# Uniform convergence of Eigenfunction decomposition on Riemannian sphere?

Let $$\{u_k\}_{k=1}^\infty$$ be a sequence of ($$L^2$$ normalized) mutually orthogonal eigenfunctions of the operator $$-\Delta$$ on the sphere $$\mathbb{S}^n$$ (here $$\Delta$$ is the Laplace Beltrami operator). Let $$u$$ be a smooth (real valued) function on the sphere. It is a well-known result that we can write $$u=\sum_{k=1}^\infty c_k u_k$$ for some (real) constants $$c_k$$. My question is: Is the convergence of this sum uniform?

I am trying to prove that the optimal constant in the Poincare inequality is $$\lambda_1=n$$. That is to say, I am trying to prove the inequlity $$\int_{\mathbb{S}^n} |\nabla u|^2 \geq n \int_{\mathbb{S}^n} |u|^2$$. Here is what I have done so far:

First, integrate by parts on the LHS so that it suffices to prove $$\int_{\mathbb{S}^n} -u\Delta u \geq n \int_{\mathbb{S}^n} |u|^2$$. Then use $$u=\sum_{k=1}^\infty c_k u_k$$ and assume that the convergence is uniform. Then we can switch the order of the sum with the derivative and integral (and use the fact that $$\{u_k\}$$ are orthonomal) so that \begin{align*} \int_{\mathbb{S}^n} -\left(\sum_{k=1}^\infty c_k u_k\right)\Delta \left(\sum_{j=1}^\infty c_j u_j\right)&= \int_{\mathbb{S}^n} -\left(\sum_{k=1}^\infty c_k u_k\right)\left(\sum_{j=1}^\infty c_j \Delta u_j\right)= \int_{\mathbb{S}^n} -\left(\sum_{k=1}^\infty c_k u_k\right)\left(\sum_{j=1}^\infty \lambda_j c_j u_j\right) \\&= \sum_{j,k}c_k c_j \lambda_j\int_{\mathbb{S}^n} u_k u_j= \sum_{j,k}c_k c_j \lambda_j \delta_{jk}=\sum_{j}c_j^2 \lambda_j\geq \lambda_1 \sum_j c_j^2. \end{align*} By the same logic, the last sum is equal to $$\int |u|^2$$.

Now obviously, this proof requires some argument showing that the sum commutes with $$\Delta$$ and the integral but I have not been able to find a reference that the sum converges uniformly. My thought is that this would follow from some basic facts in Harmonic analysis though I am no expert in that field. Would anyone be able to provide a reference for this?

• Isn't the first eigenvalue obtained by minimising the Rayleigh quotient $\int \lvert \nabla u \rvert^2 / \int u^2$ over functions $u \neq 0$ anyway? Apr 16 at 22:20
• Wow I guess I was really overthinking this... I am still interested in the uniform convergence for its own purposes Apr 16 at 22:32

## 1 Answer

I will attempt to answer your question in a more general setting. Let $$(M^n,g)$$ be a closed Riemannian manifold of dimension $$n$$, and let $$u \in C^\infty(M)$$ be an arbitrary smooth function.

We write $$(\lambda_j \mid j \in \mathbf{N})$$ for the spectrum of the Laplacian $$\Delta_g$$, with associated sequence of eigenfunctions $$(\varphi_j \mid j \in \mathbf{N})$$, normalised so that $$\lvert \varphi_j \rvert_{2} = 1$$ for all $$j$$. For each eigenvalue $$\lambda$$ we let $$V_\lambda$$ be the corresponding eigenspace, and we define the quantity $$$$A_\lambda = \sup \{ \lvert \varphi \rvert_\infty \mid \varphi \in V_\lambda, \lvert \varphi \rvert_2 = 1 \}.$$$$

In the introduction to this paper of Toth and Zelditch it is claimed that $$A_\lambda \in O(\lambda^{(n-1)/4})$$. Apparently this follows from the Weyl law; for the round sphere you should be able to read this off the spherical harmonics. The precise expression does not matter - we only need to record that $$A_\lambda$$ grows at most polynomially in $$\lambda$$: there exist $$C > 0$$ and $$N > 0$$ so that $$$$A_\lambda \leq C \lambda^N.$$$$

We decompose the given $$u \in C^\infty(M)$$ according to the eigenfunctions of $$\Delta_g$$, writing $$u = \sum_j a_j \varphi_j$$ with $$a_j \in \mathbf{R}$$. As $$u \in L^2(M)$$ we have $$\sum_j a_j^2 < \infty$$. Because also $$\Delta u \in L^2(M)$$, and in fact $$\Delta^{(k)} u = (\Delta \circ \cdots \circ \Delta) u \in L^2(M)$$, we deduce that $$\sum_j a_j^2 \lvert \lambda_j \rvert^k < \infty$$ for all $$k \in \mathbf{N}$$. In particular $$a_j$$ must decay quicker than any power of $$\lambda_j$$.

The uniform convergence follows. To be explicit, let $$\epsilon > 0$$ be given. Then for large enough $$J \in \mathbf{N}$$, $$$$\lvert \sum_{j \geq J} a_j \varphi_j \rvert_\infty \leq \sum_{j \geq J} \lvert a_j \rvert A_{\lambda_j} \leq C \sum_{j \geq J} \lvert a_j \rvert \lambda_j^N < \epsilon.$$$$