Take the 2-minute tour ×

As we know, for $1<p<\infty$, the Fourier series of $f\in L^{p}(T)$ converges to $f$ in $L^{p}$-norm. But is there any results concerning the convergence of Fourier series in $L^{\infty}$-norm? Since $L^{\infty}(T)$ is not separable, the trigonometric system fails to form a Schauder basis of $L^{\infty}(T)$, this implies that the Fourier series of $L^{\infty}(T)$-functions fails to converge in $L^{\infty}$-norm. But does the Fourier series of $f$ converge in $L^{\infty}$-norm for every $f\in C(T)$?

share|improve this question
Partial sums of Fourier series are continuous functions, so you get pointwise convergence everywhere. –  Harald Hanche-Olsen Apr 3 '11 at 12:33
You are confused. $L^\infty$ convergence of continuous functions is uniform convergence. –  Harald Hanche-Olsen Apr 3 '11 at 13:07
No one is disputing that Fourier series can diverge on a null set. Katznelson's example showed that it must be so. The point is that for a continuous function, knowing its behaviour on a set of full measure suffices to prescribe it every where. In other words, $$ \sup_{x\in T} | S_n(s) - S_m(s)| \leq \| S_n - S_m\|_\infty $$ for all partial sums $S_N$. So $L^\infty$ convergence does imply pointwise (and hence even uniform) convergence. What you should think is that for any $a > 0$, the set $A_{n,m}$ of points in $T$ such that $|S_n - S_m| > a$ decays to measure zero as $n,m\to\infty$, but –  Willie Wong Apr 3 '11 at 14:03
Short version: the canonical map from the Banach space $C({\mathbb T})$ into the Banach space $L^\infty({\mathbb T})$ really is injective. –  Yemon Choi Apr 3 '11 at 18:59
show 8 more comments

closed as too localized by Harald Hanche-Olsen, Willie Wong, Andres Caicedo, Mark Meckes, Yemon Choi Apr 3 '11 at 19:00


2 Answers

up vote 3 down vote accepted

If Fourier series of continuous functions would converge in $L^\infty$, then, by the Uniform Boundedness Principle, the operator norms in $C(\mathbb{T})$ of the partial Fourier series operators $S_Nf(t):=\sum_{n=-N}^N\hat{f}(n)e^{int}$ would be uniformly bounded. You can find, for example in Katznelson book, a proof of the fact that such norms diverge logarithmically.

share|improve this answer
add comment

I think you can find the answer in P191 "Classical Fourier Analysis",second edition, Loukas Grafakos

share|improve this answer
NO, it's not the answer to the question of mine. The result of duBois Reymond is well known and is a easy corollary of the Banach-Steinhaus theorem. But it is about pointwise convergence, not convergence in $L^\infty$-norm. –  Acky Apr 3 '11 at 13:04
but as the function and the partial sum of the Fourier series are all continuous functions, if we have $L^{\infty}$ convergence of partial sum, then we can get contradiction? –  Shaoming Guo Apr 3 '11 at 13:07
add comment
