# A Fourier transform of a continuous $L^1$ function

I know that for every function $f\in L^1(R)$ its Fourier transform has the following properties: it is continuous, it's bounded, and we have the limit $$\lim_{\omega\to\pm\infty} \hat f (\omega) = 0$$ which is the Riemann-Lebesgue lemma. Now suppose the function $f$ also has these properties (and it is in $L_1$). Will its Fourier transform be also in L1? That is, if $$f\in L^1(R), \mbox{ f is continuous and } \lim_{x\to\pm\infty} f (x) = 0$$ does it mean that $$\hat f(\omega)\in L^1(R)$$ or not? My first intuition is no, but I don't have a counter example.

Any ideas?

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I actually asked a similar question recently. The second answer here math.stackexchange.com/questions/66173/… provides the counterexample you're looking for. The function is $|x|^{-1/2} e^{-|x|}$ is continuous and integrable, but its Fourier transform is neither. –  user15464 Sep 27 '11 at 13:03
the function you gave as a counter example is not continues at x=0. –  Max Sep 27 '11 at 13:59

Let $f(x)=\frac{-1}{\log(x)+\log(1-x)}$ for $x\in(0,1)$ and $0$ for $x\not\in(0,1)$

Contour integration yields \begin{align} \int_0^1\frac{-1}{\log(x)+\log(1-x)}e^{-2\pi ix\xi}\;\mathrm{d}x &=\int_0^\infty\frac{-1}{\log(-is)+\log(1+is)}e^{-2\pi s\xi}(-i)\mathrm{d}s\\ &-\int_0^\infty\frac{-1}{\log(1-is)+\log(is)}e^{-2\pi i\xi-2\pi s\xi}(-i)\mathrm{d}s\\ &=\frac{1}{\xi}\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}e^{-2\pi s}(-i)\mathrm{d}s\\ &-\frac{e^{-2\pi i\xi}}{\xi}\int_0^\infty\frac{-1}{\log(1-is/\xi)+\log(is/\xi)}e^{-2\pi s}(-i)\mathrm{d}s\\ &\sim \frac{1-e^{-2\pi i\xi}}{2\pi i\xi\log|\xi|}\\ &=e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|} \end{align} as $|\xi|\to\infty$. $f$ is continuous and supported in $[0,1]$, so $f\in L^1$. $|\hat{f}(\xi)|\sim\left|\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|}\right|$, so $\hat{f}\not\in L^1$.

Addition: If we replace $f$ by $f^\alpha$, the same proof yields $\widehat{f^\alpha}(\xi)\sim e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi(\log|\xi|)^\alpha}$ for $\alpha\ge0$. When $\alpha=0$, $f^\alpha$ is not continuous. Otherwise, $f^\alpha$ is a counterexample for $\alpha\in(0,1]$.

Asymptotics: First a couple of estimates. Note that $$\log(-it)+\log(1+it)=\frac{1}{2}\log(t^2+t^4)+i(\tan^{-1}(t)-\frac{\pi}{2}\operatorname{sgn}(t))$$ Therefore, \begin{align} |\log(-it)+\log(1+it)| &\ge\left\{\begin{array}{} \left|\tan^{-1}(t)-\frac{\pi}{2}\operatorname{sgn}(t)\right|\ge\frac{\pi}{4}&\text{when }|t|<1\\ \left|\frac{1}{2}\log(t^2+t^4)\right|\ge\frac{1}{2}\log(2)&\text{when }|t|\ge1 \end{array}\right.\\ &\ge\frac{1}{2}\log(2)\tag{1} \end{align} for all $t$. The actual minimum is greater than $\frac{53}{62}$, but $\frac{1}{2}\log(2)$ is sufficient.

Furthermore, for $|t|<1$, we have \begin{align} |\log(-it)+\log(1+it)| &\ge|\log(|t|)|-\frac{1}{2}\log(1+t^2)\\ &\ge\log(1/|t|)-\frac{1}{2}\log(2)\\ &\ge\frac{1}{2}\log(1/|t|)\tag{2} \end{align} for $|t|<\frac{1}{2}$.

