Vtyjkyuk

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

$$lim_ntoinftyint_-π^πf_n(x)g(x)dx=int_-π^πf(x)g(x)dx.$$

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.

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 $fin L^1(-pi,pi)$, then $S_N(f)$ tends to $f$ in norm for every $fin L_1(-pi,pi)$.

This is not hard to prove, because a weakly convergent sequence is norm-bounded, hence $sup_N|S_N(f)|_1<infty$ for every $fin L^1$. 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
$$|S_N(p)-p|_1to 0quadhboxas $Ntoinfty$$$
Since the trigonometric polynomials are dense in $L^1(-pi,pi)$, given $fin L^1(-pi,pi)$, 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 $fin L^1$, as it is formulated in your question, but I have no counterexample right now.

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

$m(x)geqslant epsilon_1$, where $m$ is Lebesgue measure.

let $E_n_k:=x$

Consider $E:=limsupE_n_k$ ,then

$m(E)=m(bigcap _j=1^infty bigcup _n_k =j^inftyE_n_k)=lim_j to infty m(bigcup _n_k =j^inftyE_n_k)gt 0$

$int _E(S_n_k-f)geqslant epsilon_2 *m(E) gt 0$

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 $L^1$