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Suppose that $\left ( a_{ij} \right )_{i,j=1}^{\infty}$ is a matrix satisfying the following condition $$\sum_{i,j=1}^{\infty} \left | a_{ij} \right |^q < \infty$$ where $q>1$. For $x=\left \{ \xi_j \right \}_{j=1}^{\infty} \in l_p$ where $\frac{1}{p}+\frac{1}{q}=1$, let we define $$Ax=y := \left \{ \eta_i \right \}_{i=1}^{\infty},$$ where $$\eta_i=\sum_{j=1}^{\infty}a_{ij}\xi_j$$ for $i=1, 2, ...$ I proved that $A$ is a continuously linear operation from $l_p$ to $l_q$. I don't know how to compute the norm of $A$ that defined by $$\lVert A \rVert = \sup_{x \neq 0} \frac{\lVert Ax \rVert}{\lVert x \rVert} = \sup_{\lVert x \rVert \leq 1} \lVert Ax \rVert = \sup_{\lVert x \rVert = 1} \lVert Ax \rVert$$ How could I compute the norm of $A$ above?

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  • $\begingroup$ You only need to show that the operator norm is finite for continuity. No need to compute it explicitly. Hint: Hölder inequality $\endgroup$
    – user251257
    Commented Sep 7, 2015 at 0:10

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I don't know if there's any way to do it explicitly in general. It's certainly not going to be easy.

In the finite-dimensional case, it's NP-hard to compute the norm to within a fixed relative precision: see e.g. "Matrix P-norms are NP-hard to approximate if $p \neq 1,2,\infty$" by Julien M. Hendrickx and Alex Olshevsky.

You can think of it as an optimization problem: maximize $\sum_{i} |\eta_i|^q$ subject to $\sum_j |\xi_j|^p \le 1$. Unfortunately it's not a convex problem, so there may be lots of local maxima.

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