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 $y$?
Edit: fixed a typo in the definition of $f$. The $\|x\|_a$ needs to be squared.
The answer is No.
Otherwise you have
$$ \frac{C}{2}\|x-y\|_a^2 \leq f(y)- f(x) - \nabla f(x)^\top(y-x) \leq \frac{C'}{2}\|x-y\|_a^2$$
Now dividing both sides by $\|x-y\|_a$ and letting $y \to x$ this shows $f$ is differentiable every where for all $x \in R^n.$
But clearly $f$ is not differentiable at $x=0!$