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I have a function which is the Fourier transform of the characteristic function of the unit ball in $\mathbb{R}^3$

$$\hat{f}(\mathbf{k})=\int_B e^{-2\pi i\left<\mathbf{x},\mathbf{k}\right>}d\mathbf{x}=\frac{\sin(2\pi||\mathbf{k}||)-2\pi||\mathbf{k}||\cos(2\pi||\mathbf{k}||)}{2\pi^2||\mathbf{k}||^3}$$

I have to show that $\frac{\hat{f}(\mathbf{k})}{||\mathbf{k}||^2}$ is Lebesgue integrable. Now, I'm gonna be plain honest: I have no idea where to start. I have an integrability criterion, namely:

$f$ is integrable if and only if for every $\varepsilon>0$ there exists $g$ continuous with compact support so that $$\int_{\mathbb{R^n}}|f-g|d\mathbf{x}<\varepsilon$$

Which I have no idea how to apply in practice, and I know I can use the dominated and monotone convergence theorems, in particular, I know that $\frac{\hat{f}(\mathbf{k})}{||\mathbf{k}||^2}\leq\frac{\hat{f}(\mathbf{0})}{||\mathbf{k}||^2}$, which makes me go towards the dominated convergence theorem, the trouble is that I have to show that there exists some sequence of integrable function that converges almost everywhere towards $\frac{\hat{f}(\mathbf{k})}{||\mathbf{k}||^2}$, which I wasn't able to do. Any help will be appreciated, thank you

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  • $\begingroup$ Unless I am mistaken, Hint: Go for spherical coordinates (Motivation: the given Fourier transform as well as the norm are rotation invariant). $\endgroup$ Commented Jun 14, 2017 at 12:11
  • $\begingroup$ Go for spherical coordinates? I don't understand, I don't have to calculate the integral, I have to show it exists without calculating it. $\endgroup$
    – user438666
    Commented Jun 14, 2017 at 12:13
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    $\begingroup$ Prove that the "spherical version" of the integral converges. If you want to be rigorous, integrate (using spherical coordinates) the absolute value of your function on the set $B(0,n)-B(0,1/n)$ (- means the set difference here) and then apply the monotone convergence theorem as $n \to \infty$. $\endgroup$ Commented Jun 14, 2017 at 12:19
  • $\begingroup$ well the function is already in spherical coordinates, as it is a function only of the modulus of its argument, what you mean is to take the Fourier transform of the characteristic function of $B(0,n)-B(0,1/n)$, to obtain a sequence that converges to $\hat{f}$, right? $\endgroup$
    – user438666
    Commented Jun 14, 2017 at 12:25
  • $\begingroup$ but how do I know that the functions in the sequence are integrable? They must be to apply the convergence theorem $\endgroup$
    – user438666
    Commented Jun 14, 2017 at 12:30

1 Answer 1

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Let us first consider the function $g_n(k)=\left|\frac{\hat{f}(k)}{||k||^2} \right| \cdot \mathbb 1_{A_n}(k)$ where $A_n= B(0,n) \cap B(0,1/n)^c$.

Note that $g_n$ is non-negative, integrable (why?), increasing with respect to $n \in \mathbb N$, and that, almost surely, $g_n \to_n \left|\frac{\hat{f}(.)}{||.||^2} \right|$. Hence $$\int g_n(k)dk \to \int{ \left|\frac{\hat{f}(k)}{||k||^2}\right|dk}$$ by the monotone convergence theorem.

Now note that $g_n$ is rotation invariant (why?), hence for any $t>0$ it satisfies $\forall x_0 \in \partial B(0,t)$ we have $\frac{1}{|\partial B(0,t)|}\int_{\partial B(0,t)}g_n(x)dx = g(x_0)$. (Reminder: $\partial B(0,t)$ is the sphere of radius $t$ and $S^2= \partial B(0,1)$ ).

In what will follow, I will choose the point $(1,0,0)=x_0$, but any other point on $S^2$ can do just as well.

Let us now integrate on the spherical level sets: $$\int g_n(k)dk=\int_{0}^{\infty}\int_{S^2}r^2g_n(rw)d\sigma_{S^2}(w)dr= \frac{1}{|S^2|}\cdot\int_{0}^{\infty} r^2 g_n(r \cdot (1,0,0))dr= \\ \frac{1}{|S^2|}\cdot\int_{0}^{\infty} r^2 g_n(r \cdot (1,0,0))dr.$$

We will now split the integral: $$\int_0^1 + \int_1^\infty .$$ For the first integral, we use the fact that $r^2 g_n(r \cdot (1,0,0))= r^2 \frac{|\hat{f}(r,0,0)|}{r^2}\mathbb1_{[1/n,n]}=|\hat{f}(r,0,0)| \mathbb1_{[1/n,n]}$ which is bounded by some constant $C$ independent of $n$.

