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

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SUMMARY

The norm of a self-adjoint linear integral operator \( K \) defined by \( (Kx)(t) = \int_{a}^{b} k(t,s)x(s)ds \) does not equal its spectral radius. The operator norm is calculated as \( \sup_{x \neq 0} \frac{||Kx||}{||x||} \), while the spectral radius is determined by the largest eigenvalue of the operator. This distinction is crucial for understanding the properties of integral operators in functional analysis.

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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.
 

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