6
$\begingroup$

If a function $f$ is Lebesgue integrable and satisfies Lipschitz condition, must its Fourier transform $\hat{f} $ be Lebesgue integrable or not?

There's a theorem about Fourier series, saying the Fourier series of periodic Lipschitz function must be absolutely convergent, and I'm just thinking about if it has a similar analog in Fourier transform.

I have proven that the improper integral $$\displaystyle{\int _{- \infty }^{\infty }\hat {f} (\xi )e^{2 \pi i \xi x} d \xi }$$ is uniformly convergent for $x\in \mathbb{R}$.

My thought is to find a sequence of Lipschitz integrable functions, with $L^1$-norm and Lipschitz norm both $\le 1$, whose Fourier transform have an increasing $L^1$-norm, then by the closed graph theorem, there must exist an integrable Lipschitz function whose Fourier transform is not integrable.

But I find that kind of construction extremely difficult to be accomplished. Anyone has ideas?

$\endgroup$
2
$\begingroup$

This is a really nice question, I was surprised after thinking for a couple of hours and none of the standard examples in a harmonic analysis course fitting the description. In any case, I will try to present what can and what cannot be done to the integrability of the Fourier transform of a Lipschitz function.

Notice also that I will be employing standard analysis notation (such as $\lesssim$ and $O(g)$) as well as some standard theorems in Fourier analysis, such as van der Corput's lemma for bounding oscillatory integrals (also known as 'stationary phase method'). If you are not familiar with those, I would gladly refer you to Stein's "Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals", whose chapter VIII contains all that we shall use here.

  1. First of all, let us notice that there are a couple of things that can be proved with slightly different assumptions, or slightly more restrictive results that can be proved with the same assumptions:

1.1. If instead of $f$ being Lipschitz, one requires the different condition that $\|f(x+h)-f(x)\|_{L^2(dx)} = O(h^{\alpha}),$ for some $\alpha > 1/2,$ then the same conclusion of Bernstein's theorem holds. That is, if the above condition is satisfied and $f \in L^1(\mathbb{R}),$ then $\widehat{f} \in L^1(\mathbb{R}).$ A proof of this fact can be found, for instance, in the book "Introduction to the Theory of Fourier Integrals", by Titchmarsh.

As a by-product, notice that if $f \in L^1(\mathbb{R})$ is Lipschitz and has compact support, then the condition in Titchmarsh's theorem above is trivially satisfied from the Lipschitz condition, and thus $\widehat{f} \in L^1(\mathbb{R}).$ This shows that, if a counterexample exists, it cannot have compact support.

1.2. On the other hand, notice that $f \in L^1, f' \in L^{\infty}$ implies directly that $f \in L^{\infty},$ and thus, by interpolation, $f \in L^p, \forall p \ge 1.$ By the Hausdorff--Young inequality, we have that $\widehat{f} \in L^q, \forall q \ge 2.$ In fact, this can be significantly improved:

Claim: Let $f \in L^1(\mathbb{R})$ be Lipschitz, i.e., $f' \in L^{\infty}.$ Then $\widehat{f} \in L^p(\mathbb{R}), \forall p > 1.$

Proof: In Grafakos's "Modern Fourier Analysis", the following result can be found as a by-product of a characterization of Lipschitz spaces: 'Let $\Delta_j(f) = (\psi(\xi/2^j) \widehat{f})^{\vee},$ where $\psi$ is a smooth bump supported on $[1/2,4]$ and equal to 1 on $[1,2].$ If $f$ is Lipschitz, then

$$\sup_{j \ge 0} 2^j \|\Delta_j(f)\|_{\infty} < + \infty.'$$

We will not prove this here, but by assuming such a result, we have the following: for $\psi$ as in the statement of the Littlewood-Paley characterization of Lipschitz functions, we have

$$\int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \le \int_{\mathbb{R}} |\psi(\xi/2^j) \widehat{f}(\xi)|^p \psi(\xi/2^j) \, d \xi \lesssim 2^{(1-\frac{p}{2})j} \|\Delta_j(f)\|_2^{p},$$

where in the last inequality we used Hölder and Plancherel. Now, by interpolation we have

$$\|\Delta_j(f)\|_2 \le \|\Delta_j(f)\|_{\infty}^{1/2} \|\Delta_j(f)\|_1^{1/2}.$$