For the second integral, we apply the formula that you provided, namely $r^2g_n(r,0,0)=r^2\mathbb1_{[1/n,n]} \cdot \left| \frac{\sin(2\pi r)-2\pi r\cos(2\pi r)}{2\pi^2r^{2+3}}\right| \le r^2 \mathbb1_{[1/n,n]} \cdot \left( \frac{|\sin(2\pi r)|}{2\pi^2r^{5}}+\frac{|2\pi r\cos(2\pi r)|}{2\pi^2r^{5}}\right) \\ \le \mathbb1_{[1/n,n]} \cdot \left(\frac{c_1}{r^3} + \frac{c_2}{r^2}\right)$.

Can you take it from here and show that $\int g_n(k)dk$ is a bounded increasing sequence, hence that it converges?


Here is an alternative proof using the fact that in $\mathbb R^d$,

  • $||x||^{-r}$ is integrable on $B(0,1)$ iff $r<d$

    and

  • $||x||^{-r}$ is integrable on $B(0,1)^c$ iff $r>d$

In your case, $d=3$. Note that $ \mathbb 1_{B(0,1)}|\frac{\hat{f}(k)}{||k||^2}| \le \mathbb 1_{B(0,1)}|\frac{\hat{f}(0)}{||k||^2}|$ which is integrable.

Also note that $\mathbb 1_{B(0,1)^c}|\frac{\hat{f}(k)}{||k||^2}|= \mathbb 1_{B(0,1)^c} \left| \frac{\sin(2\pi ||k||)-2\pi r\cos(2\pi ||k||)}{2\pi^2||k||^{2+3}}\right| \le \mathbb 1_{B(0,1)^c} \left(\frac{c_1}{||k||^5} + \frac{c_2}{||k||^4}\right)$ which is also integrable.

Now you can conclude using the inequalities.

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  • $\begingroup$ Thank you for this very detailed answer, unfortunately some part of it are over my head. I don't understand what do you mean by "integrating on the spherical level sets", nor the notation $d\sigma_{S^2}(w)$. This was a question in a mock exam, we didn't have any measure theory before Lebesgue integration. $\endgroup$
    – user438666
    Commented Jun 14, 2017 at 14:36
  • $\begingroup$ The Lebesgue integral of a function on a set $X$ represents the average of the function on that set. $\int_{S^2} g(w) d\sigma_{S^2}(w)$ is the integral on the sphere $S^2$ of $g(.)$. You could write it as $d \sigma_{S^2}=\sin \theta d\theta d \phi$. The reasons for passing to spherical coordinates are: 1) for a simple proof of the convergence of the integral of $||x||^{-r}$ (depending on $r$) on $B(0,1)$ and $B(0,1)^c$ , and 2) to provide rigour on the use of the rotation invariance. $\endgroup$ Commented Jun 14, 2017 at 14:44
  • $\begingroup$ If know a proof of point 1) in my previous comment, you can simply split $f=(\mathbb1_{B(0,1)} + \mathbb1_{B(0,1)^c})f$. For $\mathbb1_{B(0,1)}f $ use the domination that you already mentioned using $|\hat{f}(0)|/||k||^2$. Proceed similarly as my proof (use the formula) for $\mathbb1_{B(0,1)^c}f$ and bound the $\sin$ and $\cos$ by $1$. $\endgroup$ Commented Jun 14, 2017 at 14:52
  • $\begingroup$ And for a proof of point one using minimal machinery, see for example [page 91, exercise 10] in Book 3 of the Princeton Lectures in Analysis. $\endgroup$ Commented Jun 14, 2017 at 14:56
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    $\begingroup$ You are forgetting the absolute value around $\hat{f}/||k||^2$. But apart from that, I am not using the dominated convergence theorem but the monotone convergence theorem in order to show that the integral of $|\hat{f}/||k||^2|$ is finite and hence that $\hat{f}(k)/||k||^2$ is integrable. Your observation on why $g_n$ are integrable is also correct, except for the fact that they are not continuous but measurable, bounded (because of the continuity of $\hat{f}/||.||^2$ on $\mathbb{R}^d-\{0\}$) and with compact support. $\endgroup$ Commented Jun 14, 2017 at 15:39

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