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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 $\alpha>1/2$.

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    $\begingroup$ 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/… ) ... $\endgroup$
    – Terry Tao
    Commented Jan 31, 2013 at 17:34
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    $\begingroup$ ... 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. ) $\endgroup$
    – Terry Tao
    Commented Jan 31, 2013 at 17:36
  • $\begingroup$ Thank you for the great diagram and explanation. Is there a reason that this problem becomes easy again for $C^{\infty}$ functions? $\endgroup$
    – Matt Jacobs
    Commented Feb 1, 2013 at 0:31

3 Answers 3

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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 $2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.

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  • $\begingroup$ I'm not sure I understand the notation $\phi_{\nu}(D_x)u$ $\endgroup$
    – Matt Jacobs
    Commented Jan 30, 2013 at 21:52
  • $\begingroup$ $D_x$ 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 $\mathbb{R}$, $\widehat{\phi(D_x)f}(\xi):=\phi(\xi)\widehat{f}(\xi)$. $\endgroup$
    – Gian Maria Dall'Ara
    Commented Jan 31, 2013 at 9:45
  • $\begingroup$ @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 $$ (\hat u)(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx. $$ $\endgroup$
    – Bazin
    Commented Jan 31, 2013 at 9:58
  • $\begingroup$ Thanks! Is there a text or paper where I can find this result? $\endgroup$
    – Matt Jacobs
    Commented Feb 1, 2013 at 0:30
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    $\begingroup$ @Matt Jacobs I would recommend the Bahouri-Chemin-Danchin book Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 343. Springer, Heidelberg, 2011. $\endgroup$
    – Bazin
    Commented Feb 1, 2013 at 16:27
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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).

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  • $\begingroup$ 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 $\alpha$. $\endgroup$
    – Matt Jacobs
    Commented Jan 30, 2013 at 21:46
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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 $\alpha$.

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  • $\begingroup$ 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? $\endgroup$
    – Matt Jacobs
    Commented Jan 30, 2013 at 21:50
  • $\begingroup$ 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. $\endgroup$
    – Daniel Spector
    Commented Jan 31, 2013 at 12:02
  • $\begingroup$ How did you get the first inequality? $\endgroup$
    – Jingeon An-Lacroix
    Commented Mar 29, 2020 at 17:59

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