# Proof to a criterion of strongly convexity

Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is a differentiable convex function, I hope to prove that $\forall x, y, \lVert \nabla f(y) - \nabla f(x) \rVert_2 \ge m \lVert y - x \rVert_2 \Longrightarrow f$ is strongly convex with parameter $m$.

I have made the case by assuming $f\in C^2$, and proved that $\lVert \nabla f(y) - \nabla f(x) \rVert_2 \ge m \lVert y - x \rVert_2 \Longrightarrow \nabla^2 f \succeq m$, where the right side is a second-order statement of a strongly convex function. This proof is not hard.

However, I hope to find a proof without using the Hessian matrix, for example, to show that $\forall x, y, \lVert \nabla f(y) - \nabla f(x) \rVert_2 \ge m \lVert y - x \rVert_2 \Longrightarrow (\nabla f(y) - \nabla f(x))^T (y-x) \ge m \lVert y-x \rVert_2^2$, where the right side is a first-order statement of a strongly convex function.

Proof of the general case: it's known from the convex analysis that $\nabla^2 f$ exists almost everywhere. At any point $x$ where $\nabla^2 f(x)$ exists, the assumption implies that $\nabla^2 f(x)\succeq mI$. Now for any fixed $x,y$, we define $g_n(t)=n(y-x)^T (\nabla f(x+t(y-x)+(y-x)/n) - \nabla f(x+t(y-x))$ for $t\in [0,1-1/n]$, and $g_n(t)=0$ for $t\in(1-1/n,1]$. Note that $g_n\ge 0$, Fatou's lemma yields \begin{align*} m\|y-x\|_2^2 &\le \int_0^1 \liminf g_n(t) dt \le \liminf \int_0^1 g_n(t)dt \\ &= (y-x)^T(\nabla f(y)-\nabla f(x)) - \limsup \int_0^{\frac{1}{n}} n(y-x)^T(\nabla f(x+t(y-x)) - \nabla f(x))\\ &\le (y-x)^T(\nabla f(y) - \nabla f(x)) \end{align*} and we are done.
Actually, $C^2$ and convex functions behave quite similarly. For example, with the help of mollifiers and convolution, any convex function $f$ can be approximated by a sequence of convex $f_k\in C_c^\infty$, where the convergence $f_k\to f$ is uniform on any compact set. We can also treat the second-order derivative of a convex function as a positive Radon measure, e.g., in the $\mathbb{R}^1$ case, we can write \begin{align*} f(x) = a_Ix + b_I + \int_I \frac{1}{2}|x-a|\mu(da) \end{align*} where $I$ is an interval. In the multivariate case the Radon measure will be matrix-valued. This point of view will be sometimes useful: for example, in stochastic analysis, Ito's original formula holds only for $C^2$ functions, but following the preceding rationale it can be generalized to the difference of two convex functions (the so-called Ito-Tanaka formula). Moreover, the important notion of local time can then be defined.