Is there an $f\in L^{1}(T)$ whose partial sums of Fourier series $S_{n}(f)$ satisfies $\S_{n}(f)\_{L^{1}(T)} \rightarrow \f\_{L^{1}(T)}$ but $S_{n}(f)$ fails to converge to $f$ in $L^1$norm ?

I was hesitating for a while whether to answer or to vote to close and to refer the OP to AoPS, but, since the question has been upvoted, here goes. Suppose that $f_k\in L^1$ converges to $f\in L^1$ in the sense of distributions and in measure. Suppose also that $\f_k\_1\to \f\_1$. Then $f_k\to f$ in $L^1$. Indeed, let $g$ be a bounded by $1$ infinitely smooth function such that $\int fg>\f\_1\delta$. Then $\int f_kg > \f\_12\delta$ for large $k$. Now, $\int_{\{f_kf<\delta\}} f_kg\ge \int_{\{f_kf<\delta\}} fg\delta\ge \int fg2\delta\ge \f\_13\delta$ for large $k$ because the integral of a fixed $L^1$ function $fg$ over a set of small measure is small. So, $\intf_k\ge \int_{\{f_kf\ge\delta\}} f_k+\f\_13\delta$ whence the first integral is at most $4\delta$ for large $k$. Thus $$ \int f_kk\le \int_{\{f_kf\ge \delta\}} (f_k+f)+\delta\le 6\delta $$ for large $k$. To apply this to $f_k=S_kf$, one only needs to check the convergence in measure. But it immediately follows from the weak type 11 bound for $S_k$ (applied to the difference of $f$ and a trigonometric polynomial approximating $f$ in $L^1$, of course). 

