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Define $T: L^2(\lbrack 0,1\rbrack) \to L^2(\lbrack 0,1\rbrack)$ by $$ Tf = \int_0^x g(x,y)f(y) dy, \quad \quad 0 \leq x \leq 1. $$ where $g$ satisfies $\int_{\lbrack 0,1 \rbrack^2} g^2(x,y) dx dy < \infty$.

I want to compute the operator norm of $T$. I didn't make much progress trying to do this directly, but I found that the condition on $g$ implies that $T$ is a Hilbert-Schmidt operator with Hilbert-Schmidt norm $||T||_{HS} = ||g||_2$. Does this also mean that the operator norm $||T|| = ||g||_2$? Or not necessarily? If not, how is this computation done?

Alex
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  • I think you can get the $L^2$ operator norm bounded by $|g|^2$. See e.g. https://math.stackexchange.com/questions/565387/norm-of-an-operator-induced-by-l2-kernel-is-bounded-by-l2-norm-of-the-ker (some books also include this computation). There are also results like https://en.wikipedia.org/wiki/Schur_test if those are of interest. – leslie townes Feb 23 '21 at 05:01

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