From the definition of $\Delta_j,$ we have $\|\Delta_j(f)\|_1 \lesssim \|f\|_1,$ and from the characterization of Lipschitz functions mentioned before, $\|\Delta_j(f)\|_{\infty} \lesssim 2^{-j}.$ Putting all those considerations together, we obtain

$$ \int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \lesssim 2^{(1-p)j},$$

and the right-hand side above is summable in $j \ge 1$ if $p>1,$ as desired. $\square.$

Thus, the Fourier transform of any such counterexample has to integrate to any order $p>1.$

  1. With these intuitions at hand, we can finally provide our counterexample: let $\psi_0(x)$ be a smooth function satisfying $\psi_0 = 1$ if $x > 20,$ and $\psi_0 = 0$ if $x < 10.$ Define then the Fourier transform of our conterexample to be

$$ \widehat{f}(x) = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x).$$

Then we need to prove that

$$f(\xi) = \int_{\mathbb{R}} \frac{e^{i\frac{x^2}{2\log x} - i\xi x}}{x \log x} \psi_0(x) \, dx$$

is Lipschitz and integrable.

2.1. $f$ is Lipschitz: it is not too hard (even though it is a bit technical) to show that $f$ is Lipschitz if and only if the integrals

$$ J_N(\xi) = \int_{10}^N \frac{e^{i\frac{x^2}{2 \log x}}}{\log x} \psi_0(x) e^{-i \xi x} \, dx $$

are uniformly bounded on $\xi, N.$ Define then the phase function $\phi(x,\xi) = \frac{x^2}{2\log x} - \xi x.$ Suppose that, for some $x_{\xi} \in \mathbb{R},$ we have $\partial_x \phi(x_{\xi},\xi) = 0.$ As $\partial_x \phi(x,\xi) = \frac{x}{\log x} + \frac{x}{2 \log^2 x} - \xi,$ then it is also not hard to show that $x_{\xi} \gtrsim |\xi| \log |\xi|.$ We split then the integral defining $J_N(\xi)$ into the one over the interval $[5,N] \setminus [\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^1(\xi),$ and the one over $[\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^2(\xi).$ Thus, $J_N(\xi) = J_N^1(\xi) + J_N^2(\xi).$

For $J_N^1,$ we use that the derivative of the phase satisfies $|\partial_x\phi(x,\xi)| \gtrsim \max\{|\xi|,\frac{x}{\log x}\},$ while for $x$ sufficiently large (10 should do for this) $\partial_{xx}^2 \phi(x,\xi) > 0,$ so we are in position to apply van der Corput's lemma, which then implies that

$$|J_N^1(\xi)| \lesssim \frac{1}{|\xi|}.$$

For the analysis of $J_N^2,$ we notice that the interval $[\xi^{1/2},\xi^{3/2}]$ contains the zero $x_{\xi}$ of $\partial_x \phi(x,\xi),$ and in that interval it also holds that $\partial_{xx}^2 \phi(x,\xi) \gtrsim \frac{1}{\log \xi}.$ Using van der Corput's lemma again, we have that

$$|J_N^2(\xi)| \lesssim \frac{(\log \xi)^{1/2}}{\log \xi} \lesssim \frac{1}{(\log \xi)^{1/2}}.$$

This finishes the case $\xi \gg 1.$ For the small $\xi$ case, the analysis here carries out even easier than what is written here. Thus, $f$ is Lipschitz.

