Convex envelope of norm 0 In class, we showed that the convex envelope of the norm 0 on vectors, defined as the number of nonzero coordinates, is the norm 1 on the unitary ball for norm 1 and $\infty$.
We also tried to prove that it is the same if we take the unitary ball in norm 2, but we failed, and we now think it is false. 
Do you know the solution for norm 2 or in general any p norm?

Edit: A convex envelope of a function $f(x)$ on a domain $D$ is a convex function $g(x)$ defined on $D$ such that  $g(x)\le f(x)$ for all $x\in D$ and for every other convex function such that $h(x)\le f(x)$, then $h(x)\le g(x)\le f(x)$.
 A: As long as $p>1$ and $n>1,$ and neither are infinite, the answer is definitely no.  
Consider this notation:
$$
f_q^p(x) = \begin{cases}
||x||_q & x \in B_p \\
+\infty &\text{else}
\end{cases},
$$
where $B_p = \{x \in \mathbb{R}^n \mid ||x||_p \le 1\}.$
Also define the epigraph
$$
\text{epi} (f) = \{(x, y) \in \mathbb{R}^{n+1} \mid y \ge f(x)\}.
$$
Then, what I will prove, which is equivalent to your statement about convex envelopes, is that, for all $p >1$ and $n \in \mathbb{N}\setminus\{1\},$ $$\text{epi} (f_1^p) \not = \text{conv}( \text{epi} (f_0^p) ).$$
proof:
Note that $(n^{-1/p} \mathbb{1}, n^{(p-1)/p}) \in \text{epi}(f_1^p).$
Now, note that $f_0^p(n^{-1/p} \mathbb{1}) = n > n^{(p-1)/p}$ for all $p, n > 1.$
Thus, $(n^{-1/p} \mathbb{1}, n^{(p-1)/p}) \not\in \text{epi}(f_0^p).$
So the question is, is $(n^{-1/p} \mathbb{1}, n^{(p-1)/p})$ a member of $\text{conv}( \text{epi} (f_0^p) )?$
The answer again is no.
To see why, note that $n^{-1/p} \mathbb{1}$ is an extreme point of $B_p$, thus is cannot be written as a convex combination of other points in $B_p$.
Therefore, it is hopeless to look for a collection of points in $\text{conv}( \text{epi} (f_0^p) )$ who's convex combination is equal to $(n^{-1/p} \mathbb{1}, n^{(p-1)/p})$.
