$L^\infty$-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.
 A: The answer to the question is no. I'm going to share some interesting things I learned from:


*

*Toth and Zelditch, Riemannian Manifolds with Uniformly Bounded Eigenfunctions: https://arxiv.org/abs/math-ph/0002038

*Garrett, Harmonic Analysis on Spheres, I: http://www-users.math.umn.edu/~garrett/m/mfms/notes_c/spheres_I.pdf - 
This is a nice set of notes building the theory of spherical harmonics in an interesting way.
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.
