1+\mu_2+...+\mu_n} and also \mu_1+\mu_{n+1}=1,y \in S
\Rightarrow f\left ( \mu_1x_1+\mu_2x_2+...+\mu_nx_n+\mu_{n+1}x_{n+1}\right ) \leq \mu f\left ( y \right )+\mu_{n+1} f\left ( x_{n+1} \right )
\Rightarrow f\left ( \mu_1x_1+\mu_2x_2+...+\mu_nx_n+\mu_{n+1}x_{n+1}\right ) \leq
\left ( \mu_1+\mu_2+...+\mu_n \right )f\left ( \frac{\mu_1x_1+\mu_2x_2+...+\mu_nx_n}{\mu_1+\mu_2+...+\mu_n} \right )+\mu_{n+1}f\left ( x_{n+1} \right )
\Rightarrow f\left ( \mu_1x_1+\mu_2x_2+...+\mu_nx_n +\mu_{n+1}x_{n+1}\right )\leq \left ( \mu_1+ \mu_2+ ...+\mu_n \right )
\left [ \frac{\mu_1}{\mu_1+ \mu_2+ ...+\mu_n}f\left ( x_1 \right )+...+\frac{\mu_n}{\mu_1+ \mu_2+ ...+\mu_n}f\left ( x_n \right ) \right ]+\mu_{n+1}f\left ( x_{n+1} \right )
\Rightarrow f\left ( \mu_1x_1+\mu_2x_2+...+\mu_nx_n+\mu_{n+1}x_{n+1}\right )\leq \mu_1f\left ( x_1 \right )+\mu_2f\left ( x_2 \right )+....
Hence Proved.
Convex Optimization - Differentiable Function
Let S be a non-empty open set in \mathbb{R}^n,then f:S\rightarrow \mathbb{R} is said to be differentiable at \hat{x} \in S if there exist a vector \bigtriangledown f\left ( \hat{x} \right ) called gradient vector and a function \alpha :\mathbb{R}^n\rightarrow \mathbb{R} such that
f\left ( x \right )=f\left ( \hat{x} \right )+\bigtriangledown f\left ( \hat{x} \right )^T\left ( x-\hat{x} \right )+\left \| x=\hat{x} \right \|\alpha \left ( \hat{x}, x-\hat{x} \right ), \forall x \in S where
\alpha \left (\hat{x}, x-\hat{x} \right )\rightarrow 0 \bigtriangledown f\left ( \hat{x} \right )=\left [ \frac{\partial f}{\partial x_1}\frac{\partial f}{\partial x_2}...\frac{\partial f}{\partial x_n} \right ]_{x=\hat{x}}^{T}
Theorem
let S be a non-empty, open convexset in \mathbb{R}^n and let f:S\rightarrow \mathbb{R} be differentiable on S. Then, f is convex if and only if for x_1,x_2 \in S, \bigtriangledown f\left ( x_2 \right )^T \left ( x_1-x_2 \right ) \leq f\left ( x_1 \right )-f\left ( x_2 \right )
Proof
Let f be a convex function. i.e., for x_1,x_2 \in S, \lambda \in \left ( 0, 1 \right )
f\left [ \lambda x_1+\left ( 1-\lambda \right )x_2 \right ]\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right )
\Rightarrow f\left [ \lambda x_1+\left ( 1-\lambda \right )x_2 \right ]\leq \lambda \left ( f\left ( x_1 \right )-f\left ( x_2 \right ) \right )+f\left ( x_2 \right )
\Rightarrow\lambda \left ( f\left ( x_1 \right )-f\left ( x_2 \right ) \right )\geq f\left ( x_2+\lambda \left ( x_1-x_2 \right ) \right )-f\left ( x_2 \right )
\Rightarrow \lambda \left ( f\left ( x_1 \right )-f\left ( x_2 \right ) \right )\geq f\left ( x_2 \right )+\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )\lambda +
\left \| \lambda \left ( x_1-x_2 \right ) \right \|\alpha \left ( x_2,\lambda\left (x_1 - x_2 \right )-f\left ( x_2 \right ) \right )
where \alpha\left ( x_2, \lambda\left (x_1 - x_2 \right ) \right )\rightarrow 0 as\lambda \rightarrow 0
Dividing by \lambda on both sides, we get −
f\left ( x_1 \right )-f\left ( x_2 \right ) \geq \bigtriangledown f\left ( x_2 \right )^T \left ( x_1-x_2 \right )
Converse
Let for x_1,x_2 \in S, \bigtriangledown f\left ( x_2 \right )^T \left ( x_1-x_2 \right ) \leq f\left ( x_1 \right )-f \left ( x_2 \right )
To show that f is convex.
Since S is convex, x_3=\lambda x_1+\left (1-\lambda \right )x_2 \in S, \lambda \in \left ( 0, 1 \right )
Since x_1, x_3 \in S, therefore
f\left ( x_1 \right )-f \left ( x_3 \right ) \geq \bigtriangledown f\left ( x_3 \right )^T \left ( x_1 -x_3\right )
\Rightarrow f\left ( x_1 \right )-f \left ( x_3 \right )\geq \bigtriangledown f\left ( x_3 \right )^T \left ( x_1 - \lambda x_1-\left (1-\lambda \right )x_2\right )
\Rightarrow f\left ( x_1 \right )-f \left ( x_3 \right )\geq \left ( 1- \lambda\right )\bigtriangledown f\left ( x_3 \right )^T \left ( x_1 - x_2\right )
Since, x_2, x_3 \in S therefore
f\left ( x_2 \right )-f\left ( x_3 \right )\geq \bigtriangledown f\left ( x_3 \right )^T\left ( x_2-x_3 \right )
\Rightarrow f\left ( x_2 \right )-f\left ( x_3 \right )\geq \bigtriangledown f\left ( x_3 \right )^T\left ( x_2-\lambda x_1-\left ( 1-\lambda \right )x_2 \right )
\Rightarrow f\left ( x_2 \right )-f\left ( x_3 \right )\geq \left ( -\lambda \right )\bigtriangledown f\left ( x_3 \right )^T\left ( x_1-x_2 \right )
Thus, combining the above equations, we get −
\lambda \left ( f\left ( x_1 \right )-f\left ( x_3 \right ) \right )+\left ( 1- \lambda \right )\left ( f\left ( x_2 \right )-f\left ( x_3 \right ) \right )\geq 0
\Rightarrow f\left ( x_3\right )\leq \lambda f\left ( x_1 \right )+\left ( 1-\lambda \right )f\left ( x_2 \right )
Theorem
let S be a non-empty open convex set in \mathbb{R}^n and let f:S \rightarrow \mathbb{R} be differentiable on S, then f is convex on S if and only if for any x_1,x_2 \in S,\left ( \bigtriangledown f \left ( x_2 \right )-\bigtriangledown f \left ( x_1 \right ) \right )^T \left ( x_2-x_1 \right ) \geq 0
Proof
let f be a convex function, then using the previous theorem −
\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )\leq f\left ( x_1 \right )-f\left ( x_2 \right ) and
\bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\leq f\left ( x_2 \right )-f\left ( x_1 \right )
Adding the above two equations, we get −
\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )+\bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\leq 0
\Rightarrow \left ( \bigtriangledown f\left ( x_2 \right )-\bigtriangledown f\left ( x_1 \right ) \right )^T\left ( x_1-x_2 \right )\leq 0
\Rightarrow \left ( \bigtriangledown f\left ( x_2 \right )-\bigtriangledown f\left ( x_1 \right ) \right )^T\left ( x_2-x_1 \right )\geq 0
Converse
Let for any x_1,x_2 \in S,\left (\bigtriangledown f \left ( x_2\right )- \bigtriangledown f \left ( x_1\right )\right )^T \left ( x_2-x_1\right )\geq 0
To show that f is convex.
