Suppose we have a function $f\in {C^1_L}$ such that ${\lVert \nabla f(x) - \nabla f(y)\rVert}_2$ $\geqslant$ $L{\lVert x - y\rVert}_2$ for some $L>0$.
We have to prove that: $$\frac{-L}{2}{\lVert(y-x)\rVert}^2 \leq f(y) - f(x) + \langle\nabla {f(x)},(y- x) \rangle \le \frac{L}{2}{\lVert(y-x)\rVert}^2 $$
I have tried in the following way:
It is known that for $f \in C^1$, $f(y)\geqslant f(x) + \nabla {f(x)^T}(y- x)$.
as $f(y) - f(x) - \nabla {f(x)^T}(y- x) \ge 0$ and the $ \frac{-L}{2}{\lVert(y-x)\rVert}^2 $ is always positive so we know that the LHS inequlity holds true?
Can anyone give hints for proving the right hand inequality?
Edit:
I have realsied that there is mistake in my solution, no where in the problem it has been mentioned that $f$ is convex. Thus the property $f(y) - f(x) - \nabla {f(x)^T}(y- x) \ge 0$ no longer holds true.
Any hints for this will be welcome. Thanks in advance!