TITLE: Prove that function $w \mapsto m \|w\|^2$ is strongly convex
QUESTION [7 upvotes]: The definition of strongly convex from Wikipedia:

It is not necessary for a function to be differentiable in order to be strongly convex. A third definition for a strongly convex function, with parameter $m$, is that, for all $x$, $y$ in the domain and $t\in [0,1]$,
$$f(tx+(1-t)y) \le t f(x)+(1-t)f(y) - \frac{1}{2} m t(1-t) \|x-y\|_2^2.$$

Prove that the $2$-norm squared $f(w) = m\|w\|^2 $ is $m$ strongly convex
I have so far tried to use the triangle inequality but I cannot derive that last negative term.

REPLY [2 votes]: If the function $f$ is twice continuously differentiable, then it is strongly convex with parameter $\mu>0$ if and only if $\nabla^2 f(x) \succeq \mu I$ for all $x$ in the domain, where $I$ is the identity and $\nabla^2f$ is the Hessian matrix, and the inequality $\succeq$ means that $\nabla^2 f(x) - \mu I$ is positive semi-definite.
The definition above is equivalent to the definition given by the OP if $f$ is twice continuously differentiable (see here for more details).
The function $f(w)=m\|w\|^2$ is twice continuously differentiable. We have that
$$
\nabla f(w)=2mw
\quad\text{and}\quad\nabla^2f(w)=2mI.
$$
Hence, the function $f$ is strongly convex with parameter $2m$.