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My professor occasionally assigns optional difficult problems which we do not turn in from Stein and Shakarchi's Complex Analysis. I am currently studying for a test in that class and try to get all of these optional problems answered. One problem he gave us is Problem 2 from Chapter 4 on page 132 which you can find here http://carlossicoli.free.fr/S/Stein_E.M.,_Shakarchi_R.-Complex_Analysis-Princeton_univ_press(2003).pdf I am currently working on part (a)

Suppose f has bounded support and is of class $C^2$. For $z \in \mathbb{C}$, let $\hat{f}(z)=\int_{-\infty}^{\infty} f(t)e^{-2\pi izt} dt$. I am supposed to observe that $\hat{f}$ is an entire function, and using integration by parts show that for fixed $a\ge 0$ then $|y|\le a$ implies that for some constant $C_a$, $|\hat{f}(x+iy)|\le \frac{C_a}{1+x^2}$. It says observe $\hat{f}$ is an entire function so I assume it is something simple but I don't see it. Maybe I will have to evaluate the integral first. Which leads me to the integration by parts. I am struggling with that without knowing the function f specifically. I tried using $f$ as $u$ and the exponential function as $dv$ but got nowhere. Thus, I am here asking for your help. Thanks!

Leo Spencer
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  • I believe this question was asked here : http://math.stackexchange.com/questions/75405/on-functions-with-fourier-transform-having-compact-support Re the entire part, this is a pretty simple function of z -- what happens when you differentiate it at any point z? – Betty Mock Oct 15 '13 at 00:02
  • no thats a completely different exercise – Leo Spencer Oct 15 '13 at 00:11
  • @BettyMock can I move the derivative under the integral and if so why? – Leo Spencer Oct 15 '13 at 00:17
  • I figured out the entire function part but am still struggling with the integration by parts to show the moderate decrease condition. – Leo Spencer Oct 15 '13 at 03:59
  • Sorry about the bad reference. I thought I had referred a fourier series question to that. – Betty Mock Oct 16 '13 at 04:18
  • Try this reference: http://math.stackexchange.com/questions/27181/bounds-for-fourier-series. In the little time I had I was not able to get farther with the approach below; and this link suggests you have to hit it with a stronger theorem. – Betty Mock Oct 16 '13 at 15:02
  • The reference above only shows an inequality in 1/z; however at least it gets something in the denominator. Your hypotheses are strong here -- you have compact support and a bound on y; so there may be a way to maneuver that answer into what you want. You probably need to see how the proof goes for what was done in 27181; or at least look up the theorem referred to. – Betty Mock Oct 16 '13 at 15:32
  • I said the wrong thing about "entire". You do need the integral to converge uniformly before switching with the derivative. But it certainly does converge uniformly because f has compact support and is $C^2$. – Betty Mock Oct 17 '13 at 17:28
  • Morera's Theorem: If $f$ is continuous on an open region, then $f$ is holomorphic iff $\int_{\Delta}f,dz=0$ for every triangle that is, along with its interior, contained in the open region. So you show that the function is continuous in $z$, and then integrate. Interchange the order of integration to put the integral onto the exponential, which gives $0$ because the exponential is holomorphic. The same kind of trick with the Cauchy integral representation allows you to move the derivative from outside to inside the integral. – Disintegrating By Parts Apr 18 '14 at 15:00

2 Answers2

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A straightforward way you can show that $\hat{f}$ is an entire function is write $$ \hat{f}(z) = \int_{-R}^{R}f(t)e^{2\pi izt}\,dt = \int_{-R}^{R}f(t)\sum_{n=0}^{\infty}\frac{(2\pi izt)^{n}}{n!}\,dt = \sum_{n=0}^{\infty}\frac{(2\pi iz)^{n}}{n!}\int_{-R}^{R}t^{n}f(t)\,dt, $$ where $f(t)=0$ outside $[-R,R]$. Interchanging integration and summation is allowed because $f$ is supported on a finite interval $[-R,R]$ and because the series for the exponential converges absolutely and uniformly on $|tz| \le r$ for any fixed $r > 0$. So the series on the right converges for each fixed $z$, which means that the series must be an entire function of $z$ with an infinite radius of convergence.

Assume that $f$ vanishes for $|t| \ge R$, and let $M$ be a bound for $f$ on $[-R,R]$. Then, for $|\Im z| \le \alpha$, $$ \left|\int_{-R}^{R}f(t)e^{2\pi itz}\,dt\right| \le \int_{-R}^{R}|f(t)|e^{2\pi\alpha|t|}\,dt \le 2RMe^{2\pi R\alpha}. $$ Let $N$ be a bound for $f''$ on $[-R,R]$. Then the far field, the bound is better. For $|\Im z| \le \alpha$ and $z\ne 0$, integration by parts twice gives $$ \left|\int_{-R}^{R}f(t)e^{2\pi izt}\,dt\right| = \left|\frac{1}{(2\pi i z)^{2}}\int_{-R}^{R}e^{2\pi itz}f''(t)\,dt\right| \le \frac{2RNe^{2\pi R\alpha}}{4\pi^{2}|z|^{2}}. $$ It's not hard to combine these into a single estimate $$ |\hat{f}(x+iy)| \le \frac{C_{\alpha}}{1+x^{2}},\;\;\; |y| \le \alpha. $$

Disintegrating By Parts
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  • When showing that you can interchange integration and summation (in the first part of your argument), don't you just need the fact that the Taylor series converges uniformly? You do have a finite measure space here so uniform convergence seems to be enough... or am I missing something? – Longti Feb 15 '18 at 01:21
  • @Longti : The power series for the exponential converges absolutely and uniformly on compact subsets. And, that's what I was pointing out in my post. – Disintegrating By Parts Feb 15 '18 at 02:22
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Yes, as you probably figured out you can do the differentiation under the integral sign; the reason would be that you are differentiating with respect to z, and all the stuff under the sign is wrt t.

I can get started here. You could write $e^{-2\pi it}$ as $e^{-2 \pi ixt}e^{2 \pi yt}$. You've got a bound on $e^{2 \pi yt} \le e^{2 \pi at}$ and the other exponential is naturally bounded. I think the integration by parts should be u(t) = f(t)$ e^{2 \pi at}$ and v'(t) = $e^{-2\pi ixt}$. It may be that you may want to integrate by parts twice and compare with the original integral. You can do that because f is in $C^2$.

See if you can get farther with this.

Betty Mock
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    I'd be careful with that logic that just because it's not the thing you're integrating you can commute the integral and $\frac{d}{dz}$. It works in a lot of cases...but not always. – jxnh Jul 24 '14 at 02:35