MHB Does the Norm of a Linear Integral Operator Equal Its Spectral Radius?

sarrah1
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Hello

A simple question.
I have a linear integral operator (self-adjoint)

$$(Kx)(t)=\int_{a}^{b} \, k(t,s)\,x(s)\,ds$$

where $k$ is the kernel. Can I say that its norm (I believe in $L^2$) equals the spectral radius of $K?$

Thanks!
Sarah
 
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No, the norm of an integral operator is not necessarily equal to its spectral radius. The norm of an integral operator is given by the supremum of its operator norm, which is defined as $\sup_{x\neq 0}\frac{||Kx||}{||x||}$. The spectral radius on the other hand is the largest eigenvalue of the operator.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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