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I am having trouble understanding why Fubini's theorem can be used in a certain proof.

We define the Fourier transform of a function $f$ as \begin{equation} \hat{f}(y):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ixy}\ dx. \end{equation}

The theorem states the following.

Theorem. Let $f\in L_1(\mathbb{R})$, let $\hat{f}(y)$ be an odd function. Then

  1. $\hat{f}(y)=\frac{-i}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\sin{xy}\ dx$.
  2. $\forall \ a>0\ \left|\int_{a}^{+\infty}\frac{\hat{f}(y)}{y}\ dy\right|\leq M$, where $M$ depends only of $f$.

Proof.

  1. I understand the proof of the first statement.
  2. Here we apply Fubini's theorem. \begin{equation} \left|\int_{a}^{+\infty}\frac{\hat{f}(y)}{y}\ dy\right|=\left| \int_{a}^{+\infty}\int_{-\infty}^{+\infty}\frac{f(x)\sin{xy}}{\sqrt{2\pi}y}\ dx\ dy \right|=\left| \int_{-\infty}^{+\infty}\int_{a}^{+\infty}\frac{f(x)\sin{xy}}{\sqrt{2\pi}y}\ dy\ dx \right|=\left| \int_{-\infty}^{+\infty}f(x)\int_{ax}^{+\infty}\frac{\sin{u}}{\sqrt{2\pi}u}\ du\ dx \right|\leq C||f||_{L_1(\mathbb{R})}, \end{equation}

since the integral $\int_{-\infty}^{+\infty}\frac{\sin{u}}{u}\ du$ converges.

It is, however, not clear to me why we can use Fubini's theorem here. In order to use it, the function $g(x,y):=\left| \frac{f(x)\sin{xy}}{y} \right|$ has to be integrable in the Lebesgue sense on $\mathbb{R}\times [a,+\infty)$. Which is not the case, since by Fubini's other theorem the integral \begin{equation} \int_{-\infty}^{+\infty}\int_{a}^{+\infty}\left| \frac{f(x)\sin{xy}}{\sqrt{2\pi}y}\right| \ dy\ dx \end{equation} would have to converge. Which is not the case, since the integral \begin{equation} \int_{b}^{+\infty}\left| \frac{\sin{u}}{u}\ \right| du \end{equation} diverges.

What is my mistake here and why can we apply Fubini's theorem in the proof?

ElectroSchOOp
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  • Is this from a book or set of lecture notes (it seems fishy) ? If you have the integrability of the Fourier transform of $f$, say, then the result would follow by regularising the integrand on the LHS. – Pantelis Tassopoulos May 08 '25 at 16:43
  • The proof of this fact is from this post https://math.stackexchange.com/questions/1828593/a-proof-of-the-fact-that-the-fourier-transform-is-not-surjective-from-mathcal – ElectroSchOOp May 08 '25 at 17:08
  • Notice that the upper boundary of the outer integral (in the answer to which you refer to) is taken first to be finite ($R$), where Fubini is justified since the function $\sin(y)/y$ has a continuous (analytic) extension to all of $\mathbb{R}$. – Pantelis Tassopoulos May 08 '25 at 18:13
  • @Pantelis Tassopoulos even in this case we need to integrate $\left| \frac{\sin{u}}{u} \right|$ from $rx$ to $Rx$, where $x$ can be any real number. But this integral cannot be uniformly bounded for all $x$. – ElectroSchOOp May 08 '25 at 18:41
  • Notice in their answer the absolute value is not brought inside, the only fact being used is the boundedness of $\int_a^b \frac{\sin u}{u}\mathrm{d}u$ for all $a, b\in \mathbb{R}$ since, as you mentioned, the improper integral converges. – Pantelis Tassopoulos May 08 '25 at 18:51
  • @Pantelis Tassopoulos yes, but the question stays the same. Why can we use Fubini's theorem even in the case discussed in that post? I do not understand why the fact that $\frac{\sin{y}}{y}$ is continuous on $\mathbb{R}$ as you have mentioned is relevant here. – ElectroSchOOp May 08 '25 at 19:06

1 Answers1

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Hint: for $0<r<R$ the function $$(x,y)\mapsto f(x)\frac{\sin xy}{y}, \quad (x,y)\in \mathbb{R}\times [r,R]$$ is in $L^1(\mathbb{R}\times [r,R])$. To see this, use the elementary inequality $|\sin (u)|\le \min(|u|,1), u\in \mathbb{R}$ which gives the pointwise bounds $$\Big|f(x)\frac{\sin xy}{y}\Big| \le |f(x)|\cdot\min(|x|, 1/|y|)\le |f(x)|\cdot\min(|x|, 1/r), (x,y)\in \mathbb{R}\times [r,R]\setminus\{0\}\,,$$ which is integrable on $\mathbb{R}\times [r,R]$, thus justifying the interchange of integrals.

Pantelis Tassopoulos
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