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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)

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    $\begingroup$ 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? $\endgroup$ – Joonas Ilmavirta Jun 22 '15 at 19:04
  • $\begingroup$ 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 $\endgroup$ – Martin Kaloe Jun 22 '15 at 19:10
  • $\begingroup$ 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. $\endgroup$ – Martin Kaloe Jun 22 '15 at 19:11
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    $\begingroup$ 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). $\endgroup$ – Christian Remling Jun 22 '15 at 19:23
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    $\begingroup$ @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. $\endgroup$ – Christian Remling Jun 22 '15 at 21:03
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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.

And you'd want there to be a dense domain stabilized by the operator, or it'd not be straightforward to define higher Sobolev-type spaces to even attempt proofs of imbedding-type theorems.

Clarify?

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  • $\begingroup$ 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? $\endgroup$ – Martin Kaloe Jun 23 '15 at 13:41
  • $\begingroup$ I don't know about a definitive result for "pointwise a.e." convergence in higher dimensions. But/and such a result by itself (even as difficult as the Carleson result is in one dimension) doesn't allow us to do much with the expansion without further hypotheses (hence, my mentioning uniform pointwise and Sobolev imbedding sorts-of-things). $\endgroup$ – paul garrett Jun 23 '15 at 13:46
  • $\begingroup$ Ah I see. Yes without uniformity one cannot do much term by term. $\endgroup$ – Martin Kaloe Jun 23 '15 at 14:13
  • $\begingroup$ In a certain sense you are being too pessimistic here. You can always differentiate term by term in the distributional sense. This is, of course, much weaker than the classical concepts but suffices for many purposes, and so, perhaps, for yours. $\endgroup$ – corserine Jul 4 '15 at 8:57

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