Since $f(x)$ is the translate of an even function, $\hat{f}(\xi)$ is an even function times $e^{i\alpha\xi}$ for some $\alpha$. Therefore, an estimate for $\xi>0$ also works for $\xi<0$. To make notation neater, we will assume $\xi>0$.

Our goal is to estimate $$\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s\tag{3}$$ asymptotically as $\xi\to\infty$. To this end, we will assume $\xi>1$ and split the domain of integration in $(3)$ into three parts: $I_1=[0,\xi^{-\epsilon})$, $I_2=[\xi^{-\epsilon},\xi^{\epsilon})$, and $I_3=[\xi^{\epsilon},\infty)$, where $\epsilon$ is chosen so that $\xi^{-\epsilon}=\epsilon$; that is, $\epsilon=\operatorname{W}(\log(\xi))/\log(\xi)\to0$ as $\xi\to\infty$. In fact, $\epsilon<\log(\log(\xi))/\log(\xi)$ for $\xi>e^e$.

$\underline{I_1=[0,\xi^{-\epsilon})}$: Since $|I_1|=\epsilon$ and $s<1$, estimate $(2)$ insures that for $\xi\ge2$, $$\int_{I_1}\left|\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\right|\;e^{-2\pi s}\;\mathrm{d}s\le\frac{2\epsilon}{\log(\xi)}\tag{4}$$

$\underline{I_2=[\xi^{-\epsilon},\xi^\epsilon)}$: Let us break up the integral over $I_2$ into a few pieces. \begin{align} &\int_{I_2}\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s\\ &=\int_{I_2}\frac{1}{\log(\xi)\left(1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}\right)}\;e^{-2\pi s}\;\mathrm{d}s\\ &=\int_{I_2}\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s+\int_{I_2}\left(\frac{1}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}-1\right)\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s\\ &=\frac{1}{2\pi\log(\xi)} - \int_{I_1+I_3}\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s + \int_{I_2}\left(\frac{\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}\right)\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s\tag{5} \end{align} Because $|I_1|=\epsilon$, we get $$\int_{I_1}\left|\frac{e^{-2\pi s}}{\log(\xi)}\right|\mathrm{d}s\le\frac{\epsilon}{\log(\xi)}\tag{6}$$ Because $I_3=[\xi^\epsilon,\infty)$, we get \begin{align} \int_{I_3}\left|\frac{e^{-2\pi s}}{\log(\xi)}\right|\mathrm{d}s &=\frac{e^{-2\pi\xi^\epsilon}}{2\pi\log(\xi)}\\ &=\frac{\epsilon e^{-2\pi/\epsilon+\log(1/\epsilon)}}{2\pi\log(\xi)}\\ &\le\frac{\epsilon}{2\pi\log(\xi)}\tag{7} \end{align} For $s\in I_2$, $\left|\frac{\log(s)}{\log(\xi)}\right|\le\epsilon$ and $s/\xi\le1$. Furthermore, if $\xi>27$, then $\epsilon<\frac{1}{3}$; therefore, $\log(\xi)=\log(1/\epsilon)/\epsilon\ge\log(3)/\epsilon$. \begin{align} \left|\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}\right| &\le\left|\frac{\log(s)}{\log(\xi)}\right|+\left|\frac{\log(-i)+\log(1+is/\xi)}{\log(\xi)}\right|\\ &\le\epsilon+\left(\frac{\pi}{2}+\frac{1}{2}\log(2)\right)\frac{\epsilon}{\log(3)}\\ &\le\frac{11}{4}\epsilon\tag{8} \end{align} Applying $(8)$ to the integral over $I_2$ in $(5)$, we get \begin{align} \int_{I_2}\left|\frac{\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}\right|\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s &\le\frac{\frac{11}{4}\epsilon}{1-\frac{11}{4}\epsilon}\frac{1}{2\pi\log(\xi)}\\ &\le\frac{33\epsilon}{2\pi\log(\xi)}\tag{9} \end{align} $\underline{I_3=[\xi^\epsilon,\infty)}$: Using estimate $(1)$, we get \begin{align} \int_{I_3}\left|\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\right|\;e^{-2\pi s}\;\mathrm{d}s &\le\frac{2}{\log(2)}\frac{e^{-2\pi\xi^\epsilon}}{2\pi}\\ &=\frac{2}{\log(2)}\frac{\epsilon e^{-2\pi/\epsilon+\log(1/\epsilon)+\log(\log(\xi))}}{2\pi\log(\xi)}\\ &=\frac{2}{\log(2)}\frac{\epsilon e^{-2\pi/\epsilon+2\log(1/\epsilon)+\log(\log(1/\epsilon))}}{2\pi\log(\xi)}\\ &\le\frac{\epsilon}{\pi\log(2)\log(\xi)}\tag{10} \end{align} Thus, combining equation $(5)$ with estimates $(4)$, $(6)$, $(7)$, $(9)$, $(10)$, we get that for $\xi>27$, \begin{align} &\left|\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s - \frac{1}{2\pi\log(\xi)}\right|\\ &\le(2+1+\frac{1}{2\pi}+\frac{33}{2\pi}+\frac{1}{\pi\log(2)})\frac{\epsilon}{\log(\xi)}\\ &\le\frac{6\epsilon}{\log(\xi)}\tag{11} \end{align} Using estimate $(11)$ with its counterpart for $-\xi$, we get that $$\left|\int_0^1\frac{-1}{\log(x)+\log(1-x)}e^{-2\pi ix\xi}\;\mathrm{d}x-e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|}\right|\le\frac{12\epsilon}{|\xi|\log|\xi|}$$

