Convex Optimization - Norm

A norm is a function that gives a strictly positive value to a vector or a variable.

Norm is a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$

The basic characteristics of a norm are −

Let $X$ be a vector such that $X\in \mathbb{R}^n$

• $\left \| x \right \|\geq 0$

• $\left \| x \right \|= 0 \Leftrightarrow x= 0\forall x \in X$

• $\left \|\alpha x \right \|=\left | \alpha \right |\left \| x \right \|\forall \:x \in X and \:\alpha \:is \:a \:scalar$

• $\left \| x+y \right \|\leq \left \| x \right \|+\left \| y \right \| \forall x,y \in X$

• $\left \| x-y \right \|\geq \left \| \left \| x \right \|-\left \| y \right \| \right \|$

By definition, norm is calculated as follows −

• $\left \| x \right \|_1=\displaystyle\sum\limits_{i=1}^n\left | x_i \right |$

• $\left \| x \right \|_2=\left ( \displaystyle\sum\limits_{i=1}^n\left | x_i \right |^2 \right )^{\frac{1}{2}}$

• $\left \| x \right \|_p=\left ( \displaystyle\sum\limits_{i=1}^n\left | x_i \right |^p \right )^{\frac{1}{p}},1 \leq p \leq \infty$

Norm is a continuous function.

Proof

By definition, if $x_n\rightarrow x$ in $X\Rightarrow f\left ( x_n \right )\rightarrow f\left ( x \right )$ then $f\left ( x \right )$ is a constant function.

Let $f\left ( x \right )=\left \| x \right \|$

Therefore, $\left | f\left ( x_n \right )-f\left ( x \right ) \right |=\left | \left \| x_n \right \| -\left \| x \right \|\right |\leq \left | \left | x_n-x \right | \:\right |$

Since $x_n \rightarrow x$ thus, $\left \| x_n-x \right \|\rightarrow 0$

Therefore $\left | f\left ( x_n \right )-f\left ( x \right ) \right |\leq 0\Rightarrow \left | f\left ( x_n \right )-f\left ( x \right ) \right |=0\Rightarrow f\left ( x_n \right )\rightarrow f\left ( x \right )$

Hence, norm is a continuous function.