3
$\begingroup$

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.
$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Not that I see how to prove it, but why do you think his version of the problem is wrong? $\endgroup$ – David C. Ullrich Aug 14 '18 at 15:23
  • $\begingroup$ I am sorry but it is indeed my homework. I know it's weird to say that weak convergence can imply strong convergence. However I can't prove or disprove it. $\endgroup$ – Luke Chen Aug 14 '18 at 16:54
  • $\begingroup$ Hello, I made some edit in the question .Maybe it's useful for you.Thank you! $\endgroup$ – Luke Chen Aug 14 '18 at 17:17
0
$\begingroup$

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.

  1. 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.

  1. {$S_n$} 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.

  1. 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 $L^1$

$\endgroup$
  • $\begingroup$ I suspect there's an error. You never used the fact that Sn is the partial Fourier sum of f. If your post were correct, this proof would apply to any weak convergent sequence, which is clearly impossible. $\endgroup$ – Giuseppe Negro Aug 16 '18 at 9:40
  • $\begingroup$ @GiuseppeNegro Well, you are right.I never used the partial sum property. But I cannot figure out where I am wrong. Step1.2.4 are obvious. I will check step 3 carefully. Thank you $\endgroup$ – Luke Chen Aug 16 '18 at 14:42
  • $\begingroup$ The error must be in Step 3. $\endgroup$ – Giuseppe Negro Aug 16 '18 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.