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.
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.
