Weak convergence implies strong convergence in $L^1$ for Fourier series?

Question

We say $\{f_n\}$ weakly converge to $f$ in $L^1[-π,π]$ if for each $g \in L^\infty[-π,π]$,

$$\lim_{n\to\infty}\int_{-π}^{π}f_n(x)g(x)dx=\int_{-π}^{π}f(x)g(x)dx.$$

There is a question in my homework which I can't prove it:

For each $f \in L^1[-π,π]$, the Fourier partial sums are denoted by $S_n$. If $S_n$ weak converge to $f$, then $S_n$ strongly converge to $f$.

I think maybe we need to find some specific characteristic functions to prove the statement while I failed to find it.

By the way, I made some effort below:

Using the property of weak convergence, we can prove the Fourier partial sum converge in measure, which means it has a pointwise almost everywhere convergence subsequence. There are two ways to prove the statement.

1. Since {$S_n$} is uniformly bounded(by weak convergence), it is obvious that the pointwise a.e convergence subsequence is also convergence in norm. But I don't know how to show the whole sequence is converge.
2. I have a lemma which guarantees if I prove the pointwise a.e convergence of {$S_n$}, then I can prove the original statement. This lemma is very difficult to prove but I can make sure it's right. However I can't prove the pointwise a.e convergence.

Answer 1

Your formulation suggests that if $S_N(f)$ tends weakly to $f$ for a specific function $f$, then $S_N(f)$ tends to $f$ in the $L^1$ norm. I don't believe this was your homework. I think your homework was this:

Prove that if $S_N(f)$ tends weakly to $f$ for every $f\in L^1(-\pi,\pi)$, then $S_N(f)$ tends to $f$ in norm for every $f\in L_1(-\pi,\pi)$.

This is not hard to prove, because a weakly convergent sequence is norm-bounded, hence $\sup_N\|S_N(f)\|_1<\infty$ for every $f\in L^1$. By the uniform boundedness principle, $\sup_N\|S_N\|$ is finite, where $S_N$ is viewed as an operator from $L^1$ to $L^1$. Now, if $p$ is a trigonometric polynomial, then clearly $$\|S_N(p)-p\|_1\to 0\quad\hbox{as $N\to\infty$}$$ Since the trigonometric polynomials are dense in $L^1(-\pi,\pi)$, given $f\in L^1(-\pi,\pi)$, and $\varepsilon>0$, there exists a trigonometric polynomial $p$ such that $\|p-f\|_1<\varepsilon$. Hence: $$\|S_N(f)-f\|= \|S_N(f)-S_N(p)+S_N(p)-p+p-f\|\leq \sup_N\|S_N\|\|p-f\|+\|S_N(p)-p\|+\|p-f\|$$ where all the norms are in $L^1(-\pi,\pi)$. The crucial point is that $\sup_N\|S_N\|$ is finite, and so we can make the l.h.s as small as we wish for sufficiently large $N$, which proves the assertion.

I am pretty sure that the result is not true for a specific, single $f\in L^1$, as it is formulated in your question, but I have no counterexample right now.

Answer 2

To prove the statement, we worked step by step.

1. {$S_n$} is bounded in $L^1$[−π,π]

This is an important property of weak convergence and I don't prove here.

2. Dunford-Pettis Theorem here

Suppose that (X,Σ,µ) is a probability space, and that $\mathscr F $ is a bounded subset of $L^1(µ)$.

$\mathscr F$ is equi-integrable if and only if $\mathscr F$ is a relatively compact subset of $L^1(µ)$ with the weak topology.

From this theorm, we can conclude that {$S_n$} is equi-integrable.

3. {$S_n$} is convergence to $f$ in measure.

Prove: If not, we have a subsequence {$S_{n_k}$} , $\epsilon_1 >0,\epsilon_2>0$, s.t

$m(\{x|S_{n_k}(x)-f(x) \geqslant \epsilon_2\})\geqslant \epsilon_1$, where $m$ is Lebesgue measure.

let $E_{n_k}:=\{x|S_{n_k}(x)-f(x) \geqslant \epsilon_2\}$

Consider $E:=limsupE_{n_k}$ ,then

$m(E)=m(\bigcap _{j=1}^{\infty} \bigcup _{{n_{k}} =j}^{\infty}E_{n_k})=\lim_{j \to \infty} m(\bigcup _{{n_{k}} =j}^{\infty}E_{n_k})\gt 0$

$\int _{E}(S_{n_k}-f)\geqslant \epsilon_2 *m(E) \gt 0$

Contradiction!

So {$S_n$} is convergence to $f$ in measure.

4. Vitali Convergence Theorem here

$f_n$ $\in$ $L^1$[−π,π], then $f_n$ convergence to $f$ in $L^1$ if and only if $f_n$ convergence to $f$ in measure and $f_n$ uniformly integrable

And equi-integrable implies uniformly integrable, so we have $S_n$ convergence to $f$ in $L^1$