25
$\begingroup$

I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e. Also, Wikipedia informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$. Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.

My first question is about the $p = 1$ case. That is:

If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?


I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere. Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.

My second question is similar:

Do there exist functions $f \in L^1$ whose Fourier series converge pointwise a.e., yet diverge in the $L^1$ metric?


(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system. I am also referring only to the Lebesgue measure on [0,1]. Of course, if anyone knows any more general results, that would be great, too.)

$\endgroup$

1 Answer 1

17
$\begingroup$

The answer to your first question is no. There is an $L^1$ function with Fourier series not converging in measure.

In the Kolmogorov example of an $L^1$ function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure $E$ and a subsequence $n_k$ such that for all $x$ in $E$, the absolute values of the partial sums $S_{n_k}$ of the Fourier series goes to infinity with $k$.

$$\forall x\in E,\ \ |S_{n_k}f(x)|\rightarrow \infty$$

This can be checked from the construction of $f$ in the original article of Kolmogorov, in its selected works.

If $S_nf$ converges in measure, then $S_{n_k}f$ must also converges in measure. This implies that there is a subsequence $n_{k_l}$ such that $S_{n_{k_l}}f(x)$ converges a.e. $x$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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