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Let $f \in C^\infty(\mathbf{R})$ be an even, smooth function such that for $|x| \geq 2$, $f(x) = 1/\text{log}|x|$. I am trying to show that $\widehat{f}$ is integrable. This must use the fact that the function is even, because odd functions agreeing with $1/\text{log}(x)$ for $x \geq 1$ do not have an integrable Fourier transform.

To try and understand this problem, we consider a decomposition in space and frequency. Define $f_j(x) = \chi_j(x) f(x)$, where $\chi_j(x) = \chi(x/2^j)$, and $\chi$ is a smooth cutoff supported on $|x| \sim 1$. We then need to understand the quantities $\| \chi_i \widehat{f_j} \|_{L^1}$ for $j \geq 0$, and $i \in \mathbf{Z}$.

Simply by calculating the derivative of $f$, and then applying Parseval's identity, I was able to show that $\| \chi_i \widehat{f_j} \|_{L^1} \lesssim 2^{-(i+j)/2} / j^2$. This gives good bounds for $i + j \geq 0$. My problem is obtaining bounds on the quantities $\| \chi_i \widehat{f_j} \|_{L^1}$ when $i + j \leq 0$. Since we haven't used the fact that $f$ is even, this must be where this property becomes useful. But I can't quite see how to do this.

Intuitively, to get good bounds we need to understand the values of $\widehat{f_j}(\xi)$ where $|\xi| \leq 1/2^j$. The fact that we have to exploit the fact that $f$ is even suggests using the identity

$$ \widehat{f_j}(\xi) = \int \frac{\chi(x/2^j)}{\log |x|} e^{2 \pi i \xi \cdot x} dx = 2 \int_0^\infty \frac{\chi(x/2^j)}{\log |x|} \cos(\xi x) dx. $$

Taking in absolute values gives $|\widehat{f_j}(\xi)| \lesssim 2^j / j$, and thus $\| \chi_i \widehat{f_j} \|_{L^1} \lesssim 2^{i+j} / j$, but this is not strong enough to sum over all $i + j \leq 0$. Nothing is really oscillating in the integral, so it does not seem like integration by parts will help us. What can we do?

Jacob Denson
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  • Thank you for fixing this. Are you assuming that $f$ is even or are you saying that $f$ is even up to a $C_0^\infty$ function? – Hans Engler Nov 08 '21 at 22:39
  • We can assume $f$ is even without loss of generality, since the Fourier transform of a function in $C_0^\infty$ is definitely integrable. – Jacob Denson Nov 08 '21 at 22:41
  • So I am not sure if my idea works but hear me out: You only need to show what you want to show for a single function, namely $f(x)=\frac1{\ln(\lvert x\rvert)} \cdot [x\not\in]-2,2[]$ (using Iverson brackets). For this $f$ I get $$\hat f(\xi) = \int_2^\infty \frac{\cos(2\pi x\xi)}{\ln(x)},\mathrm dx.$$ Now integrate by parts a bunch of times.... – Maximilian Janisch Nov 08 '21 at 23:29
  • This idea works when $\xi$ is large (you'll get factors of $1/\xi$ when you integrate by parts), but you run into problems when $\xi$ is small. This is analogous to the problem above - we have a good bound when $i + j \geq 0$ (when $\xi$ is large), but not a good bound when $i + j \leq 0$ (when $\xi$ is small). Maybe the integration by parts can be resolved when $xi$ is small, but I can't quite see how to do it. – Jacob Denson Nov 08 '21 at 23:36
  • Argh this is annoying: I wanted to use Fubini to show that $\hat f$ is integrable over $\xi\in[-1,1]$, since it is not hard to show that $$\int_2^\infty\int_{-1}^1 \frac{\cos(2\pi x\xi)}{\ln(x)},\mathrm d\xi,\mathrm dx$$ is finite. But then I notice that Fubini's Theorem need not apply to the function in question here. Maybe you can save my approach with your cutoff, on my side it is 1am so I'll go to bed and stop working on this, sorry. – Maximilian Janisch Nov 08 '21 at 23:54
  • Yes I think this works! I'll give a try writing this up. – Jacob Denson Nov 09 '21 at 00:06
  • On the other hand, it seems to show that the Fourier transform should not be integrable. Maybe I made a mistake in the calculation. – Jacob Denson Nov 09 '21 at 00:32

1 Answers1

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Credit to Maximilian Janisch for suggesting this approach.

Based on the approach suggested in this problem, it suffices to show the function

$$ g(x) = \sum_{j = 0}^\infty \sum_{i = -\infty}^{-j-10} \chi_i \widehat{f_j} $$

is integrable. Using the fact that $f_j$ is even, we find that

$$ \widehat{f_j}(\xi) = 2 \int_2^\infty \cos(2 \pi \xi x) \frac{\chi_j(x)}{\log(x)}\; dx. $$

All quantities here are non-negative for $|\xi| \leq 1/2^{-j-10}$ since then $|\xi x| \leq 2^{-10}$. Thus we can apply Tonelli's theorem to conclude that

$$ \| g \|_{L^1} \lesssim \sum_{i = -\infty}^{0} \sum_{j = i + 10}^\infty \int_2^\infty \int \chi_i(\xi) \cos(2 \pi \xi x) \frac{\chi_j(x)}{\log(x)}\; dx\; d\xi \lesssim \int_{-2^{-10}}^{2^{-10}} \int_2^{2^{-10}/|\xi|} \frac{\cos(2 \pi \xi x)}{\log(x)}\; dx\; d\xi. $$

But this integral can also be written as

$$ \int_2^\infty \frac{1}{\log(x)} \int_{-2^{-10}/x}^{2^{-10}/x} \cos(2 \pi \xi x)\; d\xi\; dx = \int_2^\infty \frac{\sin(2 \pi \cdot 2^{-10}))}{\pi x \log(x)} $$

But this seems to show that $g$ is not integrable, since these inequalities are essentially sharp, and $1/x log(x)$ is a divergent integral.

Jacob Denson
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