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Let $M$ be a compact manifold and $g_k$ be a sequence of Riemannain metrics smoothly converging to another Riemannian metric $g$.

Let $\{\lambda^k_j\}$ be the spectrum of the Laplacian of the Riemannian metric $g_k$, and $\{\lambda_j\}$ be the spectrum of the Laplacian of $g$ (with values in increasing order, i.e., $\lambda^k_{j+1} \geq \lambda^k_{j}$).

I expect that $\lambda^k_j \to \lambda_j$ as $k \to \infty$. Is this true? If yes, what is a reference? If I replace smooth convergence with $C^k$-convergence, what is the lowest $k$ such that convergence of spectrum holds?

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    $\begingroup$ The lowest $k$ for which this is true is $0$. Roughly: write down the min-max formula for $\lambda_j^k$ and $\lambda_j$, take bases of eigenfunctions $\{u_j\}$, $\{u_j^k\}$ for the two problems, and use $E = \text{span}\{u_1, \ldots, u_j\}$ as a competitor for the $\lambda_j^k$ problem and vice versa; this gives $\lambda_j^k \leq R_{g_k}(u)$ for some $u \in E$, where $R$ is the Rayleigh quotient. As this only contains the metric coefficients and volume forms if you write it out, you can estimate like $R_{g_k}(u) \leq R_g(u) + C\|g - g_k\|_{C^0}R_g(u)$. But $R_g[u] \leq \lambda_j$ as $u \in E$. $\endgroup$
    – user378654
    Sep 2 at 18:26
  • $\begingroup$ I see, so basically the finite dimensional argument works! $\endgroup$
    – Hammerhead
    Sep 2 at 23:03

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$C^0$-convergence is sufficinent.

Note that $\lambda_i$ can be defined as the least lower bound on numbers $\lambda$ such that the following property holds:

  • There is an $i$-dimensional subspace $W$ of smooth functions on $M$ such that $$ \int\limits_{(M,g)} |d f|_g^2<\lambda\cdot \int\limits_{(M,g)} f^2$$ for any $f\in W$.

Fix $\lambda>\lambda_i$ and such a subspace $W$. Note that $W$ meets the same property for $g_n$ with large $n$. It follows that $\lambda_i\geqslant \limsup_{n\to\infty} \lambda_i^n$.

Nearly the same argument shows that $\lambda_i\leqslant \liminf_{n\to\infty} \lambda_i^n$.

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  • $\begingroup$ Isn't $C^0$ sufficient? The equation you write doesn't depend on derivatives of the metric $\endgroup$
    – Otis Chodosh
    Sep 15 at 13:15
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    $\begingroup$ @OtisChodosh Yes --- my mistake; I will fix it. $\endgroup$
    – Anton Petrunin
    Sep 15 at 19:57

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