2.2. $f$ is integrable: The idea is basically the same as here, but integrating by parts first. Indeed, we need to prove some uniform estimates on the functions

$$f_N(\xi) := \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0(x) \, dx.$$

Integrating by parts with factors $u = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x)$ and $dv = e^{-i \xi x} dx.$ This gives that $f_N(\xi)$ is equivalently equal to

$$\frac{1}{i\xi}\frac{ e^{i( \frac{N^2}{2\log N} - \xi N)}}{N \log N} + $$ $$ \frac{1}{i \xi} \int_{10}^N \partial_x \left(\frac{x^2}{2\log x}\right) x \log x \frac{e^{i(\frac{x^2}{2\log x} - \xi x)} }{x^2 \log^2 x} \, \psi_0(x) dx + $$ $$ \frac{1}{ i\xi} \int_{10}^N \frac{\log x + 1}{x^2 \log^2 x} e^{i(\frac{x^2}{2 \log x} - \xi x)} \psi_0(x) \, dx + $$ $$ \frac{1}{i\xi} \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0'(x) \, dx =: \frac{1}{i\xi} (I_1(N,\xi) + I_2(N,\xi) + I_3(N,\xi) + I_4(N,\xi)).$$

One notices quite immediately that $I_1(N,\xi) \to 0$ as $N \to \infty,$ so we do not need to worry about that term when taking limits. Also, $\frac{\psi_0'(x)}{x \log x}$ has support on $[10,20]$ and is smooth there, while the derivative of the phase $\partial_x\phi(x,\xi)$ does not vanish for $\xi$ small, and neither does it on the support for $\xi$ sufficiently large. Thus, this term is bounded and decays as fast as one wishes - i.e., $|I_4(N,\xi)| = O_R(\xi^{-R})$ when $\xi \to \infty,$ for any $R > 0.$ Thus $I_4(N,\cdot) \in L^1(\mathbb{R})$ uniformly on $N \to \infty.$

We then analyze the remaining terms $I_2,I_3.$ In fact, the analysis of $I_2$ is much more delicate than that of $I_2,$ so we perform only the former and defer the latter to whoever wishes to fill in the gaps.

Rewrite $I_2(N,\xi)$ as

$$\int_{10}^N \left( \frac{x}{\log x} + \frac{x}{2 \log^2x} \right) \frac{x \log x}{x^2 \log^2 x} e^{i(\frac{x^2}{2\log x} - \xi x)} \psi_0(x) \, dx = $$ $$ = 2 \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{ (\log x)^2} \psi_0(x) dx + \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{(\log x)^3} \psi_0(x) \, dx.$$

These resemble a lot the integrals we had to estimate for proving $f$ is Lipschitz. In fact, the very same argument is applicable in these cases as well, where one obtains (after redoing the same computations with van der Corput and what not)

$$|I_2(N,\xi)| \lesssim \frac{1}{(\log |\xi|)^{3/2}}.$$

As mentioned before, the analysis of $I_3$ is in fact easier, and doing the computations yields

$$|I_3(N,\xi)| \lesssim \frac{1}{|\xi|}.$$

Notice that all such estimates were made uniformly on $N \to \infty.$ Putting all together and taking limits, one obtains that, for $|\xi| \gg 1,$

$$|f(\xi)| \lesssim \frac{1}{|\xi| (\log|\xi|)^{3/2}}.$$

As the right-hand side is integrable and $f$ was already seen to be Lipschitz, this implies $f \in L^1(\mathbb{R}),$ as desired.

$\endgroup$
4
  • $\begingroup$ Thank you very much for your hard work. The counterexample you provided is just what I desired. Also, I seem to have figured out another counterexample, which is the function $\frac{\sin(x^3)}{x^2}$ , being Lipschitz and integrable, seems to have an unintegrable Fourier transform as well. $\endgroup$
    – Antimonius
    Jan 29 at 14:33
  • $\begingroup$ Glad to help! Also, about the example you mention: I tried that one and stationary phase gives that its Fourier transform decays as $\xi^{-5/4}$ at infinity... are you sure it works? $\endgroup$ Jan 29 at 14:40
  • $\begingroup$ Maybe not, I'm not quite sure, I only tried numerical integral. $\endgroup$
    – Antimonius
    Jan 29 at 15:45
  • $\begingroup$ Hello, can you show me some brief references about Bernstein's theorem? $\endgroup$
    – liaoyulei
    Jul 9 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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