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Let $f$ be a periodic $L^1$ function, and $S_n[f]$ the $n$-th partial sum of its Fourier series. I am aware that $S_n[f]$ might not converge toward $f$ in $L^1$ (i.e., in norm). However, does it at least converge weakly? In other words, is it true that for every $L^\infty$ function $h$ (thus defining a continuous linear form on $L^1$), the integral of $S_n[f]\cdot h$ on one period converges to the integral of $f\cdot h$?

I believe the question can be rephrased as follows: if $g = f\*h$ is the convolution of an $L^1$ function and an $L^\infty$ function, is it true that the Fourier series of $g$ converges pointwise to $g$? (Clearly $g$ is a continuous function, but it is well known that this does not suffice. However, I see no reason why the convolutions of $L^1$ and $L^\infty$ functions should exhaust the continuous functions.)

If the answer is negative, is there some nice subspace of $L^\infty$ such that for all $h$ in this subspace the property holds?

Comment: More generally, one could ask, "for all functions $f$ in <some space>, and all linear forms $h$ in <some subspace of the dual space>, is it true that the Fourier series of $f$ converges to $f$ when tested against $h$?" For instance, if $f$ ranges over finite signed Borel measures on the circle and $h$ over continuous functions, the answer is negative (take $f$ to be a Dirac measure at $0$ and $h$ such that the Fourier series of $h$ does not converge at $0$); whereas if $f$ ranges over Schwartz distributions and $h$ over $C^\infty$ functions then the answer is positive (because $f\*h$ will be smooth). Is there something intelligent to be said along those lines?

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In fact, $L^1({\mathbb T}) \ast L^\infty({\mathbb T}) = C({\mathbb T})$. I don't know a proper reference, but Cohen's factorization theorem states that for any Banach algebra $A$ with a bounded approximate identity and any Banach $A$-module $X$, the subset $\lbrace ax : a\in A,\ x\in X\rbrace$ is automatically a closed subspace. –  Narutaka OZAWA Oct 29 '12 at 20:34

2 Answers 2

up vote 4 down vote accepted

No. If the partial sum projections $S_n$ converged in the weak operator topology, they would be pointwise weakly bounded hence pointwise norm bounded whence uniformly bounded. That would give convergence pointwise strongly.

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I'd say the answer to the first question is no, by the same Kolmogorov's counterexample quoted by coudy in the linked answer. Since $|S_{n_k}|\to \infty$ in a set $E$ of positive measure, by Severini-Egorov (applied to $1/|S_{n_k}|$) we also have that $|S_{n_k}|$ converges to infinity uniformly on a subset of positive measure, thus it is unbounded in $L^1$.

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