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In a recent question, we are asked to prove that a function $f\in L^1(\mathbb R)\cap C^1(\mathbb R)$ such that $f'\in L^1(\mathbb R)$ satisfies $$\tag{1} \lvert \hat{f}(\xi)\rvert\le \frac{C}{\sqrt{1+\xi^2}}, $$ for some constant $C>0$, where the Fourier transform is defined as $$ \hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-ix\xi}\, dx.$$

Question. Can the condition $f'\in L^1(\mathbb R)$ be dropped?

I expect the answer to be negative. More precisely, I conjecture the following.

Conjecture. There exists $f\in L^1(\mathbb R)\cap C^1(\mathbb R)$ such that $\xi\hat{f}(\xi)$ is unbounded as $\xi\to \infty$. Such a function must satisfy $f'\notin L^1(\mathbb R)$.

The function $$ f(x):=\frac{e^{ix^2}}{1+x^2}$$ is a candidate for this conjecture, because $f'\notin L^1(\mathbb R)$. To establish whether $\xi\hat{f}(\xi)$ is unbounded, we would need to perform an asymptotic analysis, as $\xi\to \infty$, of the oscillatory integral $$\tag{2} \hat{f}(2\xi)=\int_{-\infty}^\infty \frac{e^{i(x^2-2x\xi)}}{1+x^2}\, dx.$$ (The factor of $2$ in front of $\xi$ is just a cosmetic, that suggests to complete the square).

I don't know how to carry out this asymptotic analysis. Typically, this would be done via the principle of stationary phase; see Stein, Harmonic Analysis, Proposition 3 pag.334. However, here the phase term $x^2-2x\xi$ is not of the form $\xi\Phi(x)$, for some function $\Phi$.


Remark.

This question seems to be strongly related. It treats functions with a discontinuity in the derivative, whereas here we have a continuous, but slowly decaying, derivative. However, the following observation reduces the case of the present question to the case of the linked one. Using that, up to irrelevant constants, $$ \mathcal{F}_{x\to \xi}[e^{-ix^2}]=e^{-i\xi^2},\qquad \mathcal{F}_{x\to \xi}\left[\frac{1}{1+x^2}\right]=e^{-\lvert \xi\rvert}, $$ we can rewrite (2) as the convolution integral $$ \int_{-\infty}^\infty e^{i(\xi-\eta)^2-\lvert\eta\rvert}\, d\eta , $$ which equals $$ e^{i\xi^2}\mathcal{F}_{\eta\to \xi}\left[ e^{i\eta^2-\lvert\eta\rvert}\right](\xi).$$ We are thus in the position of asymptotically estimate the Fourier transform of a function with a single discontinuity in its first derivative. This is exactly what is done in the linked question.

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  • $\begingroup$ Do you want to have $\xi \widehat{f}(\xi)$ unbounded or diverging to $\infty$? $\endgroup$ – sometempname Mar 29 at 15:13
  • $\begingroup$ @sometempname: Unbounded would be enough. $\endgroup$ – Giuseppe Negro Mar 29 at 16:49
  • $\begingroup$ Then I think you can construct something like $$\frac{\sin^4(x)}{x^2} \cdot \sum_{k=1}^\infty \frac{\exp(2 i k x)}{\sqrt{k}} = \frac{\sin^4(x)}{x^2} \cdot \mathrm{Li}_\frac12 ( \exp(2 i x) ), $$ where the $\sin$ makes it $C^1$, and on the Fourier side it is a sequence of triangles (positive and negative) decaying like $1/\sqrt{\xi}$. $\endgroup$ – sometempname Mar 29 at 19:39
  • $\begingroup$ Btw, this specific construction doesn't actually work, but maybe it can be modified $\endgroup$ – sometempname Mar 29 at 21:04
  • $\begingroup$ @sometempname: That would be great. $\endgroup$ – Giuseppe Negro Mar 29 at 21:42
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In general you probably cannot say anything better than $\widehat{f}(\xi) \to 0$ as in the Riemann-Lebesgue lemma (even if $f$ is smooth), by density considerations. There's probably a much simpler construction, but here is a concrete example.