Let x_1,x_2 \in S, thus by mean value theorem, \frac{f\left ( x_1\right )-f\left ( x_2\right )}{x_1-x_2}=\bigtriangledown f\left ( x\right ),x \in \left ( x_1-x_2\right ) \Rightarrow x= \lambda x_1+\left ( 1-\lambda\right )x_2 because S is a convex set.
\Rightarrow f\left ( x_1 \right )- f\left ( x_2 \right )=\left ( \bigtriangledown f\left ( x \right )^T \right )\left ( x_1-x_2 \right )
for x,x_1, we know −
\left ( \bigtriangledown f\left ( x \right )-\bigtriangledown f\left ( x_1 \right ) \right )^T\left ( x-x_1 \right )\geq 0
\Rightarrow \left ( \bigtriangledown f\left ( x \right )-\bigtriangledown f\left ( x_1 \right ) \right )^T\left ( \lambda x_1+\left ( 1-\lambda \right )x_2-x_1 \right )\geq 0
\Rightarrow \left ( \bigtriangledown f\left ( x \right )- \bigtriangledown f\left ( x_1 \right )\right )^T\left ( 1- \lambda \right )\left ( x_2-x_1 \right )\geq 0
\Rightarrow \bigtriangledown f\left ( x \right )^T\left ( x_2-x_1 \right )\geq \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )
Combining the above equations, we get −
\Rightarrow \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\leq f\left ( x_2 \right )-f\left ( x_1 \right )
Hence using the last theorem, f is a convex function.
Twice Differentiable function
Let S be a non-empty subset of \mathbb{R}^n and let f:S\rightarrow \mathbb{R} then f is said to be twice differentiable at \bar{x} \in S if there exists a vector \bigtriangledown f\left (\bar{x}\right ), a \:nXn matrix H\left (x\right )called Hessian matrix and a function \alpha:\mathbb{R}^n \rightarrow \mathbb{R} such that f\left ( x \right )=f\left ( \bar{x}+x-\bar{x} \right )=f\left ( \bar{x} \right )+\bigtriangledown f\left ( \bar{x} \right )^T\left ( x-\bar{x} \right )+\frac{1}{2}\left ( x-\bar{x} \right )H\left ( \bar{x} \right )\left ( x-\bar{x} \right )
where \alpha \left ( \bar{x}, x-\bar{x} \right )\rightarrow Oasx\rightarrow \bar{x}
Sufficient & Necessary Conditions for Global Optima
Theorem
Let f be twice differentiable function. If \bar{x} is a local minima, then \bigtriangledown f\left ( \bar{x} \right )=0 and the Hessian matrix H\left ( \bar{x} \right ) is a positive semidefinite.
Proof
Let d \in \mathbb{R}^n. Since f is twice differentiable at \bar{x}.
Therefore,
f\left ( \bar{x} +\lambda d\right )=f\left ( \bar{x} \right )+\lambda \bigtriangledown f\left ( \bar{x} \right )^T d+\lambda^2d^TH\left ( \bar{x} \right )d+\lambda^2d^TH\left ( \bar{x} \right )d+
\lambda^2\left \| d \right \|^2\beta \left ( \bar{x}, \lambda d \right )
But \bigtriangledown f\left ( \bar{x} \right )=0 and \beta\left ( \bar{x}, \lambda d \right )\rightarrow 0 as \lambda \rightarrow 0
\Rightarrow f\left ( \bar{x} +\lambda d \right )-f\left ( \bar{x} \right )=\lambda ^2d^TH\left ( \bar{x} \right )d
Since \bar{x } is a local minima, there exists a \delta > 0 such that f\left ( x \right )\leq f\left ( \bar{x}+\lambda d \right ), \forall \lambda \in \left ( 0,\delta \right )
Theorem
Let f:S \rightarrow \mathbb{R}^n where S \subset \mathbb{R}^n be twice differentiable over S. If \bigtriangledown f\left ( x\right )=0 and H\left ( \bar{x} \right ) is positive semi-definite, for all x \in S, then \bar{x} is a global optimal solution.
Proof
Since H\left ( \bar{x} \right ) is positive semi-definite, f is convex function over S. Since f is differentiable and convex at \bar{x}
\bigtriangledown f\left ( \bar{x} \right )^T \left ( x-\bar{x} \right ) \leq f\left (x\right )-f\left (\bar{x}\right ),\forall x \in S
Since \bigtriangledown f\left ( \bar{x} \right )=0, f\left ( x \right )\geq f\left ( \bar{x} \right )
Hence, \bar{x} is a global optima.
Theorem
Suppose \bar{x} \in S is a local optimal solution to the problem f:S \rightarrow \mathbb{R} where S is a non-empty subset of \mathbb{R}^n and S is convex. min \:f\left ( x \right ) where x \in S.
Then:
\bar{x} is a global optimal solution.
If either \bar{x} is strictly local minima or f is strictly convex function, then \bar{x} is the unique global optimal solution and is also strong local minima.
Proof
Let \bar{x} be another global optimal solution to the problem such that x \neq \bar{x} and f\left ( \bar{x} \right )=f\left ( \hat{x} \right )
Since \hat{x},\bar{x} \in S and S is convex, then \frac{\hat{x}+\bar{x}}{2} \in S and f is strictly convex.
$\Rightarrow f\left ( \frac{\hat{x}+\bar{x}}{2} \right )
This is contradiction.
Hence, \hat{x} is a unique global optimal solution.
Corollary
Let f:S \subset \mathbb{R}^n \rightarrow \mathbb{R} be a differentiable convex function where \phi \neq S\subset \mathbb{R}^n is a convex set. Consider the problem min f\left (x\right ),x \in S,then \bar{x} is an optimal solution if \bigtriangledown f\left (\bar{x}\right )^T\left (x-\bar{x}\right ) \geq 0,\forall x \in S.
