# Uniform convergence of Fourier (orthonormal) expansion of series

Let $u \in L^2(M)$ on some closed Riemannian manifold. We can write $$u = \sum_{k \geq 0}(u,\varphi_k)\varphi_k$$ if $\varphi_k$ is some o.n basis of $L^2$ with is orthogonal in $H^1$ (eg. eigenfunctions of appropriate operator). So (I believe) for almost all $x$, $$u(x) = \sum_{k \geq 0}(u,\varphi_k)\varphi_k(x).$$ How does one know if this sum is uniformly convergent on $x \in M$? If say $u \in H^1(M)$, it would be nice to say $$\nabla u(x) = \sum_{k \geq 0}(u,\varphi_k)\nabla \varphi_k(x)$$ and this commutation of operators need uniform continuity of the RHS too. So this is why I want to know.

(this question was posted by someone some time ago in MSE and was unanswered for months but sadly I cannot find a link anymore)

• An $L^2$ orthonormal basis need not consist of smooth functions or even functions that have $L^2$ gradients, so you need to say more to make sense of your question about the convergence of derivatives. Also, I don't quite see how uniform convergence is relevant for writing gradients of $H^1$ functions. And pointwise values of the $L^2$ functions $\phi_k$ don't mean much. Can you give more details on what you want? – Joonas Ilmavirta Jun 22 '15 at 19:04
• Sorry about that, the basis should be in $H^1$. I edited the question. I wanted to take the gradient of $u$ by simply taking the term by term gradient in the infinite sum. For this you need uniform convergence in $x$ of the sequence $\sum_{i=0}^N(u,\varphi_k)\nabla \varphi_k(x)$ as $N \to \infty$. @JoonasIlmavirta – Martin Kaloe Jun 22 '15 at 19:10
• And moreover I am interested in what other senses the first infinite sum in OP holds. Obviously the equation holds in $L^2$ but pointwise a.e. should also be true I think. – Martin Kaloe Jun 22 '15 at 19:11
• Pointwise convergence a.e. does NOT hold for arbitrary ONBs; you need specific properties of the Fourier basis (this is part of the reason why Carleson's theorem is so difficult). – Christian Remling Jun 22 '15 at 19:23
• @MartinKaloe: No. Convergence a.e. on a subsequence (which, as you pointed out, is trivial) is a far cry from convergence a.e. on the original sequence. – Christian Remling Jun 22 '15 at 21:03

Following up on the comments... I think no simple general statement holds, and the very delicate Carleson-theorem type results are too much to hope for in higher dimensions. E.g., already on a multi-torus $\mathbb T^n$, with operator the usual Laplacian, the $H^1$ Sobolev space does not imbed to $C^o$: we need ${n\over 2}+\epsilon$ for uniform pointwise convergence.
Also, there are unbounded self-adjoint (densely-defined) operators that have little to do with smoothness, such as multiplication by $1+x^2$ on the real line. Being in the corresponding "formal" Sobolev spaces only assures decay, not smoothness.
• Thanks. I am only concerned about pointwise a.e. convergence, not pointwise everywhere so I think Sobolev embeddings into classes of continuous functions is not needed. From reading Theorem 1 of this note (math.uni-bielefeld.de/~tpoguntk/media/fourier_abstract.pdf), the a.e. pointwise convergence of $\sum_{1}^n (u,\varphi_k)\varphi_k(x)$ to $u(x)$ holds when working in $L^2(0,1)$. Are you saying that such a result is not true/known in higher dimensions? – Martin Kaloe Jun 23 '15 at 13:41