Let $\phi$ a positive smooth bump function supported on $\left[-\frac12,\frac12\right]$, so that $\int_{\mathbb{R}}\phi \, dx=1$, and so that $\widehat{\phi}$ is also non-negative (notice $\widehat{\phi}$ is actually a Schwartz function).

Also notice that $$\int_{\mathbb{R}}\left|\phi^{\prime}\left(x\right)\right|dx\ge\int_{-1/2}^{0}\phi^{\prime}\left(x\right)dx-\int_{0}^{1/2}\phi^{\prime}\left(x\right)dx=2\phi\left(0\right)>0.$$

Put $$\phi_{j}\left(x\right)=2^{j}\phi\left(2^{j}\left(x+j\right)\right),\quad j\in\mathbb{\mathbb{N}},$$ so that $$\int_{\mathbb{R}}\phi_{j}\left(x\right)dx =1,$$ $\phi_{j}$,$\phi_{j}^{\prime}$ have disjoint support, and $$\int_{\mathbb{R}}\left|\phi_{j}^{\prime}\left(x\right)\right|dx\ge2^{j+1}\phi\left(0\right).$$ Let $\left\{ a_{j}\right\}$ be a sequence of numbers, and take $$f\left(x\right)=\sum_{j=1}^{\infty}a_{j}\phi_{j}\left(x\right),$$ so that $f\in L^{1}$ iff $\left\{ a_{j}\right\} \in L^{1}$, also notice that $$f^{\prime}\left(x\right)=\sum_{j=1}^{\infty}a_{j}\phi_{j}^{\prime}\left(x\right),$$ is not in $L^{1}$, if for example $a_{j}=\frac{1}{j^{2}}$. Then $$\widehat{\phi_{j}}\left(\xi\right)=2^{j}\cdot2^{-j}\widehat{\phi}\left(\frac{\xi}{2^{j}}\right)\exp\left(-2\pi ij\xi\right)=\widehat{\phi}\left(\frac{\xi}{2^{j}}\right)\exp\left(-2\pi ij\xi\right),$$ and $$\widehat{f}\left(\xi\right)=\sum_{j=1}^{\infty}a_{j}\exp\left(-2\pi ij\xi\right)\widehat{\phi}\left(\frac{\xi}{2^{j}}\right).$$

Notice $\widehat{\phi}\left(0\right)=\int\phi \,dx=1$, and we can assume that $\widehat{\phi}\left(\xi\right)\ge\frac12$ on some interval $\left[-b,b\right]$, with $b>0$. Then, for $\xi\in\mathbb{N}$ if we put $$k\left(\xi\right)=\left\lceil \log_{2}\left(\frac{\xi}{b}\right)\right\rceil,$$ we see that $$\widehat{f}\left(\xi\right)=\sum_{j=1}^{\infty}a_{j}\widehat{\phi}\left(\frac{\xi}{2^{j}}\right)\ge\frac12\sum_{j\ge k\left(\xi\right)}a_{j},$$ with the choice $a_{j}=\frac{1}{j^{2}}$ we get $$\widehat{f}\left(\xi\right)\ge\frac{c}{k\left(\xi\right)}\ge\frac{c^{\prime}}{\log\left(\xi\right)},\qquad\xi\in\mathbb{N},\,\xi\ge2,$$ for some constants $c, c^\prime>0$.

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  • $\begingroup$ Brilliant. Thank you. It may be that $O(\frac{1}{\log \xi})$ is a lower bound for the decay rate of any $L^1$ function. On this, there is an exercise on Grafakos "Classical Fourier analysis", it is 2.2.9 (c), pag.117 of the 3rd edition. I still have to completely understand this, though. $\endgroup$ – Giuseppe Negro Mar 30 at 21:48

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