Do equivalent 2-norms $||x||$ and $||y||$ imply equivalent results in vector-matrix multiplication for $Ax$ and $Ay$?

Question

The solution for an exercise I was doing includes the following statement:

$\max_{||x||_2 = ||y||_2 = 1}|y^HAx|$ $\geq \max_{||x||_2 = 1}|x^HAx|$

I don't understand how this result is obtained. That is, why does $x = y$ in this case? I can see that it has something to do with the restriction on $x$ and $y$ in determining the maximum but cannot understand why equivalent 2-norms would imply equivalent results in vector-matrix multiplication. Is this a general result or specific to the case where $||x|| = 1$?

Answer 1

Let $S_1:=\{|x^HAx| :||x||_2 =1 \}$ and $S_2:=\{|y^HAx| :||x||_2= ||y||_2=1 \}$, then

$$ S_1 \subseteq S_2.$$

Hence $\max S_1 \le \max S_2.$