Recently I read that there are $L^1(\mathbb{T})$ functions such that their Fourier series does not converge in the norm of $L^1$, that is if $S_N(f)(x)$ represents the $N$-nth partial sum of the Fourier series of $f$, there exists $f$ such that:
$$\int_{-\pi}^{\pi} |S_N(f)(x)-f(x)| dx $$
does not approaches $0$ as $N$ approaches infinity. However, it is widely known that:
$$\int_{-\pi}^{\pi} |S_N(f)(x)-f(x)|^2 dx \rightarrow0 $$
as $N\rightarrow\infty$. Why can´t I use the Cauchy-Schwarz inequality, that states if two functions $F,G$ are in $L^2(\mathbb{T})$:
$$\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}F(x) \overline{G(x)} dx\right|^2\leq\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}|F(x)|^2dx\right) \left(\frac{1}{2\pi}\int_{-\pi}^{\pi}|G(x)|^2dx\right)$$
by taking $F(x)=|S_N(f)(x)-f(x)|$ and $G(x)=1$ so that:
$$\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}|S_N(f)(x)-f(x)|1 dx\right|^2\leq\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}|S_N(f)(x)-f(x)|^2dx\right) \left(\frac{1}{2\pi}\int_{-\pi}^{\pi}1dx\right) =\frac{1}{2\pi}\int_{-\pi}^{\pi}|S_N(f)(x)-f(x)|^2$$
and then take $N\rightarrow\infty$?
I´m realy struggling to see what hypothesis I´m assuming that should prevent me to do that