Does the Fourier series of an $L^1$ function converge to the function *weakly* in $L^1$?

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?

• 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

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.
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 $L^1$.