How to Estimate the Operator Norm ||A||_2 for a Difference Operator?

[Max Fleiss]

Greetings everyone!

I have a set of tasks I need to solve using using operator norms, inner product... and have some problems with the task in the attachment. I would really appreciate your help and advice.

This is what I have been thinking about so far:
I have to calculate a non trivial upper bound, so maybe it could be done by:
$$b=max( ||A||_1,||A||_2,||A||_\infty )$$

Since $$A$$ is a difference operator I estimated the following:
$$||A||_1= 4$$
$$||A||_\infty= 2$$
But how can I estimate $$||A||_2=?$$
If I know that abs row sum is 2 (besides 0 there appears only one 1 and one -1 in the rows) and abs column sum is 4 (it is two times the size of row length dim(A)=2mn x mn). Can I estimate $$||A||_2$$ by:
$$||A||_2=\sqrt{rows^2+columns^2 }=\sqrt{(2 \cdot 2mn)^2+(4 \cdot mn)^2}$$$$=4 \sqrt{(mn)^2+(mn)^2}=4 \sqrt{2} \sqrt{m^2n^2}$$ since $$mn$$ are positive $$||A||_2=4 \sqrt{2} mn$$
So I would say $$b=max(L_1,L_2,L_\infty)=L_2=4 \sqrt{2} mn$$

Is my conclusion, approximation of a non trivial upper bound b right?

Thank you in advance for your help!

 

[fresh_42]

It is not clear why you use different norms here. In any case, you need to investigate $||Ax||$ for any $x$. Application of $A$ on $x$ should give you an idea for un upper bound, i.e. you need to use the definition of $A$ somewhere, since not all operators are bounded.