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A smooth convex functions with $C^1$ has not always a Lipschitz continuous gradient. Please see the answer.

If $F$ is convex and has a Lipschitz continuous gradient with modulus L, then $F^*$ is $1/L$-strongly convex.

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No, the answer and solution you edited into the question are not correct.

  1. The function $F(x)=x^4$ is convex and very smooth, but its gradient is not Lipschitz on $\mathbb R$. The gradient is locally Lipschitz, but not globally Lipschitz. The conjugate function is $F^*(x)=c|x|^{4/3}$ (for some constant $c>0$) which is not strongly convex on $\mathbb R$.

  2. Even locally, a $C^1$-smooth convex function does not have a Lipschitz gradient in general. Consider $G(x)=|x|^{4/3}$: it is convex, $G'(x)=\frac43 x^{1/3}$ is continuous, but is not locally Lipschitz. The conjugate function is of the form $G^*(x)=c\,x^4$ for some $c>0$. This is not strongly convex, even locally.

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  • $\begingroup$ Hi, thank you for your answer. I changed my post. $\endgroup$ – Vivian Aug 12 '13 at 7:52

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