# Improper Fourier transform

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 $$c and the limit $$\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?

• Of course "$f$ is integrable and twice differentiable" should be "$f$, $f'$ and $f''$ are integrable". – David C. Ullrich Aug 25 '19 at 13:46
• You're correct, I forgot I was speaking about an unbounded integration domain – Sim Aug 25 '19 at 13:48
• It is an old exam question from some university. It was also the "even-ness" that confused me on how to start. – Sim Aug 25 '19 at 13:51

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 $$\lim_k\Lambda_k x=0$$.

• Maybe you can upvote this question with your status to invite other people to think about this statement? – Sim Aug 25 '19 at 16:15
• @Sim Ok, now it's your turn to upvote and "accept" the answer... – David C. Ullrich Aug 25 '19 at 16:22
• @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... – David C. Ullrich Aug 25 '19 at 17:10
• @Sim I added some details... – David C. Ullrich Aug 25 '19 at 17:37
• @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 $\int_{-a}^b$. – David C. Ullrich Aug 25 '19 at 19:21