The $\pi$ be an $n$-dimensional vector on the probability simplex, that is $\pi_{i} \ge 0$ and $\sum_{i} \pi_{i} = 1$.

It is trivial to show that $\|\pi\|_{2} \ge \|\pi\|_{\infty}$. Let us instead compare the squared 2-norm to the infinity norm. Without loss of generality, assume that $\pi_{1}$ is the largest element.

Overall we have $\|\pi\|_{2}^{2} \le \|\pi\|_{\infty} \le \|\pi\|_{2}$.

This also provides a sufficient (and probably necessary) condition for equality: $\pi_{i} = 0$ or $\pi_{i} = \pi_{1}$ for all $i$.