Fourier series: proving that the limit is zero

Question

Let $f: \mathbb{R}\to \mathbb{C}$ be a $2\pi$ periodic function that satisfies:

$f(t)=\frac{1}{t^{\frac{1}{3}}}$ for every $t\in (0,2\pi]$.

Show that: $\;\lim_{n\to \infty} \int_0^{2\pi} |f(t)-(S_n(f))(t)|^2 dt=0$.

We notice that $\;\lim_{n\to \infty} \int_0^{2\pi} |f(t)-(S_n(f))(t)|^2 dt= \lim_{n\to \infty} \|f(t)-(S_n(f))(t)\|_2^2$.
So it is sufficient to prove that:

$\lim_{n\to \infty} \|f(t)-(S_n(f))(t)\|_2=0$

$(S_n(f))(t)=(D_n*f)(t)$ where $D_n(x)=\frac{1}{2\pi}\sum_{k=-n}^{n}\exp(ikx)$ is Dirichlet Kernel (which is not an approximate identity! Therefore we cannot use the approximation theorem for approximate identities).

I tried to use Dirichlet's theorem but I am not sure, how to apply it here. I will highly appreciate your guide.

Answer 1

This is a consequence of Plancherel's theorem: Consider $g_n:= f-S_nf$, then up to a constant (depending on your normalization of the Fourier coefficients) we have $$ \| g_n\|_{L^2}^2= \sum_{k\in \mathbb{Z}} |\hat{g}_n(k)|^2. $$ Now notice that $$ \hat{g}_n(k)= \begin{cases} 0 & \text{ if } & |k|\leq n,\\ \hat{f}(k) & \text{ if } & |k|>n, \end{cases} $$ so that $$ \| g_n\|_{L^2}^2 = \sum_{|k|>n} |\hat{f}_n(k)|^2. $$ Now since $f\in L^2$ we have $\sum_{k\in \mathbb{Z}}|\hat{f}(k)|^2<\infty$, then the sum in the last display tends to zero as $n\to \infty$.