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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?

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    $\begingroup$ 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? $\endgroup$
    – Leo Moos
    Apr 16 at 22:20
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    $\begingroup$ Wow I guess I was really overthinking this... I am still interested in the uniform convergence for its own purposes $\endgroup$ Apr 16 at 22:32
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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 \begin{equation} A_\lambda = \sup \{ \lvert \varphi \rvert_\infty \mid \varphi \in V_\lambda, \lvert \varphi \rvert_2 = 1 \}. \end{equation}

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 \begin{equation} A_\lambda \leq C \lambda^N. \end{equation}

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 $J \in \mathbf{N}$, \begin{equation} \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. \end{equation}

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