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

Suppose we are given the Fourier coefficients of an $L^2$ function on the circle. Are there necessary and sufficient conditions on the coefficients that allow us to determine that $f$ is Hölder continuous of order $\alpha$?

Note that the necessary condition $|\hat{f}(n)| \leq C_f|n|^{-\alpha}$ is not sufficient. For example if $\hat{f}(n)=|n|^{-2/3}$ for all $n$ then $f$ is an $L^2$ function whose Fourier series does not converge absolutely. Therefore $f$ cannot be Holder continuous of order $\alpha>1/2$.

share|improve this question
    
As a general rule, the further away the integrability exponent $p$ of a physical space-based function space is from the Hilbert exponent $2$, the harder it is to control the norm via the magnitude of the Fourier coefficients (basically because analogues of the Plancherel identity, such as the Hanner inequalities or the Hausdorff-Young inequalities, become less and less efficient as one moves further away from 2). The Holder classes have exponent $\infty$ (as depicted for instance in this diagram of mine: terrytao.wordpress.com/2010/03/11/… ) ... –  Terry Tao Jan 31 '13 at 17:34
1  
... and so one cannot hope for really sharp criteria based only on the magnitude of individual Fourier coefficients. (However, thanks to Littlewood-Paley theory, which has much better L^p stability properties than the Fourier transform, one can get good control in terms of Littlewood-Paley components of the function, as Bazin points out below. ) –  Terry Tao Jan 31 '13 at 17:36
    
Thank you for the great diagram and explanation. Is there a reason that this problem becomes easy again for $C^{\infty}$ functions? –  Matt Jacobs Feb 1 '13 at 0:31
add comment

3 Answers 3

up vote 5 down vote accepted

There is an excellent characterization of Hölder spaces via the Fourier transform, using Besov spaces. Let $\alpha\in (0,1)$: a function $u$ defined on $\mathbb R^n$ belongs to $L^\infty\cap C^\alpha$ if and only if it belongs to $B^\alpha_{\infty,\infty}$, i.e. $$ \sup_{\nu\in \mathbb N}2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty}<+\infty,\quad\text{i.e. the sequence} (2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty})_{\nu\in \mathbb N} \in \ell^\infty. $$

Here $\phi_\nu$ stands for a Littlewood-Paley decomposition: $$ 1=\sum_{\nu\in \mathbb N}\phi_\nu(\xi), $$ $\phi_0$ is compactly supported and for $\nu\ge 1$, $\phi_\nu(\xi)=\phi(2^{-\nu}\xi)$ where $\phi$ is supported in the ring $1/2\le \vert\eta\vert\le 2$ so that $\phi_\nu$ is supported in the ring $2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.

share|improve this answer
    
I'm not sure I understand the notation $\phi_{\nu}(D_x)u$ –  Matt Jacobs Jan 30 '13 at 21:52
    
$D_x$ should stand for $i\partial_x$ (the $i$ is to make it self-adjoint, i.e. to make $e^{i\xi x}$ an eigenfunction of $D_x$ with real eigenvalue). If $\phi$ is a measurable function on $\mathbb{R}$, $\widehat{\phi(D_x)f}(\xi):=\phi(\xi)\widehat{f}(\xi)$. –  Gian Maria Dall'Ara Jan 31 '13 at 9:45
    
@Matt Jacobs As said in the previous comment, $f(D)u$ is the function whose Fourier transform is $f(\xi)\hat u(\xi)$. The operator $f(D)$ is called a Fourier multiplier for this reason. An integral representation is $$ (f(D)u)(x)=\int e^{2i\pi x\cdot \xi} f(\xi) \hat u(\xi) d\xi, $$ with $$ (\hat u)(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx. $$ –  Bazin Jan 31 '13 at 9:58
    
Thanks! Is there a text or paper where I can find this result? –  Matt Jacobs Feb 1 '13 at 0:30
    
@Matt Jacobs I would recommend the Bahouri-Chemin-Danchin book Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 343. Springer, Heidelberg, 2011. –  Bazin Feb 1 '13 at 16:27
add comment

I suspect, as Daniel Spector said, that to find a characterization of Holder continuity in terms of Fourier coefficients is very hard. A nice necessary condition is a theorem by Bernstein asserting that if $f$ is $\alpha$-Holder for $\alpha>\frac{1}{2}$ then its Fourier coefficients are in $\ell^1(\mathbb{Z})$ (see Katznelson "An introduction to harmonic analysis" for a proof).

share|improve this answer
    
Yes I used Bernstein's theorem to construct my counterexample above. Do you know of any functions that both satisfy $f\in \mathcal{l}^1(\ZZ)$ and $|\hat{f}(n)|\leq C_f|n|^{-\alpha}$ for some $\alpha>1/2$ but $f$ is not Holder continuous of order $\alpha$. –  Matt Jacobs Jan 30 '13 at 21:46
add comment

I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An Introduction) which is one place to understand the subtleties of this issue.

One thing I will mention is that the Sobolev embedding theorem implies sufficient conditions for Holder continuity. If, for example, $n^2 |\hat{f}(n)|^2$ is summable ($f \in H^1$), then $f$ is $C^{0,\alpha}$ for $\alpha<\frac{1}{2}$. More generally, you can find conditions based on the following idea:

$|f|_\alpha \leq \sum_n |\hat{f}(n)| |n|^\alpha = \sum_n |n|^{\alpha+\frac{1}{2}+\epsilon}|\hat{f}(n)| \frac{1}{|n|^{\frac{1}{2}+\epsilon}} \leq \sum_n |n|^{2\alpha+1+2\epsilon}|\hat{f}(n)|^2 \sum_n \frac{1}{|n|^{1+2\epsilon}}$

Therefore if $|n|^{2\alpha+1}|\hat{f}(n)|^2$ is summable then $f$ is Holder continuous of any order strictly less than $\alpha$.

share|improve this answer
    
That is a nice sufficient condition. I own Stein and Shakarchi, as far as I am aware they never tackled such complicated questions, have I skimmed over some section? –  Matt Jacobs Jan 30 '13 at 21:50
    
Not that you have skimmed over a section - only that the development of the representation of functions as Fourier series was a hard problem that occupied the best minds for a long time (Euler was skeptical, for instance). They do not mention the Holder case, as far as I remember, though the discussion of what can be said about necessary and sufficient conditions for continuity should be analogous. –  Daniel Spector Jan 31 '13 at 12:02
add comment

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.