# $L^\infty$-bounds on eigenfunctions of Laplace-Beltrami opeator

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.

• 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). Commented Jun 23, 2015 at 17:28
• @Chappers Yes let us take the eigenfunctions to be orthonormal with respect to the $L^2$ inner product.
– Upin
Commented Jun 23, 2015 at 17:58
• 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. Commented Jun 24, 2015 at 3:40
• 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 Commented May 22, 2020 at 3:06

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 $$$$\tag{\ast} \label{ineq} \frac{||F_x||_{L^\infty}}{||F_x||_{L^2}} \geq \frac{F_x(x)}{||F_x||_{L^2}} \geq ||F_x||_{L^2} = \sqrt{F_x(x)},$$$$ and this holds for every $$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.

• 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. Commented Oct 27, 2020 at 9:51