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Can you please show how did you get to the approximation of these integrals? And another thing, are you sure the function you got isn't in L1? (it has a sin in the numerator which can make troubles) –  Max Sep 27 '11 at 21:20
@Max: sorry it took so long, but I've added a rather lengthy justification for the asymptotic approximation. –  robjohn Sep 29 '11 at 21:40
@Max: $\|f\|_{L^1}=\int|f(x)|\;\mathrm{d}x$ so cancellation doesn't help. –  robjohn Sep 29 '11 at 21:52
+1 just for the effort –  user13838 Sep 29 '11 at 21:52
robjohn, thanks for the detailed explanation. I think i yet dont see it. The function $$\frac{\epsilon(\xi)}{|\xi | log | \xi |}$$ - I'm not sure about the epsilon in the numerator, but I think its not in L1, therefore it cannot help you to check if $$\hat f$$ is integrable on R. –  Max Oct 1 '11 at 23:28

Let $a\in(0,1)$. The function $$f(x)=(1-\log(1-x^2))^{-a}\text{ if }|x|<1, 0\text{ otherwise}$$ is continuous with compact support and $\ \hat f\$ decays at infinity like $|\xi|^{-1}(\log|\xi|)^{-a}$; in particular is not integrable. You can see an outline of the proof in T. Tao's answer to my question in MO. I have not checked the details. The reason why this example works is the singularity of $f'$ at $x=\pm1$.

For functions without compact support I would try something like $$\frac{1/\log|x|}{1+x^2}\quad\text{or}\quad \frac{\sin x^3}{1+x^2}.$$

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Do you mean it decays like $1/(|\xi|(\log|\xi|)^\alpha)$? –  robjohn Sep 27 '11 at 16:08
@robjon. Yes. I will edit my answer. –  Julián Aguirre Sep 27 '11 at 16:51
@JuliánAguirre why $\xi$ but not $z$ for example ? :) –  user13838 Sep 27 '11 at 17:50
I believe that my answer, which has the same sort of singularities at $0$ and $1$ that yours does at $-1$ and $1$, indicates that your example will work for $\alpha=1$, as well. –  robjohn Sep 27 '11 at 18:17
@percusse Almost every book I used in graduate school (Stein, Stein&Weiss, Katznelson, Rudin,...), and almost every paper I read, used $\xi$ as the "Fourier" variable. Had I been in physics, I would have probably used $\omega$. But $z$? Never. $z$ is a complex variable, and sometimes the third component of a point in 3-dimensional space. –  Julián Aguirre Sep 27 '11 at 19:48