Euclidean norm second derivative

Question

I really need Your help.

I need to prove that Euclidean norm is strictly convex. I know that a function is strictly convex if $f ''(x)>0$. Can I use it for Euclidean norm and how? $||x||''=\frac{||x||^2-x^2}{||x||^3}$

Thank You!

Answer 1

The function $$ \lVert \mbox{ }\rVert:\mathbb{R}^n\to \mathbb{R},\, f(x)=\lVert x\rVert $$ is certainly convex. In fact, given $t\in [0,1]$ and $x,y\in \mathbb{R}^n$ we have: $$ \lVert tx+(1-t)y\rVert=\lVert tx+(1-t)y\rVert\le \lVert tx\rVert+\lVert (1-t)y\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert. $$ But is it STRICTLY CONVEX? i.e. if $t\in (0,1)$ and $x,y \in \mathbb{R}^n$, with $x\ne y$, then $$ \lVert tx+(1-t)y\rVert<t\Vert x\rVert+(1-t)\lVert y\rVert. $$ The answer is NO. In fact, if $t\in (0,1)$, and $y=sx$, with $0<s \ne 1$, and $x\in \mathbb{R}^n$, we certainly have $x\ne y$. But \begin{eqnarray} \lVert tx+(1-t)y\rVert&=&\lVert tx+(1-t)sx\rVert=\lVert (t+(1-t)s)x\rVert=(t+(1-t)s)\lVert x\rVert\\ &=&t\lVert x\rVert+(1-t)\lVert sx\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert \end{eqnarray} Hence the function $\lVert\mbox{ } \rVert$ is CONVEX but NOT STRICTLY CONVEX.

Answer 2

By the definition of convexity and the triangle inequality we have

$$ ||ax+(1-a)y|| \leq |a| ||x||+ |1-a|||y||,\quad 0\leq a\leq1. $$

Answer 3

It is easier to work with the definition that a function f is convex if for any $\lambda \in [0,1]$ and for all $x,y \in (a,b)$: $$ f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y)$$

Then:

$$ ||(1-\lambda x) + \lambda y ||^2 = \sum_{i} |(1-\lambda x_i) + \lambda y_i|^2 \leq \sum_{i} (1-\lambda) |x_i|^2 + \lambda |y_i|^2 = (1-\lambda)||x||^2 + \lambda ||y||^2 $$

Answer 4

Here's a second-derivative proof for $\mathbb{R}^n$. The gradient and Hessian do not exist at the origin, but everywhere else they are given by $$\nabla f(x) = \|x\|^{-1} x, \quad \nabla^2 f(x) = \|x\|^{-1} I - \|x\|^{-3} xx^T$$ Strict convexity requires that the Hessian be positive definite; that is, $v^T(\nabla^2 f(x))v>0$ for all $v\neq 0$. But suppose $v=\alpha x$, where $\alpha$ is a scalar. Then $$\begin{aligned} \nabla^2 f(x) \cdot v &= \alpha \left( \|x\|^{-1} I - \|x\|^{-3} xx^T \right) x \\ &= \alpha \|x\|^{-1} x - \alpha \|x\|^{-3} (xx^T) x = \alpha \|x\|^{-1} x - \alpha \|x\|^{-3} x (x^T x) \\ &= \alpha\|x\|^{-1} x - \alpha\|x\|^{-3} \|x\|^2 x = \alpha\|x\|^{-1} x - \alpha\|x\|^{-1}x = 0. \end{aligned}$$ Thus $v^T (\nabla^2 f(x) ) v = 0$, and the Hessian is not positive definite as required. This establishes that the norm is not strictly convex anywhere other than the origin. It's not strictly convex at the origin, either, but one must appeal to the fundamental definition for that instead, as Mercy offered. (That answer should be the one accepted, not this one, because it is complete.)