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Suppose $u_n$ is an orthonormal basis of smooth functions on $S^1$.

Does there exist a smooth function $u$ such that the generalised Fourier series $$u=\sum_{n\in\mathbb{N}} \langle u,u_n\rangle u_n $$ does not converge uniformly?

We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.

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    $\begingroup$ I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p \in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) \ne 0$ then the series will not even converge pointwise. $\endgroup$ – Nate Eldredge May 16 at 3:40
  • $\begingroup$ Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious? $\endgroup$ – Goonfiend May 16 at 3:46
  • $\begingroup$ Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, \dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$. $\endgroup$ – Nate Eldredge May 16 at 3:48
  • $\begingroup$ Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it. $\endgroup$ – Goonfiend May 16 at 3:52
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The series need not converge uniformly, nor even pointwise everywhere.

For instance, fix $x \in S^1$ and consider the space $C^\infty_x$ of smooth functions that vanish at $x$. This is a dense subspace of $L^2(S^1)$, so by choosing a countable dense subset of $C^\infty_x$ and applying Gram-Schmidt, we can find a sequence $u_1, u_2, \dots \in C^\infty_x$ which is an orthonormal basis for $L^2(S^1)$. Then if $u(x) \ne 0$, the series evaluated at $x$ converges to 0, not to $u(x)$.

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