Because the squared euclidean $\ell_2$ norm is self-dual (see why here), we have $S_2(y) \equiv \frac{1}{2}\|y\|_2^2$ and so $\
abla S_2(y) \equiv y$.
In general, $S_p(y)$ will not be differentiable at the origin. Indeed, extending the above idea for the $p=2$ case, it's not difficult to show (e.g see this wikipedia page, in french) that
> $$(\frac{1}{2}\|.\|_p^2)^* = \frac{1}{2}\|.\|_q^2, $$ where $q > 1$ is the _harmonic conjugate_ of $p$, i.e $1/p + 1 / q = 1$.
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Thus, in case $p \in (1, 2)$, your problem is reduced to studying the differentiability of a squared norm $\|.\|_q^2$ , with $q := p/(p-1) > 2$. Note that this is differentiable everywhere except perhaps at the origin, where you'll need to administer special treatment...