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 take a function $g(x) = \frac{L}{2} x^Tx -f(x)$ we need to prove that g(x) is a convex function.
I tried to prove it in the following manner:
We know that to prove that a function is convex we need to prove that for $0<\alpha<1$ the given inequality holds true $g(\alpha x +(1-\alpha)y) \le \alpha g(x) + (1 - \alpha)g(y)$.
We take the LHS of the inequality and get $$ \frac{L}{2}[\alpha x + (1 - \alpha)y]^T[\alpha x + (1 - \alpha)y] - f(\alpha x + (1 - \alpha)y)$$
on solving it for 2-3 steps we get $$ \frac{L}{2}[\alpha \lVert x \rVert^2 + (1-\alpha)\lVert y \rVert^2)] - f(\alpha x + (1 - \alpha)y)$$
similarly, for the RHS term we get $$ \frac{L}{2}[\alpha \lVert x \rVert^2 + (1-\alpha)\lVert y \rVert^2)] - [f(x)+(1-\alpha)f(y)]$$
the norm inequality gives us that f is monotonically increasing and thus convex, so RHS $\ge$ LHS and hence $g(x)$ is convex.
IS my proof correct? If not what else I need to do? any hints or suggestions will be helpful. Thanks in advance!