let $x \in \mathbb{R}^n$ where $f(x) = (1 + ||x||^2)^{1/2}$. Prove that it is convex. As of right now, we define a convex function to be a function with a positive semi definite second derivative. So how do I prove that the second derivative of this function is positive semi definite?
I tried to use Taylors theorem where
$f(a+h) = f(a) + f'(a)*h + R(t), t \in (0, 1)$ I'm trying to figure out a way to prove that the remainder is always non negative.
Just compute the second derivative, we have, by the chain rule $$\def\norm#1{\left|#1\right|} Df(x)h = \frac 1{(1 + \norm{x}^2)^{1/2}} \cdot \def\<#1>{\left<#1\right>}\<x, h> $$ Hence, $$ D^2f(x)[h,k] = -\frac 1{(1 + \norm x^2)^{3/2}}\<x,h>\<x,k> + \frac{1}{(1 + \norm x^2)^{1/2}}\<h,k> $$ So, we have \begin{align*} D^2 f(x)[h,h] &= - \frac 1{\def\op{(1 + \norm x^2)}\op^{3/2}}\<x,h>^2 + \frac{\op}{\op^{3/2}}\norm h^2\\ &= \frac{\norm h^2 + \norm h^2\norm x^2 - \<x,h>^2}{\op^{3/2}}\\ &\ge \frac{\norm h^2}{\op^{3/2}} \qquad\text{by Cauchy-Schwarz} \end{align*} So $D^2 f(x)$ is positive definite.