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Let $(T_n)$ be a sequence of elliptic operators defined in $H^2(\Omega)\cap H_0^1(\Omega)$ to $L^2(\Omega)$, with $\Omega$ being a bounded domain with smooth boundary. All of them have a smallest eigenvalue $\lambda_n$ to which we associate a positive eigenfunction $\varphi_n$, with norm equal to 1.

Now, let $T_n \to T$ ($T$ is an elliptic operator as well). Let $\lambda$ be the smallest eigenvalue of $T$, and $\varphi$ the positive eigenfunction with norm equal to 1 associated to $\lambda$. Show that $\varphi_n\to\varphi$ as $n\to \infty$.

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In what sense do the operators $T_n$ converge? – Jose27 Jan 25 '12 at 18:45

1 Answer 1

I suppose $T_n \to T$ in the sense, that the resolvents $\mathcal R_n(\mathrm i)\to \mathcal R(\mathrm i)$ converge in norm. $$R_n \phi_n =\frac{1}{\lambda_n-i} \phi_n.$$ Note that the embedding $H_0^1(\Omega)\hookrightarrow L^2(\Omega)$ is compact and hence $\phi_n$ admits a subsequence such that $\phi_n \rightharpoonup \phi$ weakly in $L^2(\Omega)$ and $\mathcal R(i)\phi_n \to \mathcal R(i)\phi$ in $L^2(\Omega)$. Similarly $\lambda_n \to \lambda$ passing to a subsequence. Then a simple $2\varepsilon$ argument shows $$\lim_{n\to \infty} R_n \phi_n = R\phi.$$ Obviously, also $\frac{1}{\lambda_n -i} \phi_n \to \frac{1}{\lambda-i} \phi$ and thus $\mathcal R(\mathrm i) \phi= \frac{1}{\lambda -i} \phi $. This leads to $T\phi = \lambda \phi$. That $\lambda$ is the first eigenvalue, then follows by the nonnegativity of each of the eigenfunction that the eigenfunction obtained this way is in fact the first.

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