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The most common way to verify if the Fourier transform of a function $f$ is integrable $(\hat f\in L^1(\mathbb{R}))$ is by proving that the function $f$ is integrable and $f'$, $f''$ are also differentiable. Now, I came across a weaker condition for the Fourier tranform to be improper integrable. The statement was as follows:

If $f:\mathbb{R} \to \mathbb{C}$ is a differentiable, integrable function and even, i.e. $f(x)=f(-x)$, then the Fourier transform $\hat{f}: \mathbb{R} \to \mathbb{C}$ will be an improper integral, i.e. $\hat{f}\chi_{[c,d]}$ is integrable for all $c<d$ and the limit $$\lim_{c \to -\infty}\lim_{d\to +\infty} \int_c^d \hat{f}$$ will exist. Furthemore, would there be a formula for this improper integral in terms of $f$?

Can anyone provide me with a hint for this problem? I have no idea how to start?

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  • $\begingroup$ Of course "$f$ is integrable and twice differentiable" should be "$f$, $f'$ and $f''$ are integrable". $\endgroup$ – David C. Ullrich Aug 25 '19 at 13:46
  • $\begingroup$ You're correct, I forgot I was speaking about an unbounded integration domain $\endgroup$ – Sim Aug 25 '19 at 13:48
  • $\begingroup$ It is an old exam question from some university. It was also the "even-ness" that confused me on how to start. $\endgroup$ – Sim Aug 25 '19 at 13:51
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This seems to be true, which I find very surprising. (Edit: I found it surprising because I was forgetting a classical result - see the note a few paragraphs down.)

First note that $f$ being even is irrelevant to the principal value $\lim_{c\to\infty}\int_{-c}^c\hat f$:

If $\lim_{c\to\infty}\int_{-c}^c\hat f$ exists for every even function $f\in L^1$ with $f'\in L^1$ then the same limit exists for every $f\in L^1$ with $f'\in L^1$.

Proof: Given $f,f'\in L^1$, let $$g(t)=\frac12(f(t)+f(-t)).$$ Then $\hat g$ is even and $$\hat g(\xi)=\frac12(\hat f(\xi)+\hat f(-\xi)),$$so $$\int_{-c}^c\hat f=\int_{-c}^c\hat g,$$qed.

It doesn't follow that evenness is irrelevant to the problem; if $f$ is even then $$\lim_{x\to\infty}\lim_{b\to\infty}\int_{-a}^b\hat f=\lim_{c\to\infty}\int_{-c}^c\hat f.$$

Here's the part that surprises me - I would have thought it would sound familiar if true:

Note: No, it's not surprising at all. The corresponding fact for Fourier series follows in half a line from the fact that if $f$ is periodic and has bounded variation then the Fourier series converges to $f$ at every point of continuity. I may as well leave the rest of this here:

Thm. If $f\in L^1$ is absolutely continuous and $f'\in L^1$ then $\lim_{c\to\infty}\int_{-c}^c\hat f=f(0).$

Note of course that's not quite right, there should be a $\sqrt{2\pi}$ somewhere. Anyway,

Let $X$ be the Banach space of all absolutely continuous integrable $f$ with $f'\in L^1$, with norm $$||f||_X=||f||_1+||f'||_1.$$For $c>0$ let $$\Lambda_cf=\int_{-c}^c\hat f.$$

There exists a bounded function $S$ with $$S'(t)=\frac{\sin(t)}t.$$If $f\in X$ then Fubini's theorem plus an integration by parts show that, again omitting irrelevant constants, $$\Lambda_cf=\int f(t)\frac{\sin(ct)}t=\int f(t/c)\frac{\sin(t)}t=-\int\frac{f'(t/c)}c S(t).$$

So $$|\Lambda_cf|\le||S||_\infty\int\frac{|f'(t/c)|}{|c|}=||S||_\infty||f'||_1\le||S||_\infty||f||_X.$$In particular, $$||\Lambda_c||_{X^*}\le||S||_\infty.$$Now if $f\in X$ then $f(0)=\int_{-\infty}^0f'$, so $f\mapsto f(0)$ is a bounded linear functional on $X$. Since $\Lambda_cf\to f(0)$ for all $f$ in a dense subspace and $||\Lambda_c||_{X^*}$ is bounded it follows that $\Lambda_cf \to f(0)$ for all $f\in X$.

