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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 $L^p(\Omega)$?

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Just to make sure, is the norm you are using on $L^p(\Omega,\mathbb{R}^{n\times n})$ given by $$ \|A\|_{L^p(\Omega,\mathbb{R}^{n\times n})}=\| \|A\|_{\mathcal{L}(\mathbb{R}^n)}\|_{L^p(\Omega,\mathbb{R})} \quad \forall A \in L^p(\Omega,\mathbb{R}^{n\times n})? $$ –  Mercy Jun 19 '12 at 13:43
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At 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. –  Davide Giraudo Jun 19 '12 at 14:01
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For 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. –  Willie Wong Jun 19 '12 at 14:23
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I 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 $L^p$. –  Willie Wong Jun 19 '12 at 14:38
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@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 of the eigenvalue difference. –  Willie Wong Jun 19 '12 at 15:14

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