Proof
Let \bar{x} is an optimal solution, i.e, f\left (\bar{x}\right )\leq f\left (x\right ),\forall x \in S
\Rightarrow f\left (x\right )=f\left (\bar{x}\right )\geq 0
f\left (x\right )=f\left (\bar{x}\right )+\bigtriangledown f\left (\bar{x}\right )^T\left (x-\bar{x}\right )+\left \| x-\bar{x} \right \|\alpha \left ( \bar{x},x-\bar{x} \right )
where \alpha \left ( \bar{x},x-\bar{x} \right )\rightarrow 0 as x \rightarrow \bar{x}
\Rightarrow f\left (x\right )-f\left (\bar{x}\right )=\bigtriangledown f\left (\bar{x}\right )^T\left (x-\bar{x}\right )\geq 0
Corollary
Let f be a differentiable convex function at \bar{x},then \bar{x} is global minimum iff \bigtriangledown f\left (\bar{x}\right )=0
Examples
-
f\left (x\right )=\left (x^2-1\right )^{3}, x \in \mathbb{R}.
\bigtriangledown f\left (x\right )=0 \Rightarrow x= -1,0,1.
\bigtriangledown^2f\left (\pm 1 \right )=0, \bigtriangledown^2 f\left (0 \right )=6>0.
f\left (\pm 1 \right )=0,f\left (0 \right )=-1
Hence, f\left (x \right ) \geq -1=f\left (0 \right )\Rightarrow f\left (0 \right ) \leq f \left (x \right)\forall x \in \mathbb{R}
-
f\left (x \right )=x\log x defined on S=\left \{ x \in \mathbb{R}, x> 0 \right \}.
{f}'x=1+\log x
{f}''x=\frac{1}{x}>0
Thus, this function is strictly convex.
f \left (x \right )=e^{x},x \in \mathbb{R} is strictly convex.
Quasiconvex and Quasiconcave functions
Let f:S \rightarrow \mathbb{R} where S \subset \mathbb{R}^n is a non-empty convex set. The function f is said to be quasiconvex if for each x_1,x_2 \in S, we have f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq max\left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \},\lambda \in \left ( 0, 1 \right )
For example, f\left ( x \right )=x^{3}
Let f:S\rightarrow R where S\subset \mathbb{R}^n is a non-empty convex set. The function f is said to be quasiconvex if for each x_1, x_2 \in S, we have f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\geq min\left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \}, \lambda \in \left ( 0, 1 \right )
Remarks
- Every convex function is quasiconvex but the converse is not true.
- A function which is both quasiconvex and quasiconcave is called quasimonotone.
Theorem
Let f:S\rightarrow \mathbb{R} and S is a non empty convex set in \mathbb{R}^n. The function f is quasiconvex if and only if S_{\alpha} =\left ( x \in S:f\left ( x \right )\leq \alpha \right \} is convex for each real number \alpha$
Proof
Let f is quasiconvex on S.
Let x_1,x_2 \in S_{\alpha} therefore x_1,x_2 \in S and max \left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \}\leq \alpha
Let \lambda \in \left (0, 1 \right ) and let x=\lambda x_1+\left ( 1-\lambda \right )x_2\leq max \left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \}\Rightarrow x \in S
Thus, f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq max\left \{ f\left ( x_1 \right ), f\left ( x_2 \right ) \right \}\leq \alpha
Therefore, S_{\alpha} is convex.
Converse
Let S_{\alpha} is convex for each \alpha
x_1,x_2 \in S, \lambda \in \left ( 0,1\right )
x=\lambda x_1+\left ( 1-\lambda \right )x_2
Let x=\lambda x_1+\left ( 1-\lambda \right )x_2
For x_1, x_2 \in S_{\alpha}, \alpha= max \left \{ f\left ( x_1 \right ), f\left ( x_2 \right ) \right \}
\Rightarrow \lambda x_1+\left (1-\lambda \right )x_2 \in S_{\alpha}
\Rightarrow f \left (\lambda x_1+\left (1-\lambda \right )x_2 \right )\leq \alpha
Hence proved.
Theorem
Let f:S\rightarrow \mathbb{R} and S is a non empty convex set in \mathbb{R}^n. The function f is quasiconcave if and only if S_{\alpha} =\left \{ x \in S:f\left ( x \right )\geq \alpha \right \} is convex for each real number \alpha.
Theorem
Let f:S\rightarrow \mathbb{R} and S is a non empty convex set in \mathbb{R}^n. The function f is quasimonotone if and only if S_{\alpha} =\left \{ x \in S:f\left ( x \right )= \alpha \right \} is convex for each real number \alpha.
Differentiable Quasiconvex Function
Theorem
Let S be a non empty convex set in \mathbb{R}^n and f:S \rightarrow \mathbb{R} be differentiable on S, then f is quasiconvex if and only if for any x_1,x_2 \in S and f\left ( x_1 \right )\leq f\left ( x_2 \right ), we have \bigtriangledown f\left ( x_2 \right )^T\left ( x_2-x_1 \right )\leq 0
Proof
Let f be a quasiconvex function.
