Let $x \in R^m$. It is known that $\|x\|_{\infty}\leq \|x\|_2\leq \sqrt m\|x\|_{\infty}$.
What the difference between above inequality and if we are saying that $\|x\|_2\sim \sqrt m\|x\|_{\infty}$?
(we say that $f\sim g$, if $cf \leq g \leq Cf$ for some absolute constants $c,C>0$. Note, f and g may have dimensional dependence).
Why equivalence $\|x\|_2\sim \sqrt m\|x\|_{\infty}$ is not true if vector $x$ is compressible?
(Vector $x$ is k-sparse if the support of the vector $|supp(x)|\leq k$ . Vector is (k,d)-compressible if it is withing Euclidean distance $d$ from the set of all sparse vectors.)