According to Prof. Vandenberg's lecture notes ECE236C (Spring 2019) - Gradient Method a function is called Lipschitz continuous gradient when
$$ \|\nabla f(y)-\nabla f(x)\|_2 \leq \alpha\|y-x\|_2 $$
Note that the definition does not assume that $f$ is a convex function.
However, if $f$ is a convex function we have $$ f(y) \leq f(x) + \langle \nabla f(x),y-x \rangle + \frac{\alpha}{2}\|y-x\|_2^2 $$
Can we prove the reverse, i.e., if we have
$$ f(y) \leq f(x) + \langle \nabla f(x),y-x \rangle + \frac{\alpha}{2}\|y-x\|_2^2 $$ then $$ \|\nabla f(y)-\nabla f(x)\|_2 \leq \alpha\|y-x\|_2 $$
Hint: using $ f(y) \leq f(x) + \langle \nabla f(x),y-x \rangle + \frac{\alpha}{2}\|y-x\|_2^2 $ we can conclude that $g(x)=\frac{\alpha}{2}\|x\|_2^2-f(x)$ is convex. Since $g(x)$ is convex, monotonicity of the gradient of $g$ results in
$$ \langle \nabla f(x) - f(y),y-x \rangle \leq \alpha\|y-x\|_2^2 $$ but how is possible to get back to the following? $$ \|\nabla f(y)-\nabla f(x)\|_2 \leq \alpha\|y-x\|_2 $$ I am wondering if the above method is right way to prove it. In addition, is there any other condition that can be applied to the convex function $f(x)$ to have Lipschitz continuous gradient?
No. Consider the indicator function for the non-negative real numbers. Then I(x) = 0 if x is >= 0, and $\infty$ otherwise. (Here, I is defined on the augmented real numbers.)
I is convex, but I is not Lipscitz since the derivative at 0 is unbounded.