Let x_1,x_2 \in S such that f\left ( x_1 \right ) \leq f\left ( x_2 \right )
By differentiability of f at x_2, \lambda \in \left ( 0, 1 \right )
f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )=f\left ( x_2+\lambda \left (x_1-x_2 \right ) \right )=f\left ( x_2 \right )+\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )
+\lambda \left \| x_1-x_2 \right \|\alpha \left ( x_2,\lambda \left ( x_1-x_2 \right ) \right )
\Rightarrow f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )-f\left ( x_2 \right )-f\left ( x_2 \right )=\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )
+\lambda \left \| x_1-x_2 \right \|\alpha \left ( x2, \lambda\left ( x_1-x_2 \right )\right )
But since f is quasiconvex, f \left ( \lambda x_1+ \left ( 1- \lambda \right )x_2 \right )\leq f \left (x_2 \right )
\bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right )+\lambda \left \| x_1-x_2 \right \|\alpha \left ( x_2,\lambda \left ( x_1,x_2 \right ) \right )\leq 0
But \alpha \left ( x_2,\lambda \left ( x_1,x_2 \right )\right )\rightarrow 0 as \lambda \rightarrow 0
Therefore, \bigtriangledown f\left ( x_2 \right )^T\left ( x_1-x_2 \right ) \leq 0
Converse
let for x_1,x_2 \in S and f\left ( x_1 \right )\leq f\left ( x_2 \right ), \bigtriangledown f\left ( x_2 \right )^T \left ( x_1,x_2 \right ) \leq 0
To show that f is quasiconvex,ie, f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )\leq f\left ( x_2 \right )
Proof by contradiction
Suppose there exists an x_3= \lambda x_1+\left ( 1-\lambda \right )x_2 such that $f\left ( x_2 \right )
For x_2 and x_3,\bigtriangledown f\left ( x_3 \right )^T \left ( x_2-x_3 \right ) \leq 0
\Rightarrow -\lambda \bigtriangledown f\left ( x_3 \right )^T\left ( x_2-x_3 \right )\leq 0
\Rightarrow \bigtriangledown f\left ( x_3 \right )^T \left ( x_1-x_2 \right )\geq 0
For x_1 and x_3,\bigtriangledown f\left ( x_3 \right )^T \left ( x_1-x_3 \right ) \leq 0
\Rightarrow \left ( 1- \lambda \right )\bigtriangledown f\left ( x_3 \right )^T\left ( x_1-x_2 \right )\leq 0
\Rightarrow \bigtriangledown f\left ( x_3 \right )^T \left ( x_1-x_2 \right )\leq 0
thus, from the above equations, \bigtriangledown f\left ( x_3 \right )^T \left ( x_1-x_2 \right )=0
Define U=\left \{ x:f\left ( x \right )\leq f\left ( x_2 \right ),x=\mu x_2+\left ( 1-\mu \right )x_3, \mu \in \left ( 0,1 \right ) \right \}
Thus we can find x_0 \in U such that x_0 = \mu_0 x_2= \mu x_2+\left ( 1- \mu \right )x_3 for some \mu _0 \in \left ( 0,1 \right ) which is nearest to x_3 and \hat{x} \in \left ( x_0,x_1 \right ) such that by mean value theorem,
\frac{f\left ( x_3\right )-f\left ( x_0\right )}{x_3-x_0}= \bigtriangledown f\left ( \hat{x}\right )
\Rightarrow f\left ( x_3 \right )=f\left ( x_0 \right )+\bigtriangledown f\left ( \hat{x} \right )^T\left ( x_3-x_0 \right )
\Rightarrow f\left ( x_3 \right )=f\left ( x_0 \right )+\mu_0 \lambda f\left ( \hat{x}\right )^T \left ( x_1-x_2 \right )
Since x_0 is a combination of x_1 and x_2 and $f\left (x_2 \right )
By repeating the starting procedure, \bigtriangledown f \left ( \hat{x}\right )^T \left ( x_1-x_2\right )=0
Thus, combining the above equations, we get:
f\left ( x_3\right )=f\left ( x_0 \right ) \leq f\left ( x_2\right )
\Rightarrow f\left ( x_3\right )\leq f\left ( x_2\right )
Hence, it is contradiction.
Examples
Step 1 − f\left ( x\right )=X^3
Let f \left ( x_1\right )\leq f\left ( x_2\right )
\Rightarrow x_{1}^{3}\leq x_{2}^{3}\Rightarrow x_1\leq x_2
\bigtriangledown f\left ( x_2 \right )\left ( x_1-x_2 \right )=3x_{2}^{2}\left ( x_1-x_2 \right )\leq 0
Thus, f\left ( x\right ) is quasiconvex.
Step 2 − f\left ( x\right )=x_{1}^{3}+x_{2}^{3}
Let \hat{x_1}=\left ( 2, -2\right ) and \hat{x_2}=\left ( 1, 0\right )
thus, $f\left ( \hat{x_1}\right )=0,f\left ( \hat{x_2}\right )=1 \Rightarrow f\left ( \hat{x_1}\right )\setminus
Thus, \bigtriangledown f \left ( \hat{x_2}\right )^T \left ( \hat{x_1}- \hat{x_2}\right )= \left ( 3, 0\right )^T \left ( 1, -2\right )=3 >0
Hence f\left ( x\right ) is not quasiconvex.
Strictly Quasiconvex Function
Let f:S\rightarrow \mathbb{R}^n and S be a non-empty convex set in \mathbb{R}^n then f is said to be strictly quasicovex function if for each x_1,x_2 \in S with f\left ( x_1 \right ) \neq f\left ( x_2 \right ), we have $f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )
Remarks
- Every strictly quasiconvex function is strictly convex.
- Strictly quasiconvex function does not imply quasiconvexity.
- Strictly quasiconvex function may not be strongly quasiconvex.
- Pseudoconvex function is a strictly quasiconvex function.
Theorem
Let f:S\rightarrow \mathbb{R}^n be strictly quasiconvex function and S be a non-empty convex set in \mathbb{R}^n.Consider the problem: min \:f\left ( x \right ), x \in S. If \hat{x} is local optimal solution, then \bar{x} is global optimal solution.
Proof
Let there exists \bar{x} \in S such that f\left ( \bar{x}\right )\leq f \left ( \hat{x}\right )
Since \bar{x},\hat{x} \in S and S is convex set, therefore,
\lambda \bar{x}+\left ( 1-\lambda \right )\hat{x}\in S, \forall \lambda \in \left ( 0,1 \right )
Since \hat{x} is local minima, f\left ( \hat{x} \right ) \leq f\left ( \lambda \bar{x}+\left ( 1-\lambda \right )\hat{x} \right ), \forall \lambda \in \left ( 0,\delta \right )
Since f is strictly quasiconvex.
$$f\left ( \lambda \bar{x}+\left ( 1-\lambda \right )\hat{x} \right )
Hence, it is contradiction.
Strictly quasiconcave function
Let f:S\rightarrow \mathbb{R}^n and S be a non-empty convex set in \mathbb{R}^n, then f is saud to be strictly quasicovex function if for each x_1,x_2 \in S with f\left (x_1\right )\neq f\left (x_2\right ), we have
f\left (\lambda x_1+\left (1-\lambda\right )x_2\right )> min \left \{ f \left (x_1\right ),f\left (x_2\right )\right \}.
Examples
-
f\left (x\right )=x^2-2
It is a strictly quasiconvex function because if we take any two points x_1,x_2 in the domain that satisfy the constraints in the definition $f\left (\lambda x_1+\left (1- \lambda\right )x_2\right )
-
f\left (x\right )=-x^2
It is not a strictly quasiconvex function because if we take take x_1=1 and x_2=-1 and \lambda=0.5, then f\left (x_1\right )=-1=f\left (x_2\right ) but f\left (\lambda x_1+\left (1- \lambda\right )x_2\right )=0 Therefore it does not satisfy the conditions stated in the definition. But it is a quasiconcave function because if we take any two points in the domain that satisfy the constraints in the definition f\left ( \lambda x_1+\left (1-\lambda\right )x_2\right )> min \left \{ f \left (x_1\right ),f\left (x_2\right )\right \}. As the function is increasing in the negative x-axis and it is decreasing in the positive x-axis.
