In the process of proving that the fourier transform $L^1(\mathbb R) \rightarrow \mathcal C_0(\mathbb R)$ is not surjective, I am asked to show that given an odd integrable function $f$ on $\mathbb R$, the integral $$\int_0^{+\infty} \frac{\hat{f}(t)}{t} dt$$ is convergent, with the convention $\hat{f}(t) = \int_{-\infty}^{+\infty}f(x)\exp(-ixt)dx$.

Because $f$ is odd, we may also write $\hat{f}(t) = -2i\int_{0}^{+\infty}f(x)\sin(xt)dx$.

In order to prove the convergence, we consider a positive real number $A$ and we would like to write the following

$$\int_0^{A} \frac{\hat{f}(t)}{t} dt = -2i\int_0^{A} \int_{0}^{+\infty} \frac{f(x)\sin(xt)}{t} dxdt = -2i\int_0^{+\infty} \int_{0}^{A} \frac{f(x)\sin(xt)}{t} dtdx$$

My problem is that I can't justify the last equality properly.

I would like to use Fubini-Lebesgue theorem in order to exchange both integrals, but I can't prove that the function inside is integrable in both $t$ and $x$ on the domain $[0,+\infty]\times[0,A]$.
Could someone please explain why it is possible to exchange the integrals ? I thank you very much in advance.

  • $\begingroup$ Not that I've worked it out, but $|\sin(xt)|/t\le x$ is bound to come in somewhere.... $\endgroup$ – David C. Ullrich Apr 3 '19 at 14:49
  • $\begingroup$ @DavidC.Ullrich I thought about this too, but this would make a term $x|f(x)|$ appear inside the integral, and since $f$ is only an element of $L^1(\mathbb R)$, we can't say anything about the integrability of $xf(x)$. $\endgroup$ – Suzet Apr 3 '19 at 15:32

As already mentioned (see previously below) you can't apply Fubini directly to $\int_0^A\hat f(t)/t\,dt$, because the hypotheses need not be satisfied.

Edit: The answer is yes or no, depending on what sort of "convergence" we want for the integral.

Yes: Sure enough, looking at $\int_\delta^A$ leads very easily to at least a partial result:

If $f$ is an odd integrable function on $\Bbb R$ then $\lim_{\delta\to0,A\to\infty}\int_\delta^A\frac{\hat f(t)}{t}\,dt$ exists.

The proof is straightforward. For $0<\delta<A$ define $$K_{\delta,A}(t)=\frac1t\chi_{[-A,-\delta]\cup[[\delta,A]}(t).$$

Since $f$ and $K_{\delta,A}$ are both integrable it's immediate from Fubini that $$\int_\delta^A\frac{\hat f(t)}{t}\,dt=\frac12\int\hat f K_{\delta,A}=\frac12\int f\widehat{K_{\delta,A}}.$$It's easy to see that $$\lim_{\delta\to0,A\to\infty}\widehat{K_{\delta,A}}(x) =m(x)=\text{sgn}(x)\int_0^\infty\frac{\sin(t)}t\,dt,$$and the fact that $\int_0^A\frac{\sin(t)}t$ is bounded shows, by a little change of variable, that $$\left|\widehat{K_{\delta,A}}\right|\le c \quad(0<\delta<A, x\in\Bbb R),$$ so the existence of the limit follows from Dominated Convergence.

No: Otoh there exists an odd integrable $f$ such that the integral does not converge absolutely.

First note that if $\mu=\delta_1-\delta_{-1}$ then (ignoring irrelevant constants) $\hat\mu(t)=\sin(t)$, so $\int_0^\infty\frac{|\hat\mu(t)|}{t}=\infty$.

Now say $f_n$ is a sequence of odd integrable functions with $f_n \to\mu$ in an appropriate weak* topology. Then $||f_n||_1$ is bounded, but $\widehat{f_n}\to\hat \mu$ pointwise, so Fatou's lemma implies that $\int_0^\infty|\widehat{f_n}(t)|/t\to\infty$. The Closed Graph Theorem now shows that the integral is infinite for some odd integrable $f$.

Previously: The detail you ask about in the argument you started can't be fixed: Define $$I(A)=\int_0^A\frac{|\sin(t)|}{t}\,dt.$$ Then $$\int_0^A\frac{|\sin(xt)|}{t}\,dt=I(xA),$$hence $$\int_0^{+\infty} \int_{0}^{A} \frac{|f(x)\sin(xt)|}{t} dtdx=\int_0^\infty|f(x)|I(xA)\,dx,$$which need not be finite, since $I$ is unbounded.

So you can't apply Fubini, at leat not in such a straighforward way.. Maybe you get somewhere starting with $\int_\delta^A$ instead of $\int_0^A$, or maybe the result is simply false.

  • $\begingroup$ The textbook I am using (French book "Calcul intégral" by Jacques Faraut, written for (under)graduate students) asks, as the first question of the exercise, to prove the equality $\int_0^{A} \frac{\hat{f}(t)}{t} dt = 2i\int_0^{+\infty} \int_{0}^{A} \frac{f(x)\sin(xt)}{t} dtdx$ (the absence of the minus sign must be a typo in my opinion). This boils down to justify that we can switch the integrals. If we can't do this using Fubini's theorem, then I wonder if there is an alternative method in order to conclude in this specific case. $\endgroup$ – Suzet Apr 4 '19 at 6:41
  • $\begingroup$ @Suzet Got a partial result - see edit. Does the problem specify inn what sense we're supposed to the the integral is "convergent"? $\endgroup$ – David C. Ullrich Apr 9 '19 at 15:54
  • $\begingroup$ @Suzet Finished it. Sure enough the answer depends on what we mean by "convergent". $\endgroup$ – David C. Ullrich Apr 21 '19 at 23:26
  • $\begingroup$ I apologize for not answering sooner, and I thank you very much for all your efforts. Your answer really is fantastic. By "convergent", It believe that the problem means "the limit exists", and nothing more. This is what you managed to prove, so thank you very much. $\endgroup$ – Suzet May 9 '19 at 20:53

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