[Math] Weak convergence implies strong convergence in $L^1$ for Fourier series

fourier seriesreal-analysisstrong-convergenceweak-convergence

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.

Best Answer

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.

Related Question