Strongly Quasiconvex Function
Let f:S\rightarrow \mathbb{R}^n and S be a non-empty convex set in \mathbb{R}^n then f is strongly quasiconvex function if for any x_1,x_2 \in S with \left ( x_1 \right ) \neq \left ( x_2 \right ), we have $f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )
Theorem
A quasiconvex function f:S\rightarrow \mathbb{R}^n on a non-empty convex set S in \mathbb{R}^n is strongly quasiconvex function if it is not constant on a line segment joining any points of S.
Proof
Let f is quasiconvex function and it is not constant on a line segment joining any points of S.
Suppose f is not strongly quasiconvex function.
There exist x_1,x_2 \in S with x_1 \neq x_2 such that
f\left ( z \right )\geq max\left \{ f\left ( x_1 \right ), f\left ( x_2 \right ) \right \}, \forall z= \lambda x_1+\left ( 1-\lambda \right )x_2, \lambda \in \left ( 0,1 \right )
\Rightarrow f\left ( x_1 \right )\leq f\left ( z \right ) and f\left ( x_2 \right )\leq f\left ( z \right )
Since f is not constant in \left [ x_1,z \right ] and \left [z,x_2 \right ]
So there exists u \in \left [ x_1,z \right ] and v=\left [ z,x_2 \right ]
\Rightarrow u= \mu_1x_1+\left ( 1-\mu_1\right )z,v=\mu_2z+\left ( 1- \mu_2\right )x_2
Since f is quasiconvex,
\Rightarrow f\left ( u \right )\leq max\left \{ f\left ( x_1 \right ),f \left ( z \right ) \right \}=f\left ( z \right )\:\: and \:\:f \left ( v \right ) \leq max \left \{ f\left ( z \right ),f\left ( x_2 \right ) \right \}
\Rightarrow f\left ( u \right )\leq f\left ( z \right ) \:\: and \:\: f\left ( v \right )\leq f\left ( z \right )
\Rightarrow max \left \{ f\left ( u \right ),f\left ( v \right ) \right \} \leq f\left ( z \right )
But z is any point between u and v, if any of them are equal, then f is constant.
Therefore, max \left \{ f\left ( u \right ),f\left ( v \right ) \right \} \leq f\left ( z \right )
which contradicts the quasiconvexity of f as z \in \left [ u,v \right ].
Hence f is strongly quasiconvex function.
Theorem
Let f:S\rightarrow \mathbb{R}^n and S be a non-empty convex set in \mathbb{R}^n. If \hat{x} is local optimal solution, then \hat{x} is unique global optimal solution.
Proof
Since a strong quasiconvex function is also strictly quasiconvex function, thus a local optimal solution is global optimal solution.
Uniqueness − Let f attains global optimal solution at two points u,v \in S
\Rightarrow f\left ( u \right ) \leq f\left ( x \right ).\forall x \in S\:\: and \:\:f\left ( v \right ) \leq f\left ( x \right ).\forall x \in S
If u is global optimal solution, f\left ( u \right )\leq f\left ( v \right ) and f\left ( v \right )\leq f\left ( u\right )\Rightarrow f\left ( u \right )=f\left ( v\right )
$$f\left ( \lambda u+\left ( 1-\lambda\right )v\right )
which is a contradiction.
Hence there exists only one global optimal solution.
Remarks
- A strongly quasiconvex function is also strictly quasiconvex fucntion.
- A strictly convex function may or may not be strongly quasiconvex.
- A differentiable strictly convex is strongly quasiconvex.
Pseudoconvex Function
Let f:S\rightarrow \mathbb{R} be a differentiable function and S be a non-empty convex set in \mathbb{R}^n, then f is said to be pseudoconvex if for each x_1,x_2 \in S with \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\geq 0, we have f\left ( x_2 \right )\geq f\left ( x_1 \right ), or equivalently if f\left ( x_1 \right )>f\left ( x_2 \right ) then $\bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )
Pseudoconcave function
Let f:S\rightarrow \mathbb{R} be a differentiable function and S be a non-empty convex set in \mathbb{R}^n, then f is said to be pseudoconvex if for each x_1, x_2 \in S with \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\geq 0, we have f\left ( x_2 \right )\leq f\left ( x_1 \right ), or equivalently if f\left ( x_1 \right )>f\left ( x_2 \right ) then \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )>0
Remarks
If a function is both pseudoconvex and pseudoconcave, then is is called pseudolinear.
A differentiable convex function is also pseudoconvex.
-
A pseudoconvex function may not be convex. For example,
f\left ( x \right )=x+x^3 is not convex. If x_1 \leq x_2,x_{1}^{3} \leq x_{2}^{3}
Thus,\bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )=\left ( 1+3x_{1}^{2} \right )\left ( x_2-x_1 \right ) \geq 0
And, f\left ( x_2 \right )-f\left ( x_1 \right )=\left ( x_2-x_1 \right )+\left ( x_{2}^{3} -x_{1}^{3}\right )\geq 0
\Rightarrow f\left ( x_2 \right )\geq f\left ( x_1 \right )
Thus, it is pseudoconvex.
A pseudoconvex function is strictly quasiconvex. Thus, every local minima of pseudoconvex is also global minima.
Strictly pseudoconvex function
Let f:S\rightarrow \mathbb{R} be a differentiable function and S be a non-empty convex set in \mathbb{R}^n, then f is said to be pseudoconvex if for each x_1,x_2 \in S with \bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )\geq 0, we have f\left ( x_2 \right )> f\left ( x_1 \right ),or equivalently if f\left ( x_1 \right )\geq f\left ( x_2 \right ) then $\bigtriangledown f\left ( x_1 \right )^T\left ( x_2-x_1 \right )
Theorem
Let f be a pseudoconvex function and suppose \bigtriangledown f\left ( \hat{x}\right )=0 for some \hat{x} \in S, then \hat{x} is global optimal solution of f over S.
Proof
Let \hat{x} be a critical point of f, ie, \bigtriangledown f\left ( \hat{x}\right )=0
Since f is pseudoconvex function, for x \in S, we have
\bigtriangledown f\left ( \hat{x}\right )\left ( x-\hat{x}\right )=0 \Rightarrow f\left ( \hat{x}\right )\leq f\left ( x\right ), \forall x \in S
Hence, \hat{x} is global optimal solution.
Remark
If f is strictly pseudoconvex function, \hat{x} is unique global optimal solution.
Theorem
If f is differentiable pseudoconvex function over S, then f is both strictly quasiconvex as well as quasiconvex function.
Remarks
The sum of two pseudoconvex fucntions defined on an open set S of \mathbb{R}^n may not be pseudoconvex.
Let f:S\rightarrow \mathbb{R} be a quasiconvex function and S be a non-empty convex subset of \mathbb{R}^n then f is pseudoconvex if and only if every critical point is a global minima of f over S.
Let S be a non-empty convex subset of \mathbb{R}^n and f:S\rightarrow \mathbb{R} be a function such that \bigtriangledown f\left ( x\right )\neq 0 for every x \in S then f is pseudoconvex if and only if it is a quasiconvex function.
