Let $A_n(x)$ be a sequence of symmetric matrix functions that converges in $L^p(\Omega)$ to $A(x)$. Is it true that the eigenvalues of $A_n(x)$, or a subsequence of these, converge to the eigenvalues of $A(x)$ in $L^p(\Omega)$?
Does matrix convergence in $L^p$ imply convergence of the eigenvalues in $L^p$?
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linear-algebra
real-analysis
matrices
functional-analysis
eigenvalues-eigenvectors
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0$\Omega$ is an open subset of $\mathbb{R}^m$. Sorry, I should have been a little more precise with my notation. – 2012-06-19
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1At least, it's true if you work pointwise, for a subsequence. You will use the fact that each matrix is diagonalizable, and that the orthogonal group is compact. – 2012-06-19
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0By this you mean: for every $x$ there is a subsequence such that the eigenvalues of $A_{n_{k}}(x)$ converge to the eigenvalues of $A(x)$? Or do you mean a.e. convergence? – 2012-06-19
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0I mean the first, maybe the second is true if we assume continuity of $A_n$ and $A$. – 2012-06-19
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0Just to make sure, is the norm you are using on $L^p(\Omega,\mathbb{R}^{n\times n})$ given by $$ \|A\|_{L^p(\Omega,\mathbb{R}^{n\times n})}=\| \|A\|_{\mathcal{L}(\mathbb{R}^n)}\|_{L^p(\Omega,\mathbb{R})} \quad \forall A \in L^p(\Omega,\mathbb{R}^{n\times n})? $$ – 2012-06-19
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0I was thinking $$||A||_{L^p(\Omega,\mathbb{R}^{n\times n})}=(\sum_{i,j=1}^n||a_{i,j}||_{L^p(\Omega,\mathbb{R})}^p)^{1/p},$$ if $a_{i,j}$ are the entries of the matrix. I'm not sure if this is the same. – 2012-06-19
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1For finite dimensional vector spaces, all norms are comparable. So I think convergence in your norm is equivalent to convergence in that of @Deltapsi. Note that you can commute the $\ell_p$ sum over entries with the $L^p$ integral. – 2012-06-19
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0If the answer to the question is "Yes" it should be true, even if the matrices are not continuous. I think one of the problems is that we do not know a lot about the eigenvectors, except that we might for example choose them to lie in the sphere and can hence find (for fixed $x$) a converging subsequence (which I guess is just another way of rephrasing your earlier argument). Also, there is a subsequence of eigenvectors converging weakly in $L^p$, but I cannot really see how to use this. Thanks, by the way, for you guys asking and commenting! – 2012-06-19
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3I don't think you should use eigenvectors: the map from matrix to eigenvectors is non-unique and unstable when you have repeated eigenvalues. Note that @Davide's argument plus Chebyshev should already give you convergence of the eigenvalue functions in weak $L^p$. – 2012-06-19
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0@Willie Wong: Sorry, I didn't quite get this, isn't Chebyshev usually used to show a.e. convergence or convergence in measure? – 2012-06-19
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1@sepp Convergence in $L^p$ gives _quantitative_ upper bound on the measure of the set over which pointwise $A_n$ deviates from $A$ more than $\lambda$ (this is Chebyshev). The pointwise difference $A_n - A$ can be converted into a pointwise upper bound on the difference of eigenvalues. This gives a bound on the [weak $L^p$ norm](http://en.wikipedia.org/wiki/Lp_space#Weak_Lp) of the eigenvalue difference. – 2012-06-19
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0@Willie Wong: Now I got it. Thanks a lot for explaining this to me! Although I now remember hearing about weak $L^p$ spaces at some time, I completely forgot about it. So convergence in weak $L^p$ is not the same as weak convergence in $L^p$, I think that is what got me confused (-; . – 2012-06-19