$L$-Lipschitz gradient of $f$ implies inverse strongly monotone of $\nabla f$

Question

Let $f:\; \mathcal{H} \to R$ be a continuously differentiable convex function such that $$\|\nabla f(x) -\nabla f(y)\|\leq L\|x-y\|.$$

Prove that the mapping $\nabla f$ is $1/L$ inverse strongly monotone, that is $$\langle \nabla f(x) -\nabla f(y), x-y\rangle\geq \dfrac{1}{L}\|\nabla f(x)-\nabla f(y)\|^2.$$

Thanks for your help.

Answer 1

This is false. Take $\mathcal{H}=R$, $f(x)=\sin x$. Then $|\cos x-\cos y|\le |x-y|$, and your inequality becomes $(\cos x-\cos y)(x-y)\ge (\cos x-\cos y)^2$, which is obviously false, because the left-hand side can be negative.