TITLE: Norm preservation properties of a unitary matrix
QUESTION [6 upvotes]: Tried to prove the following facts:

If $B \in M_{n\times n}(\mathbb{C})$ is unitary  (i.e. $ B^{-1} = B^{*}$), then:

$\forall x\in \mathbb{C}^n :\| Bx \|_2 = \| x \|_2  $
$\forall A\in M_{n\times n}(\mathbb{C})$ :  $ \| BA \|_F = \| A \|_F  $ (where $ \| . \|_F $ denotes Frobenius norm).

$\forall A,B\in M_{n\times n}(\mathbb{C}): \| AB \|_{ \infty} \leq \| A \|_{ \infty}\| B \|_{ \infty}   $

In (1),tried to conduct direct algebraic manipulations from the definitions of the norms, but I obtained no results. 
  In (2) the use of definition $ \| B \|_{ \infty} := max\{\frac{\|Bx\|}{\|x\|}: x\in \mathbb{C}^n \setminus \{0\} \}$ and the inequality $\|AB\| \leq \|A\|\|B\|$ gives straightforward result. However, the use of definition $\|B\|\ _{\infty} :=max\{ \sum_{j=1}^{n}|b_{ij}|: i\in \{1,...n\}\}$ does not give direct results.
I would be thankful for hints/advices!

REPLY [2 votes]: $$
||Bx||_2 = \sum_i\left(\sum_j B_{ij}x_j) \right)^* \left(\sum_k B_{ik}x_k) \right) $$
$$
\\
\left(\sum_j B_{ij}x_j) \right)^* = \sum_j B_{ij}^*x_j^*  =  \sum_j (B^T_{ji})^*x_j^* =  \sum_j B^{-1}_{ji}x_j^* 
$$
$$
||Bx||_2 = \sum_i\left(\sum_j B^{-1}_{ji}x_j^*  \right)^* \left(\sum_k B_{ik}x_k) \right) \\
= \sum_j\sum_k \left(  \sum_iB^{-1}_{ji}  B_{ik}  \right) x_j^*x_k
=\sum_j\sum_k \delta_{jk} x_j^*x_k = \sum_j  x_j^*x_k = ||x||_2
$$
The equality for the Frobenius norm follows directly since for any arbitrary $x$ the sum of the quares for each column of $BAx$ matches the sum of the squares for that same column of $Ax$.