Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there an $f\in L^{1}(T)$ whose partial sums of Fourier series $S_{n}(f)$ satisfies $\|S_{n}(f)\|_{L^{1}(T)} \rightarrow \|f\|_{L^{1}(T)}$ but $S_{n}(f)$ fails to converge to $f$ in $L^1$-norm ?

share|improve this question
4  
@Peter Humphries: but, this is the converse of what the OP wants to know about! Your inequality only shows that convergence of $S_n(f)$ implies convergence of $\| S_n(f) \|$. –  Zen Harper May 9 '11 at 3:19
4  
A probably-useless observation: since pointwise a.e. convergence + ‘norm convergence’ in the weak sense of your question imply $L^1$-convergence, you need to look for functions whose Fourier series don't convergence pointwise. See en.wikipedia.org/wiki/… for some starting points. –  L Spice May 9 '11 at 3:54
6  
@Peter Humphries - sorry to chip in again, but you say "apply Bolzano-Weierstrass". This is incorrect; the Bolzano-Weierstrauss theorem (that bounded sequences have convergent subsequences) is true in $R$ or $R^k$, but definitely NOT true in $L^1$, or indeed in any infinite-dimensional Banach space (the closed unit ball of a Banach space is sequentially compact if and only if the space is finite-dimensional). I think you need the detailed structure of Fourier series for this question; I seriously doubt that general Banach space theory will be able to resolve this question either way. –  Zen Harper May 9 '11 at 5:36
2  
This is probably a crazy suggestion, but do you know much about random Fourier series, i.e. series $\sum_n Y_n e^{in \theta}$ with the $Y_j$ being random variables? From what I understand, this is a very technical, difficult topic; but, on the other hand, if you're trying to construct complicated counterexamples, it may be that probabilistic methods succeed where "deterministic" methods fail! There are many known conditions for almost sure norm convergence, etc. Unfortunately I don't have access to very much literature, but if you have a good library then you might be able to find something. –  Zen Harper May 9 '11 at 8:05
2  
Well, of course there are the standard books: A. Zygmund, Trigonometric series; Kahane and/or Katznelson, Introduction to Harmonic analysis, etc. etc. - but this is from memory only. Tom Korner might have some stuff, but I think most of his books are at a lower level. If you're "junior" (whatever you mean by that) without access to literature, then all this will be difficult; online stuff can be useful, but there are too many gaps. Fourier series done rigorously is an advanced, tricky topic with all sorts of difficult and obscure results only available in old, rare books. –  Zen Harper May 9 '11 at 9:23

1 Answer 1

up vote 6 down vote accepted

I was hesitating for a while whether to answer or to vote to close and to refer the OP to AoPS, but, since the question has been upvoted, here goes.

Suppose that $f_k\in L^1$ converges to $f\in L^1$ in the sense of distributions and in measure. Suppose also that $\|f_k\|_1\to \|f\|_1$. Then $f_k\to f$ in $L^1$.

Indeed, let $g$ be a bounded by $1$ infinitely smooth function such that $\int fg>\|f\|_1-\delta$. Then $\int f_kg > \|f\|_1-2\delta$ for large $k$. Now, $\int_{\{|f_k-f|<\delta\}} f_kg\ge \int_{\{|f_k-f|<\delta\}} fg-\delta\ge \int fg-2\delta\ge \|f\|_1-3\delta$ for large $k$ because the integral of a fixed $L^1$ function $fg$ over a set of small measure is small.

So, $\int|f_k|\ge \int_{\{|f_k-f|\ge\delta\}} |f_k|+\|f\|_1-3\delta$ whence the first integral is at most $4\delta$ for large $k$. Thus $$ \int |f_k-k|\le \int_{\{|f_k-f|\ge \delta\}} (|f_k|+|f|)+\delta\le 6\delta $$ for large $k$.

To apply this to $f_k=S_kf$, one only needs to check the convergence in measure. But it immediately follows from the weak type 1-1 bound for $S_k$ (applied to the difference of $f$ and a trigonometric polynomial approximating $f$ in $L^1$, of course).

share|improve this answer
    
I think this is a valuable answer. While I might have been taught (and since forgotten) the result you prove, I certainly hadn't though of applying this to convolution with the Dirichlet kernel –  Yemon Choi May 13 '11 at 20:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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