Consider the function $f \in C_{st}$ which on the interval $]-\pi,\pi[$ is equal to the function $x \cos(x)$. Then I have to argue whether or not the Fourier series for $f$ converges uniformly or in $L^2$.
I have said that as $x \cos(x)$ is a well known continuous function on $\mathbb{R}$ and that the limits $\lim_{x \rightarrow \pi^-} x \cos(x)$ and $ \lim_{x \rightarrow -\pi^+} x \cos(x)$ exists $f$ is especially continous. Futhermore that $f \in C^1_{st}$ as $x \cos(x)$ is $C^1$ on $\mathbb{R}$ and the derivative of f $ \cos(x) - x\sin(x)$is a well known continous function and $C^1$ in the inverval $]-\pi,\pi[$. Thus the Fourier series for $f$ converges uniformly on $\mathbb{R}$ towards $f$.
However I am not sure about the converges in $L^2$. From a definition in my book a sequence $\{f_n\} \in C_{st}$ is said to converge towards $f in C_{st}$ in $L_2$ if $||f-f_n|| \rightarrow 0$ for $n \rightarrow \infty$. But I am not sure how to prove or disprove that? Can you help me?
Let $S_n$ be the fourier series of $f$. You said that $(S_n)$ converges uniformly to $f$, that is : \begin{align} \forall \varepsilon >0, \exists N >0, \forall n > N, \| f-S_n\|_{\infty} < \varepsilon \end{align}
Now, look at $|f(t)-S_n(t)|^2$ : if n > N, then it is always less than $\varepsilon^2$. So \begin{align} \int_0^{2\pi} |f-S_n|^2 < 2\pi \varepsilon^2 \end{align}
and now you can conclude