Convex Optimization - Programming Problem
There are four types of convex programming problems −
Step 1 − min \:f\left ( x \right ), where x \in S and S be a non-empty convex set in \mathbb{R}^n and f\left ( x \right ) is convex function.
Step 2 − min \: f\left ( x \right ), x \in \mathbb{R}^n subject to
g_i\left ( x \right ) \geq 0, 1 \leq m_1 and g_i\left ( x \right ) is a convex function.
g_i\left ( x \right ) \leq 0,m_1+1 \leq m_2 and g_i\left ( x \right ) is a concave function.
g_i\left ( x \right ) = 0, m_2+1 \leq m and g_i\left ( x \right ) is a linear function.
where f\left ( x \right ) is a convex fucntion.
Step 3 − max \:f\left ( x \right ) where x \in S and S be a non-empty convex set in \mathbb{R}^n and f\left ( x \right ) is concave function.
Step 4 − min \:f\left ( x \right ), where x \in \mathbb{R}^n subject to
g_i\left ( x \right ) \geq 0, 1 \leq m_1 and g_i\left ( x \right ) is a convex function.
g_i\left ( x \right ) \leq 0, m_1+1 \leq m_2 and g_i\left ( x \right ) is a concave function.
g_i\left ( x \right ) = 0,m_2+1 \leq m and g_i\left ( x \right ) is a linear function.
where f\left ( x \right ) is a concave function.
Cone of feasible direction
Let S be a non-empty set in \mathbb{R}^n and let \hat{x} \in \:Closure\left ( S \right ), then the cone of feasible direction of S at \hat{x}, denoted by D, is defined as D=\left \{ d:d\neq 0,\hat{x}+\lambda d \in S, \lambda \in \left ( 0, \delta \right ), \delta > 0 \right \}
Each non-zero vector d \in D is called feasible direction.
For a given function f:\mathbb{R}^n \Rightarrow \mathbb{R} the cone of improving direction at \hat{x} is denoted by F and is given by
F=\left \{ d:f\left ( \hat{x}+\lambda d \right )\leq f\left ( \hat{x} \right ),\forall \lambda \in \left ( 0,\delta \right ), \delta >0 \right \}
Each direction d \in F is called an improving direction or descent direction of f at \hat{x}
Theorem
Necessary Condition
Consider the problem min f\left ( x \right ) such that x \in S where S be a non-empty set in \mathbb{R}^n. Suppose f is differentiable at a point \hat{x} \in S. If \hat{x} is a local optimal solution, then F_0 \cap D= \phi where $F_0=\left \{ d:\bigtriangledown f\left ( \hat{x} \right )^T d
Sufficient Condition
If F_0 \cap D= \phi f is a pseudoconvex function at \hat{x} and there exists a neighbourhood of \hat{x},N_\varepsilon \left ( \hat{x} \right ), \varepsilon > 0 such that d=x-\hat{x}\in D for any x \in S \cap N_\varepsilon \left ( \hat{x} \right ), then \hat{x} is local optimal solution.
Proof
Necessary Condition
Let F_0 \cap D\neq \phi, ie, there exists a d \in F_0 \cap D such that d \in F_0 and d\in D
Since d \in D, therefore there exists \delta_1 >0 such that \hat{x}+\lambda d \in S, \lambda \in \left ( 0,\delta_1 \right ).
Since d \in F_0, therefore $\bigtriangledown f \left ( \hat{x}\right )^T d
Thus, there exists \delta_2>0 such that $f\left ( \hat{x}+\lambda d\right )
Let \delta=min \left \{\delta_1,\delta_2 \right \}
Then $\hat{x}+\lambda d \in S, f\left (\hat{x}+\lambda d \right )
But \hat{x} is local optimal solution.
Hence it is contradiction.
Thus F_0\cap D=\phi
Sufficient Condition
Let F_0 \cap D\neq \phi nd let f be a pseudoconvex function.
Let there exists a neighbourhood of \hat{x}, N_{\varepsilon}\left ( \hat{x} \right ) such that d=x-\hat{x}, \forall x \in S \cap N_\varepsilon\left ( \hat{x} \right )
Let \hat{x} is not a local optimal solution.
Thus, there exists \bar{x} \in S \cap N_\varepsilon \left ( \hat{x} \right ) such that $f \left ( \bar{x} \right )
By assumption on S \cap N_\varepsilon \left ( \hat{x} \right ),d=\left ( \bar{x}-\hat{x} \right )\in D
By pseudoconvex of f,
$$f\left ( \hat{x} \right )>f\left ( \bar{x} \right )\Rightarrow \bigtriangledown f\left ( \hat{x} \right )^T\left ( \bar{x}-\hat{x} \right )
$\Rightarrow \bigtriangledown f\left ( \hat{x} \right) ^T d
\Rightarrow d \in F_0
hence F_0\cap D \neq \phi
which is a contradiction.
Hence, \hat{x} is local optimal solution.
Consider the following problem:min \:f\left ( x\right ) where x \in X such that g_x\left ( x\right ) \leq 0, i=1,2,...,m
f:X \rightarrow \mathbb{R},g_i:X \rightarrow \mathbb{R}^n and X is an open set in \mathbb{R}^n
Let S=\left \{x:g_i\left ( x\right )\leq 0,\forall i \right \}
Let \hat{x} \in X, then M=\left \{1,2,...,m \right \}
Let I=\left \{i:g_i\left ( \hat{x}\right )=0, i \in M\right \} where I is called index set for all active constraints at \hat{x}
Let $J=\left \{i:g_i\left ( \hat{x}\right )
Let $F_0=\left \{ d \in \mathbb{R}^m:\bigtriangledown f\left ( \hat{x} \right )^T d
Let $G_0=\left \{ d \in \mathbb{R}^m:\bigtriangledown gI\left ( \hat{x} \right )^T d
or $G_0=\left \{ d \in \mathbb{R}^m:\bigtriangledown gi\left ( \hat{x} \right )^T d
Lemma
If S=\left \{ x \in X:g_i\left ( x\right ) \leq 0, \forall i \in I\right \} and X is non-empty open set in \mathbb{R}^n. Let \hat{x}\in S and g_i are different at \hat{x}, i \in I and let g_i where i \in J are continuous at \hat{x}, then G_0 \subseteq D.
Proof
Let d \in G_0
Since \hat{x} \in X and X is an open set, thus there exists \delta_1 >0 such that \hat{x}+\lambda d \in X for \lambda \in \left ( 0, \delta_1\right )
Also since g_\hat{x}0, $g_i\left ( \hat{x}+\lambda d\right )
Since d \in G_0, therefore, $\bigtriangledown g_i\left ( \hat{x}\right )^T d 0, g_i\left ( \hat{x}+\lambda d\right )
Let \delta=min\left \{ \delta_1, \delta_2, \delta_3 \right \}
therefore, $\hat{x}+\lambda d \in X, g_i\left ( \hat{x}+\lambda d\right )
\Rightarrow \hat{x}+\lambda d \in S
\Rightarrow d \in D
\Rightarrow G_0 \subseteq D
Hence Proved.
