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According to Prof. Vandenberg's lecture notes [http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf], page 11, 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?

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No. Intuitively, a function is Lipschitz means that the derivative is continuous. (It's a bit stronger, even.) Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=|x|$. This function is convex, but its derivative is not continuous, therefore it is not Lipschitz.

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