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I asked this question in math.stackexchange but got no answers (link: https://math.stackexchange.com/questions/2323520/strong-smoothness-of-lp-norm). So I decided to ask this question here. Hope I didn't break any rules :)

Define $f(x)=\|x\|_a^2$ for some $1<a<2$. It is well-known that (for example, Theorem 16 of http://ttic.uchicago.edu/~shai/papers/KakadeShalevTewari09.pdf) $f$ is strongly convex w.r.t. $\|\cdot\|_a$, meaning that there exists $C>0$ depending only on $a$ such that $f(y)\geq f(x)+\nabla f(x)^\top(y-x) + \frac{C}{2}\|x-y\|_a^2$ holds for all $x$ and $y$.

I am wondering whether $f$ is also strongly smooth: does there exist $C'>0$ depending only on $a$, such that $f(y)\leq f(x)+\nabla f(x)^\top(y-x)+\frac{C'}{2}\|x-y\|_a^2$ holds for all $x$ and $y$?

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    $\begingroup$ No. Try something like $x=(1,0)$, $y=(1,\epsilon)$, $\epsilon$ small. $\endgroup$
    – Christian Remling
    Commented Jul 8, 2017 at 3:53
  • $\begingroup$ No for $f$ but Yes for it's convex conjugate $f^*$. Indeed, take $C' = 1/C$. BTW, this is an iff. $\endgroup$
    – dohmatob
    Commented Jul 10, 2017 at 6:19

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