Theorem
Necessary Condition
Let f and g_i, i \in I, are different at \hat{x} \in S, and g_j are continous at \hat{x} \in S. If \hat{x} is local minima of S, then F_0 \cap G_0=\phi.
Sufficient Condition
If F_0 \cap G_0= \phi and f is a pseudoconvex function at \left (\hat{x}, g_i 9x \right ), i \in I are strictly pseudoconvex functions over some \varepsilon - neighbourhood of \hat{x}, \hat{x} is a local optimal solution.
Remarks
Let \hat{x} be a feasible point such that \bigtriangledown f\left(\hat{x} \right)=0, then F_0 = \phi. Thus, F_0 \cap G_0= \phi but \hat{x} is not an optimal solution
But if \bigtriangledown g\left(\hat{x} \right)=0, then G_0=\phi, thus F_0 \cap G_0= \phi
-
Consider the problem : min f\left(x \right) such that g\left(x \right)=0.
Since g\left(x \right)=0, thus $g_1\left(x \right)=g\left(x \right)
Let \hat{x} \in S, then g_1\left(\hat{x} \right)=0 and g_2\left(\hat{x} \right)=0.
But \bigtriangledown g_1\left(\hat{x} \right)= - \bigtriangledown g_2\left(\hat{x}\right)
Thus, G_0= \phi and F_0 \cap G_0= \phi.
Convex Optimization - Fritz-John Conditions
Necessary Conditions
Theorem
Consider the problem − min f\left ( x \right ) such that x \in X where X is an open set in \mathbb{R}^n and let g_i \left ( x \right ) \leq 0, \forall i =1,2,....m.
Let f:X \rightarrow \mathbb{R} and g_i:X \rightarrow \mathbb{R}
Let \hat{x} be a feasible solution and let f and g_i, i \in I are differentiable at \hat{x} and g_i, i \in J are continuous at \hat{x}.
If \hat{x} solves the above problem locally, then there exists u_0, u_i \in \mathbb{R}, i \in I such that u_0 \bigtriangledown f\left ( \hat{x} \right )+\displaystyle\sum\limits_{i\in I} u_i \bigtriangledown g_i \left ( \hat{x} \right )=0
where u_0,u_i \geq 0,i \in I and \left ( u_0, u_I \right ) \neq \left ( 0,0 \right )
Furthermore, if g_i,i \in J are also differentiable at \hat{x}, then above conditions can be written as −
u_0 \bigtriangledown f\left ( \hat{x}\right )+\displaystyle\sum\limits_{i=1}^m u_i \bigtriangledown g_i\left ( \hat{x} \right )=0
u_ig_i\left (\hat{x} \right )=0
u_0,u_i \geq 0, \forall i=1,2,....,m
\left (u_0,u \right ) \neq \left ( 0,0 \right ), u=\left ( u_1,u_2,s,u_m \right ) \in \mathbb{R}^m
Remarks
u_i are called Lagrangian multipliers.
The condition that \hat{x} be feasible to the given problem is called primal feasible condition.
The requirement u_0 \bigtriangledown f\left (\hat{x} \right )+\displaystyle\sum\limits_{i=1}^m u-i \bigtriangledown g_i\left ( x \right )=0 is called dual feasibility condition.
The condition u_ig_i\left ( \hat{x} \right )=0, i=1, 2, ...m is called complimentary slackness condition. This condition requires u_i=0, i \in J
Together the primal feasible condition, dual feasibility condition and complimentary slackness are called Fritz-John Conditions.
Sufficient Conditions
Theorem
If there exists an \varepsilon-neighbourhood of \hat{x}N_\varepsilon \left ( \hat{x} \right ),\varepsilon >0 such that f is pseudoconvex over N_\varepsilon \left ( \hat{x} \right )\cap S and g_i,i \in I are strictly pseudoconvex over N_\varepsilon \left ( \hat{x}\right )\cap S, then \hat{x} is local optimal solution to problem described above. If f is pseudoconvex at \hat{x} and if g_i, i \in I are both strictly pseudoconvex and quasiconvex function at \hat{x},\hat{x} is global optimal solution to the problem described above.
Example
-
min \:f\left ( x_1,x_2 \right )=\left ( x_1-3 \right )^2+\left ( x_2-2 \right )^2
such that x_{1}^{2}+x_{2}^{2} \leq 5, x_1+2x_2 \leq 4, x_1,x_2 \geq 0 And \hat{x}=\left ( 2,1 \right )
Let g_1\left (x_1,x_2 \right )=x_{1}^{2}+x_{2}^{2} -5,
g_2\left (x_1,x_2 \right )=x_1+2x_2-4,
g_3\left (x_1,x_2 \right )=-x_1 and g_4\left ( x_1, x_2 \right )= -x_2.
Thus the above constraints can be written as −
g_1\left (x_1,x_2 \right )\leq 0,
g_2\left (x_1,x_2 \right )\leq 0,
g_3\left (x_1,x_2 \right )\leq 0 and
g_4\left (x_1,x_2 \right )\leq 0 Thus, I = \left \{1,2 \right \} therefore, u_3=0,u_4=0
\bigtriangledown f \left (\hat{x} \right )=\left (2,-2 \right ),\bigtriangledown g_1\left (\hat{x} \right )=\left (4,2 \right ) and \bigtriangledown g_2\left (\hat{x} \right )=\left (1,2 \right )
Thus putting these values in the first condition of Fritz-John conditions, we get −
u_0=\frac{3}{2} u_2, \:\:u_1= \frac{1}{2}u_2, and let u_2=1, therefore u_0= \frac{3}{2},\:\:u_1= \frac{1}{2}
Thus Fritz John conditions are satisfied.
-
min f\left (x_1,x_2\right )=-x_1.
such that x_2-\left (1-x_1\right )^3 \leq 0,
-x_2 \leq 0 and \hat{x}=\left (1,0\right )
Let g_1\left (x_1,x_2 \right )=x_2-\left (1-x_1\right )^3,
g_2\left (x_1,x_2 \right )=-x_2
Thus the above constraints can be wriiten as −
g_1\left (x_1,x_2 \right )\leq 0,
g_2\left (x_1,x_2 \right )\leq 0,
Thus, I=\left \{1,2 \right \}
\bigtriangledown f\left (\hat{x} \right )=\left (-1,0\right )
\bigtriangledown g_1 \left (\hat{x} \right )=\left (0,1\right ) and g_2 \left (\hat{x} \right )=\left (0, -1 \right )
Thus putting these values in the first condition of Fritz-John conditions, we get −
u_0=0,\:\: u_1=u_2=a>0
Thus Fritz John conditions are satisfied.
