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Suppose $F$ has discrete Fourier transform $(a_n)$ where $a_n=0$ unless $n=2^k$ for some $k > 0$, in which case $a_n=1/k$ (or $a_n=1/k^2$ if you want: I'm happy with anything polynomial). What sort of regularity conditions does $F$ have? Is it Holder continuous, or not?

To be explicit:

$$ F(x)=\sum_{k=1}^\infty k^{-2} \exp(ix2^k) $$

for example.

More generally, I'm interested in two dimensional (discrete) Fourier transforms: is there a good reference for this sort of thing?

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3 Answers 3

up vote 7 down vote accepted

If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}_\alpha$ only if the Fourier coefficients satisfy $a_n = {\rm O}( n^{-\alpha})$; this is in Katznelson's book (Chapter I, Corollary 4.6) for instance.

[EDIT (2013-07-10): at the time I thought this was "iff" but a comment points out that I misremembered; in any case, for lacunary series such as the one in the question, a lot more is known than in the general case; see e.g. Katznelson Chapter V for the basics.]

So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff.

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Great! Will have a look at Katznelson, but at the very least, this gives me an intuition about what's going on. –  Matthew Daws Nov 8 '09 at 14:32
1  
@Yemon, could you tell me on which page I can find the proof of the statement you stated at the beginning? I found the statement that Lip$_\alpha$ implies Fourier coefficient decays like $O(1/n^{\alpha})$, but I couldn't find the converse. Thank you. –  Syang Chen Jul 15 '12 at 8:25
    
@SyangChen my mistake - thanks for pointing this out –  Yemon Choi Jul 10 '13 at 22:25

One of the first examples (historically) of nowhere differentiable continuous functions was given by $a_{2^n} = 2^n$ and $0$ otherwise. Taking tensor powers of this function you get very irregular functions of the kind you want. By very irregular here I mean nowhere differentiable (and so at least not in $\operatorname{Lip}_1$, but maybe you can get much more). In any case these Fourier series are called lacunary (à la Hadamard) and there should be a lot of literature about them.

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@AndrewStacey I took the liberty of fixing your fix and operatornaming something while at it –  Yemon Choi Jul 10 '13 at 17:06

Gian Maria Dall'Ara's comment is the solution. This function that you describe is a (typical) example of a continuous function that's nowhere differentiable. In fact, suppose that you have an integrable function $F$ such that $\hat F(n) = a _ n $ whenever $n=\lambda _ k$ and zero otherwise, where we assume that the sequence $\lambda_k$ is lacunary in the sense of Hadamard (i.e ${\lambda _ {k+1}} / {\lambda _ k}\geq c$). If the function $F$ is differentiable at some point then $a _ {\lambda _k}=o(\frac{1}{\lambda _ k})$ (actually i have the impression that the proof of this fact uses the weaker assumption that $F$ is Lipschitz continuous at some point). More generally, if you replace differentiability of the function $F$ with $\alpha$-Hölder continuity (in a neighborhood of zero say) for $0<\alpha <1$ then you conclude that $a _ {\lambda _k}=O(\frac{1}{\lambda _ k ^\alpha})$. So your function is not $\alpha$- Hölder either.

Remark 1: The contrary is also true since $a _ {\lambda _k}=o(\frac{1}{\lambda _ k})$ implies that $F$ is differentiable at any point of the circle where the partial sums converge to the function. I have some doubts about the precise hypothesis needed here. I'm not sure if you need your function to have only positive spectrum, but your function here does anyways.

Remark 2: You can look in Grafakos book for example, or of course, in Zygmund's trigonometric series (that would be my first reference for this type of problems). Katznelson has also a lot of information. But I know that Grafakos book contains these results for sure.

Remark 3: So your function is nowhere differentiable and is not Hölder continuous either. However it has other nice properties. For example, it belongs to any $L^p$ for $1\leq p <\infty$ and the $L^p$ norm is comparable to the $L^2$ norm (here note that the lacunary gaps force the Littlewood-Paley pieces of the function to behave as independent random variables). On top of that, using kolmogorov's result on lacunary Fourier series you get an easy a.e convergence result of the partial sums to the function (something which is still true for $L^2$ functions in general, but several scales deeper and more difficult to prove).

Remark 4: Finally, your function has only positive frequences and belongs to $L^p$ on the circle, hence it belongs to the Hardy space $H^p$ on the circle. I don't know if you can use that in your problem, but it is a strong property.

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