Let $f$ be a periodic $L^1$ function, and $S\_n[f]$ the $n$-th partial sum of its Fourier series. I [am aware](https://mathoverflow.net/questions/28428/convergence-of-fourier-series-of-l1-functions) 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?



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


