Let $f: \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable such that $f$ is $m$-strongly convex and $\nabla f$ is $M$-Lipschitz continuous. In other words$$ | \nabla f(x) - \nabla f(y) | \geq m | x - y | $$$$ | \nabla f(x) - \nabla f(y) | \leq M | x - y |$$for all $x,y \in \mathbb{R}^d$. I need to define an operator $H$ such that$$ \nabla f(x) - \nabla f(y) = H \cdot (x-y)$$without assuming that $f$ is twice differentiable. Clearly if it is, then one can simple define $H$ as$$ H = \int^1_0 \nabla^2 f(x + r (y-x) ) dt $$by the fundamental theorem of calculus. Is there a way of bounding $H$ implicitly? Something of the form:$$ a| H z| \leq \begin{cases} m |z| && \text{if} && a = -1 \\ M |z| && \text{if} && a = 1 \end{cases}$$for any $z \in \mathbb{R}^d$. I do not need $H$ to be written in closed form, so if it is defined as the above inequality, is this correct?
Edit: Clarification that $H$ need not be a function of $x$ and $y$. The desired result is an upper bound on the norm/absolute value of $H$.
