Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n\to \infty$.

share|cite|improve this question
In what sense do the operators $T_n$ converge? – Jose27 Jan 25 '12 at 18:45

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.