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Suppose that $f$ is continuous on $[0,1]$. Thus, $f\in L^1(\mathbb{R})$ and its Fourier transform exists, as $$ \hat{f}(\xi) := \int_\mathbb{R} e^{-2\pi i x \xi} f(x)dx, $$ which can also be written as $$ \hat{f}(\xi) = -\int_\mathbb{R} e^{-2\pi i \xi (x-\frac{1}{2\xi})} f(x)dx = \int_\mathbb{R} e^{-2\pi i x \xi} (f(x)-f(x+\frac{1}{2\xi}))dx.$$ Therefore, if $f$ satisfies a Hölder condition $f(x)-f(x+h)=O(h^a)$ for small $h$, where $O$ is independent to $x$, it holds $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$.

My question is whether we may obtain $f$ satisfies some Hölder condition if we know that $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$. If not, what other condition should be restricted to $f$ or $\hat{f}$ to get the result?

Jukka Kohonen
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Watheophy
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1 Answers1

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Even more generally, on $\mathbb{R}^d$, for $s>d/2$, we have a continuous embedding $ \dot{H}^s(\mathbb{R}) \rightarrow C^{k,\alpha}$, where the left hand side is the homogeneous Sobolev space of degree $s$, and $k=[s-d/2]$ and $\alpha$ is the fractional part of $s-d/2$. These orders are exactly compatible with the Sobolev embedding theorem, specified to the $C^k$ function space cases.

And clearly, being in the (homogeneous) Sobolev spaces is a decay condition on the Fourier side.

For more details, see page 37, Theorem 1.50 of the textbook "Fourier analysis and nonlinear partial differential equations" by H. Bahouri, J.-Y. Chemin, R. Danchin.

Alex M.
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