Karush-Kuhn-Tucker Optimality Necessary Conditions
Consider the problem −
min \:f\left ( x \right ) such that x \in X, where X is an open set in \mathbb{R}^n and g_i \left ( x \right )\leq 0, i=1, 2,...,m
Let S=\left \{ x \in X:g_i\left ( x \right )\leq 0, \forall i \right \}
Let \hat{x} \in S and let f and g_i,i \in I are differentiable at \hat{x} and g_i, i \in J are continuous at \hat{x}. Furthermore, \bigtriangledown g_i\left ( \hat{x} \right), i \in I are linearly independent. If \hat{x} solves the above problem locally, then there exists u_i,i \in I such that
\bigtriangledown f\left ( x\right)+\displaystyle\sum\limits_{i\in I} u_i \bigtriangledown g_i\left ( \hat{x} \right)=0, \:\:u_i \geq 0, i \in I
If g_i,i \in J are also differentiable at \hat{x}. then \hat{x}, then
\bigtriangledown f\left ( \hat{x}\right)+\displaystyle\sum\limits_{i= 1}^m u_i \bigtriangledown g_i\left ( \hat{x} \right)=0
u_ig_i\left ( \hat{x} \right)=0, \forall i=1,2,...,m
u_i \geq 0 \forall i=1,2,...,m
Example
Consider the following problem −
min \:f\left ( x_1,x_2 \right )=\left ( x_1-3\right )^2+\left ( x_2-2\right )^2
such that x_{1}^{2}+x_{2}^{2}\leq 5,
x_1,2x_2 \geq 0 and \hat{x}=\left ( 2,1 \right )
Let g_1\left ( x_1, x_2 \right)=x_{1}^{2}+x_{2}^{2}-5,
g_2\left ( x_1, x_2 \right)=x_{1}+2x_2-4
g_3\left ( x_1, x_2 \right)=-x_{1} and g_4\left ( x_1,x_2 \right )=-x_2
Thus the above constraints can be written as −
g_1 \left ( x_1,x_2 \right)\leq 0, g_2 \left ( x_1,x_2 \right) \leq 0
g_3 \left ( x_1,x_2 \right)\leq 0, and g_4 \left ( x_1,x_2 \right) \leq 0 Thus, I=\left \{ 1,2 \right \} therefore, u_3=0,\:\: u_4=0
\bigtriangledown f \left ( \hat{x} \right)=\left ( 2,-2 \right), \bigtriangledown g_1 \left ( \hat{x} \right)= \left ( 4,2 \right) and
\bigtriangledown g_2\left ( \hat{x} \right ) =\left ( 1,2 \right )
Thus putting these values in the first condition of Karush-Kuhn-Tucker conditions, we get −
u_1=\frac{1}{3} and u_2=\frac{2}{3}
Thus Karush-Kuhn-Tucker conditions are satisfied.
Algorithms for Convex Problem
Method of Steepest Descent
This method is also called Gradient method or Cauchy's method. This method involves the following terminologies −
x_{k+1}=x_k+\alpha_kd_k
d_k= - \bigtriangledown f\left ( x_k \right ) or d_k= -\frac{\bigtriangledown f\left ( x_k \right )}{\left \| \bigtriangledown f\left ( x_k \right ) \right \|}
Let \phi \left (\alpha \right )=f\left ( x_k +\alpha d_k\right )
By differentiating \phi and equating it to zero, we can get \alpha.
So the algorithm goes as follows −
Initialize x_0,\varepsilon_1,\varepsilon_2 and set k=0.
Set d_k=-\bigtriangledown f\left ( x_k \right ) or d_k=-\frac{\bigtriangledown f\left (x_k \right )}{\left \|\bigtriangledown f\left (x_k \right ) \right \|}.
find \alpha_k such that it minimizes \phi\left ( \alpha \right )=f\left ( x_k+\alpha d_k \right ).
Set x_{k+1}=x_k+\alpha_kd_k.
If $\left \| x_{k+1-x_k} \right \|
The optimal solution is \hat{x}=x_{k+1}.
Newton Method
Newton Method works on the following principle −
f\left ( x \right )=y\left ( x \right )=f\left ( x_k \right )+\left ( x-x_k \right )^T \bigtriangledown f\left ( x_k \right )+\frac{1}{2}\left ( x-x_k \right )^T H\left ( x_k \right )\left ( x-x_k \right )
\bigtriangledown y\left ( x \right )=\bigtriangledown f\left ( x_k \right )+H\left ( x_k \right )\left ( x-x_k \right )
At x_{k+1}, \bigtriangledown y\left ( x_{k+1} \right )=\bigtriangledown f\left ( x_k \right )+H\left ( x_k \right )\left ( x_{k+1}-x_k \right )
For x_{k+1} to be optimal solution \bigtriangledown y\left ( x_k+1 \right )=0
Thus, x_{k+1}=x_k-H\left ( x_k \right )^{-1} \bigtriangledown f\left ( x_k \right )
Here H\left ( x_k \right ) should be non-singular.
Hence the algorithm goes as follows −
Step 1 − Initialize x_0,\varepsilon and set k=0.
Step 2 − find H\left ( x_k \right ) \bigtriangledown f\left ( x_k \right ).
Step 3 − Solve for the linear system H\left ( x_k \right )h\left ( x_k \right )=\bigtriangledown f\left ( x_k \right ) for h\left ( x_k \right ).
Step 4 − find x_{k+1}=x_k-h\left ( x_k \right ).
Step 5 − If $\left \| x_{k+1}-x_k \right \|
Step 6 − The optimal solution is \hat{x}=x_{k+1}.
Conjugate Gradient Method
This method is used for solving problems of the following types −
min f\left ( x \right )=\frac{1}{2}x^T Qx-bx
where Q is a positive definite nXn matrix and b is constant.
Given x_0,\varepsilon, compute g_0=Qx_0-b
Set d_0=-g_0 for k=0,1,2,...,
Set \alpha_k=\frac{g_{k}^{T}g_k}{d_{k}^{T}Q d_k}
Compute x_{k+1}=x_k+\alpha_kd_k
Set g_{k+1}=g_k+\alpha_kd_k
Compute \beta_k=\frac{g_{k}^{T}g_k}{d_{k}^{T}Qd_k}
Compute x_{k+1}=x_k+\alpha_kd_k
Set g_{k+1}=x_k+\alpha_kQd_k
Compute \beta_k=\frac{g_{k+1}^{T}g_{k+1}}{g_{k}^{T}gk}
Set d_{k+1}=-g_{k+1}+\beta_kd_k.