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

Click For Summary
The norm of a self-adjoint linear integral operator does not equal its spectral radius. The operator norm is defined as the supremum of the ratio of the norm of the output to the norm of the input. In contrast, the spectral radius is determined by the largest eigenvalue of the operator. Therefore, while both concepts are related to the behavior of the operator, they are distinct and can yield different values. Understanding this difference is crucial for analyzing linear integral operators in functional analysis.
sarrah1
Messages
55
Reaction score
0
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
 
Physics news on Phys.org
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.

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
Replies
1
Views
2K
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K