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Suppose we are on a closed Riemannian manifold $M$. Any function $f\in C^\infty(M)$ may be decomposed as $$f = \sum_{j = 0}^\infty f_j\phi_j,$$ where $\phi_j\in C^\infty(M)$ are the Laplace eigenfunctions with associated eigenvalues $\lambda_j$ ($||{\phi_j}||_{L^2} = 1$) and $f_j:=\langle f,\phi_j\rangle_{L^2}$.

I was wondering, is there a way to prove that for any $\epsilon>0$ there exists an $C_\epsilon > 0$ such that $$||\phi_j||_{L^\infty}\leq C_\epsilon \lambda_j^{n/4+\epsilon}$$ for all $j\geq0$. Moreover, I'd like to conclude something like $$\forall f\in C^\infty(M),\forall K>0,\exists C_K>0:|f_j|\leq \frac{C_K}{\lambda_j^K}.$$

How is this possible/straightforward?

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  • $\begingroup$ I think you can get this from bounds on Sobolev norms and the Sobolev embedding theorem. $\endgroup$
    – Michael Renardy
    Commented Nov 9 at 16:55
  • $\begingroup$ Could you maybe elaborate a bit? $\endgroup$
    – Mathematics enthusiast
    Commented Nov 9 at 18:00
  • $\begingroup$ The numerology here looks a little wrong, doesn't it? I think Sogge's result has (in your notation) $\|\phi_j\|_\infty\lesssim\lambda_j^{\frac{n-1}{4}}$ $\endgroup$
    – Ben Johnsrude
    Commented Nov 9 at 19:02
  • $\begingroup$ Maybe this is where the $\epsilon$ comes into play? $\endgroup$
    – Mathematics enthusiast
    Commented Nov 9 at 19:08
  • $\begingroup$ The estimate $\|\phi_j\|_\infty\leq C_\varepsilon\lambda_j^{\frac{n}{4}+\varepsilon}$ is weaker than Sogge's result $\|\phi_j\|_\infty\leq C\lambda_j^{\frac{n}{4}-\frac{1}{4}}$ $\endgroup$
    – Ben Johnsrude
    Commented Nov 9 at 19:53

1 Answer 1

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You can do it via Sobolev embedding, as in the comment by @Michael Renardy (estimate the $W^{2,2}$-norm by $\lambda_j$ and then, if $2>n/2$, apply Sobolev embedding with the right scaling; if $2 \leq n/2$, apply Sobolev iteratively until you reach $\infty$). This is a bit boring.

A faster way is to apply $L^2-L^\infty$ estimates for the heat semigroup $e^{t \Delta}$ (and this is where all Sobolev embeddings are hiddenn). One has $\|e^{t \Delta}\|_{2 \to \infty} \leq Ct^{-n/4}$. If $-\Delta \phi_j=\lambda_j \phi_j$, then $e^{t \Delta} \phi_j=e^{-\lambda_j t} \phi_j$ and then $\|e^{-\lambda_j t} \phi_j\|_\infty \leq Ct^{-n/4}$. Choosing $t=\lambda_j^{-1}$ you get the estimate you want, even without $\epsilon$.

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  • $\begingroup$ The first approach is known as Moser iteration. $\endgroup$
    – Deane Yang
    Commented Nov 10 at 4:28
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    $\begingroup$ @DeaneYang Actually I had in mind $L^p$-regualrity since the coefficients are smooth and I was moving at steps of 2 derivatives, rather 1 as in Moser. As in the answer one reachs a bound in $L^{p_1}$ with $1/p_1=1/2-2/n$ and then in $W^{2,p_1}$ by the equation and elliptic regularity and so on until such a $p$ is bigger of $n/2$. But I agree that with Moser's iteration one gets a proof in a wider class of operators. $\endgroup$
    – Giorgio Metafune
    Commented Nov 10 at 14:51
  • $\begingroup$ Thanks for the clarification. I like Moser iteration because it's low tech (Sobolev inequality, integration by parts, Holder iinequality) and you get an estimate of the constant in terms of the constant in the Sobolev inequality. $\endgroup$
    – Deane Yang
    Commented Nov 10 at 18:30
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    $\begingroup$ @DeaneYang You are right and you need Sobolev only for $p=2$! $\endgroup$
    – Giorgio Metafune
    Commented Nov 10 at 18:43

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