Of course there's nothing special about $0$:

Cor. If $f\in X$ then $\lim_{c\to\infty}\int_{-c}^c\hat f(\xi)e^{it\xi}\,d\xi=f(t)$.

Hint: $\hat f(\xi)e^{it\xi}=\hat g(\xi)$ if $g=???$


Addendum For the benefit of anyone unhappy about the functional analysis above:

Standard Exercise. Suppose $E$ is a Banach space, $\Lambda_n\in E^*$, and $||\Lambda_n||$ is bounded. If $\Lambda_nx\to0$ for all $x$ in some dense subspace of $E$ then $\Lambda_nx\to 0$ for all $x\in E$.

(Apply this with $\tilde\Lambda_cf=\Lambda_cf-f(0)$ above; $\Lambda_cf\to f(0)$ is the same as $\tilde\Lambda_cf\to0$.)

This is nothing but epsilons and deltas. It's also the same as the proof that a uniform limit of continuous functions is continuous (and in fact it follows from that result if you look at it right). Anyway:

Solution Say $||\Lambda_n||\le c$ for all $n$, and say $S$ is the dense subspace in question.

Suppose $x\in E$. Let $\epsilon>0$. Choose $y\in S$ with $$||x-y||<\frac\epsilon{2c}.$$Since $y\in S$ there exists $N$ so $$|\Lambda_ny|<\frac\epsilon2\quad(n>N).$$Now if $n>N$ we have $$|\Lambda_nx|\le|\Lambda_n y|+|\Lambda_n(x-y)|<\frac\epsilon2+||\Lambda_n||\,||x-y||\le\frac\epsilon2+c\frac\epsilon{2c}=\epsilon.$$


The exercise really is a simple consequence of the fact that a uniform limit of continuous functions is continuous (although the argument may be a bit "abstract").

Interesting Solution: Let $K$ be the one-point compactification of $\Bbb N$: $$K=\Bbb N\cup\{\infty\}.$$

Suppose $||\Lambda_n||\le c$ for all $n$ and let $S$ be the given dense subspace of $E$ on which $\Lambda_n\to0$.

Suppose $x\in E$. Choose $y_n\in S$ with $||x-y_n||\to0$. Define $f_n:K\to\Bbb C$ by $$f_n(k)=\begin{cases}\Lambda_ky_n,\quad(k\in\Bbb N),\\0,&(k=\infty).\end{cases}$$Now the fact that $\lim_k\Lambda_ky_n=0$ says precisely that $f_n$ is continuous. If $f$ is defined as $f_n$ was except with $x$ in place of $y_n$ then $$|f_n(k)-f(k)|\le c||y_n-x||\quad(k\in K).$$So $f_n\to f$ uniformly on $K$, hence $f$ is continuous, which says $\lim_k\Lambda_k x=0$.

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  • $\begingroup$ Maybe you can upvote this question with your status to invite other people to think about this statement? $\endgroup$ – Sim Aug 25 '19 at 16:15
  • $\begingroup$ @Sim Ok, now it's your turn to upvote and "accept" the answer... $\endgroup$ – David C. Ullrich Aug 25 '19 at 16:22
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    $\begingroup$ @Sim Not much functional analysis. The only fnctionnal analysis I see is an exercise: If $E$ is a Banach space, $\Lambda_n\in E^*$, $||\Lambda_n||$ is bounded and $\Lambda_n x\to0$ for all $x$ in a dense subspace then $\Lambda_n x\to0$ for all $x\in E$. (Hint: Start with $\epsilon>0$...) I mean you need to know a little functional analysis or you're not going to get anywhere with the Fourier transform... $\endgroup$ – David C. Ullrich Aug 25 '19 at 17:10
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    $\begingroup$ @Sim I added some details... $\endgroup$ – David C. Ullrich Aug 25 '19 at 17:37
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    $\begingroup$ @Sim Of course an exam question that required all that would be crazy. I think II got what they had in mind: They expect the student to simply know the Thm above (search for "Fourier series bounded variation" and you should find what I'm getting at) - then the point to the question was to note how $f$ even gets you from $\int_{-c}^c$ to $\int_{-a}^b$. $\endgroup$ – David C. Ullrich Aug 25 '19 at 19:21

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