Prove the equivalence of the matrix $2$-norm and $\infty$-norm. That is, for $A \in \mathbb{C}^{m \times n}$ prove that there exist constants $c_1,c_2$ such that $c_1\|A\|_\infty \leq \|A\|_2 \leq c_2\|A\|_\infty$.
proof [right side inequality]
First note that for $y \in \mathbb{C}^n$ that
$$ \|y\|_2 = \sqrt{\sum_{j =1}^n |y_j|^2 } \leq \sqrt{\sum_j^n \left( \max_{1\leq j \leq n}|y_j|^2 \right)} = \sqrt{\sum_j^n \left( \max_{1\leq j \leq n} |y_j| \right)^2 } = \sqrt{\sum_j^n \|y\|_\infty ^2 } = \sqrt{n \|y\|_\infty ^2 } = \sqrt{n} \|y\|_\infty$$
This shows that for arbitrary $x\in \mathbb{C}^n$ that $||Ax||_2 \leq \sqrt{n}\|Ax\|_\infty$. Now, here's where I'm running into issues. I need to show that
$$\sup_{\|x\|_2 =1} \|Ax\|_2 \leq \sqrt{n} \sup_{\|x\|_\infty=1} \|Ax\|_\infty$$
But the sets over which we are taking the supremum is not the same set. I am wondering, is it always true that if
$$f(x) \leq c_1g(x)$$
for all $x$ is it also true that
$$ \sup_{x \in A} f(x) \leq c_1 \sup_{x \in B}$$
But anyway, I know this for a fact
$$\sup_{\|x\|_2 =1} \|Ax\|_2 \leq \sqrt{n} \sup_{\|x\|_2=1} \|Ax\|_\infty$$
But I don't know how to justify the supremum being montone over a unit ball.
proof [left inequality]
Here I'm having similar issues. I can show that for a vector that $\|Ax\|_\infty \leq \|Ax\|_2$ so I have pretty much the same situation (I believe), and that I'm stuck on relating the supremums. I obviously have these two
$$\sup_{\|x\|_2 =1} \|Ax\|_\infty \leq \sup_{\|x\|_2=1} \|Ax\|_2$$
or
$$\sup_{\|x\|_\infty =1} \|Ax\|_\infty \leq \sup_{\|x\|_\infty=1} \|Ax\|_2$$
but not
$$\sup_{\|x\|_\infty =1} \|Ax\|_\infty \leq \sup_{\|x\|_2=1} \|Ax\|_2$$