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For a Hilbert space $L^2$ we have the notion of an orthonormal basis $\{f_j\}$ being a sequence of orthonormal elements such that any element $f$ in $L^2$ can be approximated by partial sums in terms of this basis $$f = \sum_{j=1}^\infty \langle f, f_j \rangle f_j$$ Here the sum converges wrt the $L^2$ norm. This is what I mean by generalized Fourier series.

I have been reading about Carleson's Theorem that says specifically for Fourier series, the series converges pointwise almost everywhere to the approximated function. I have also read that this is not true for a general orthonormal basis. I was hoping someone would be able to provide me with an example demonstrating that statement on a finite measure space, maybe $L^2([0,1])$: A function whose partial sums in terms of the basis do not convergence pointwise almost everywhere.

muaddib
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    Here is something similar to yours: http://mathoverflow.net/q/209900/75968 – Svetoslav Jul 27 '15 at 21:57
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    Of interest is Definition 8 and Theorem 11 from section 7 of this paper. – David Mitra Jul 28 '15 at 12:12
  • Section 9 of the paper linked above gives an example of what you want, I think. – David Mitra Jul 28 '15 at 12:22
  • The paper recommended by @DavidMitra says in page 602 that Zahorski in 1960 asserted the existence of an $L^2$ function whose Fourier series could be re-arranged to diverge a.e. If you take $f$ to be that function and the $f_j$ that particular rearrangement of the Fourier base, then I think you have your counter-example. – Enredanrestos Jul 28 '15 at 13:12
  • You can find classical results about this in Garcia Topics in Almost Everywhere Convergence. – David C. Ullrich Jul 28 '15 at 16:57

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Thank you for the comments. I was able to use them to provide an answer to When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?. I'll restate the result from there as it provides an example:

Consider the measure space $L^2([0,1])$ with uniform lebesgue measure. Define the functions (Haar functions) $f_{2^i + k}$ by $$f_{2^i + k} = 2^{i/2}\chi_{\left[\frac{k}{2^i}, \frac{k+1/2}{2^i}\right]} - 2^{i/2}\chi_{\left[\frac{k + 1/2}{2^i}, \frac{k+1}{2^i}\right]}$$ for $i \geq 1, 0 \leq k < 2^i$.

On pg. 598 of the paper mentioned in the comments: Topics in Orthogonal Functions - Price the author states the Haar functions defined above form a complete orthonormal basis. Then pg. 603 of the paper states

For every complete orthonormal system $\Phi$ there is an $L^2$ function $g$ whose $\Phi$-Fourier series can be rearranged to diverge almost everywhere.

So we can finish as follows. Take the Haar functions as above (a complete orthonormal set) and the function $g$ defined above. We write this function's $\Phi$-Fourier series as $$\sum_{n=1}^\infty \langle g, f_n \rangle f_n$$ Now there exists a rearrangement $\sigma$ of the indices of the sum such that it diverges almost everywhere. Now let $g_n = \langle g, f_{\sigma(n)} \rangle f_{\sigma(n)}$ be this rearrangement of terms. The sum $\sum g_n$ diverges almost everywhere.

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muaddib
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  • One might mention that in the linked paper, a specific permutation of the Haar basis, and a sketch of a proof that that it "works", is given. – David Mitra Jul 28 '15 at 18:28
  • What does your notation $\chi_{[a,b]}$ mean? And what is "the function $g$ defined above"? – tparker Jan 26 '18 at 03:05
  • $\chi_{[a, b]}$ is the indicator function of the set $[a,b]$. It is $1$ for $x$ in the set and $0$ otherwise. $g$ is defined the highlighted text - "an $L^2$ function whose $\Phi$-Fourier can be rearranged to diverge almost everywhere." – muaddib Jan 26 '18 at 14:07
  • Ah, thanks, I missed the definition of $g$. – tparker Jan 26 '18 at 15:08
  • So the function sequence of partial sums $h_n(x)$ is Cauchy w.r.t. the $L^2$ norm, but the real sequence $h_n(x_0)$ is not Cauchy for any fixed $x_0$? – tparker Jan 26 '18 at 15:16
  • Regardless of the rearrangement, the $L^2$ norm exists and is the same - so yes, it is cauchy wrt to that. If by $h_n$ you mean the sequence of partial sums given by rearrangement such that it "diverges almost everywhere" then your second statement is almost accurate. There is a measure zero set such that the sequence does converge pointwise. – muaddib Jan 27 '18 at 17:28