Let $w_k$ be the eigenfunctions of the Laplace-Beltrami operator on a compact manifold $M$ without boundary. We assume that $\{w_k\}$ are orthonormal, thus $\|w_k \|_{L^2} = 1$.

We know $w_k$ are smooth functions. Is such a bound true: $$\lVert w_k \rVert_{L^\infty(M)} \leq C$$ for all $k$? i.e. are all the eigenfunctions bounded above p/w a.e. by a single constant? Can we remove the a.e. part?

In 1D domains the eigenfunctions are sine and cosine functions which are nice of course.

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    $\begingroup$ Are you assuming these eigenfunctions have some specific scaling (like the one that makes them orthonormal)? Otherwise, the question isn't well-defined: since continuous functions on a compact manifold are bounded, you can just rescaled them all to make this work (or, indeed, to not work). $\endgroup$
    – Chappers
    Commented Jun 23, 2015 at 17:28
  • $\begingroup$ @Chappers Yes let us take the eigenfunctions to be orthonormal with respect to the $L^2$ inner product. $\endgroup$
    – Upin
    Commented Jun 23, 2015 at 17:58
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    $\begingroup$ I suspect such a bound is not true in general, but I don't know of a counterexample. I think a Sobolev inequality should give a bound roughly of the form $||w||_{L^\infty} \leq C \lambda^{n/4} ||w||_{L^2}$, where $n$ is the dimension and $w$ is an eigenfunction with eigenvalue $\lambda$. This does not rule out that the eigenfunctions may grow pointwise as the eigenvalue increases. Coming up with an example to show this is sharp is another question. $\endgroup$ Commented Jun 24, 2015 at 3:40
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    $\begingroup$ The first few sentences in this paper of Toth and Zelditch state that the $L^\infty$ norm of the $L^2$-normalized $\lambda$-eigenfunction is $O(\lambda^{(n-1)/4})$, and that the round sphere shows that this is sharp. Their main theorem is that under a completely integrable geodesic flow assumption, the manifolds with uniformly bounded eigenfunctions are flat. arxiv.org/abs/math-ph/0002038 $\endgroup$ Commented May 22, 2020 at 3:06

1 Answer 1


The answer to the question is no. I'm going to share some interesting things I learned from:

Toth and Zelditch prove a theorem of the form: Under a "completely integrable geodesic flow assumption", if there is a uniform $L^\infty$-bound on all $L^2$-normalized eigenfunctions, then $(M, g)$ is flat. (This is outside my area of expertise and I have not attempted to understand the assumption or to read their proof.)

Remark: The statement "there exists an orthonormal basis of eigenfunctions with uniformly bounded $L^\infty$-norm" is strictly weaker than "there is a uniform bound on the $L^\infty$-norm that applies to every orthonormal basis of eigenfunctions". For example, on the torus $\mathbb{R}^n/\mathbb{Z}^n$, the former is true, but not the latter.

Claim: Let $E_\lambda$ be the $\lambda$-eigenspace. There exists $w \in E_\lambda$ such that \begin{align*} \frac{||w||_{L^\infty}}{||w||_{L^2}} \geq \sqrt{\frac{\dim E_\lambda}{\operatorname{volume}(M)}}. \end{align*}

Corollary: If there exists a sequence of eigenvalues of unbounded multiplicity, then there exists a sequence of $L^2$-normalized eigenfunctions with unbounded $L^\infty$-norm.

Note that the standard torus and the standard sphere both have a sequence of eigenvalues of unbounded multiplicity.

Proof of Claim: For $x \in M$, consider this functional on $E_\lambda$: $f \in E_\lambda \mapsto f(x) \in \mathbb{C}$. By the Riesz representation theorem, there exists $F_x \in E_\lambda$ such that for every $f \in E_\lambda$, $f(x) = \langle f, F_x \rangle$.

In particular, $F_x(x) = \langle F_x, F_x \rangle = ||F_x||_{L^2}^2$, so $||F_x||_{L^2} = \sqrt{F_x(x)}$ and \begin{equation} \tag{$\ast$} \label{ineq} \frac{||F_x||_{L^\infty}}{||F_x||_{L^2}} \geq \frac{F_x(x)}{||F_x||_{L^2}} \geq ||F_x||_{L^2} = \sqrt{F_x(x)}, \end{equation} and this holds for every $x \in M$. (We'll try to find $x_0$ where $F_{x_0}(x_0)$ is large.)

Let $\{w_i\}$ be an orthonormal basis for $E_\lambda$. Expand $F_x$ in terms of the basis: $F_x = \sum_i \langle F_x, w_i \rangle w_i = \sum_i \overline{w_i(x)} w_i$. Now evaluate at $x$: \begin{align*} F_x(x) = \sum_i \overline{w_i(x)} w_i(x) = \sum_i | w_i(x)|^2. \end{align*} Integrate over $M$: \begin{align*} \int_M F_x(x)\, \operatorname{dvol}(x) &= \sum_i \int_M | w_i(x)|^2 \operatorname{dvol}(x) \\ &= \sum_i 1 \\ &= \dim E_\lambda. \end{align*} Thus there exists $x_0 \in M$ such that $F_{x_0}(x_0) \geq \dim E_\lambda/\operatorname{volume}(M)$. Using \eqref{ineq}, this shows that \begin{align*} \frac{||F_{x_0}||_{L^\infty}}{||F_{x_0}||_{L^2}} \geq \sqrt{\frac{\dim E_\lambda}{\operatorname{volume}(M)}}. \end{align*}

End Proof of Claim.

Remark: There's nothing special about $E_\lambda$, and in fact we could replace $E_\lambda$ by the span of any finite set of continuous functions.

  • $\begingroup$ Note that while the statement on the orthonormal basis is strictly weaker, the stronger statement may hold. For example, on a torus defined from an irrational lattice, multiplicity is bounded so that eigenfunctions are uniformly bounded. $\endgroup$ Commented Oct 27, 2020 at 9:51

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