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I want to solve the problem #8 of Stein's book Complex analysis in Chapter 4, for the first part I've got the following:

We know that the coefficients of a series are:

$$a_n=\frac{f^{(n)}(z_0)}{n!}$$

Since the series is centered at $z_0=0$ and using the inverse Fourier transform we get :

$$f^{(n)}=(2 \pi i )^{n} \int_{-\infty}^{\infty}\hat{f}(\zeta)\zeta^{n}e^{2\pi ix\zeta}d\zeta=(2 \pi i )^{n} \int_{-M}^{M}\hat{f}(\zeta)\zeta^{n}e^{2\pi ix\zeta}d\zeta$$

and evaluating at zero and dividing by the factorial we have:

$$a_n=\frac{(2 \pi i )^{n}}{n!} \int_{-M}^{M}\hat{f}(\zeta)\zeta^{n}d\zeta$$

but then I don't know how to bound this to get that

$$\limsup_{n-> \infty}(n!|a_n|)^{1/n} \le 2 \pi M$$

I was trying to do the following:

$$|a_n|\le \frac{(2 \pi )^n}{n!}\int_{-M}^{M}|\hat{f}(\zeta)||\zeta^{n}|d\zeta$$

the thing is that I don't know how to ensure that the integral is less than $M^n$

Another way I thought of was to use Cauchy's inequalities then I will get the following

$$|a_n| \le \frac{n!}{R^n}||f||_{C}$$

for an appropiate election of the radius of the circle, so I want to pick the circle with center in zero and of radius $M$ but then How can I bound the integral? and I don't think this could take me to the result.

For the second part I don't understand if I can use that $\hat{f}$ is compact supported, if I can then I use Paley- Weiner theorem but I don't know how modify it to get the result since that $\epsilon$ is annoying me. If I can't use it then I can hardly think in a way to prove this.

Can someone help me with this issue please? Thanks in advance.

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  • $\begingroup$ If $\hat{f}$ is continuous and compactly supported, then it is bounded, right? $\endgroup$ – carmichael561 Mar 26 '16 at 19:36
  • $\begingroup$ Right :) it is correct, but how do you know is continuous? $\endgroup$ – user162343 Mar 26 '16 at 19:37
  • $\begingroup$ What are the hypotheses on $f$? The Fourier transform of any reasonable function is continuous. $\endgroup$ – carmichael561 Mar 26 '16 at 19:39
  • $\begingroup$ Just that $f$ is a series :) $\endgroup$ – user162343 Mar 26 '16 at 19:40
  • $\begingroup$ Presumably $f$ must be integrable on $\mathbb{R}$, or else how is the Fourier transform defined? And the Fourier transform of any $L^1$ function is continuous. $\endgroup$ – carmichael561 Mar 26 '16 at 19:41
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If $f\in L^1(\mathbb{R})$ then $\hat{f}$ is continuous, and since it is supported in $[-M,M]$, it must be bounded, by some $C>0$. Therefore $$ n!a_n\leq (2\pi)^n\int_{-M}^M|\hat{f}(t)||t|^n\;dt \leq C(2\pi)^n\int_{-M}^M|t|^n\;dt =\frac{2CM^{n+1}}{n+1}(2\pi )^n$$

Now taking $n$th roots of both sides yields

$$ (n!a_n)^{\frac{1}{n}}\leq \frac{(2CM)^{\frac{1}{n}}}{(n+1)^{\frac{1}{n}}}2\pi M$$

and since $\lim_{n\to\infty}K^{\frac{1}{n}}=1=\lim_{n\to\infty}(n+1)^{\frac{1}{n}}$ for any constant $K$, it follows that $$ \limsup_{n\to\infty}(n!a_n)^{\frac{1}{n}}\leq2\pi M $$

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  • $\begingroup$ How do you take limsup ? I mean which is the calculation of limsup? $\endgroup$ – user162343 Mar 26 '16 at 19:47
  • $\begingroup$ What about the second part? $\endgroup$ – user162343 Mar 26 '16 at 19:49
  • $\begingroup$ I think is trickier right? $\endgroup$ – user162343 Mar 26 '16 at 19:50
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    $\begingroup$ I think the radius of convergence is bigger. For instance, what if $a_n=\frac{1}{n!}$? $\endgroup$ – carmichael561 Mar 26 '16 at 19:56
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    $\begingroup$ Yes, which has radius of convergence $\infty$. $\endgroup$ – carmichael561 Mar 26 '16 at 19:57
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You were already on the right track, but stopped a little too soon. You showed $$ a_n = \frac{(2\pi i)^{n}}{n!}\int_{-M}^{M}\hat{f}(\zeta)\zeta^n d\zeta $$ Therefore, \begin{align} (n! |a_n|)^{1/n}&=2\pi\left|\int_{-M}^{M}\hat{f}(\zeta)\zeta^n d\zeta\right|^{1/n} \\ &\le 2\pi\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta M^{n}\right)^{1/n} \\ &= 2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} \end{align} The limit of the right side exists as $n\rightarrow\infty$ and, assuming $\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta \ne 0$, that limit is $2\pi M$. Therefore, \begin{align} \limsup_{n} (n!|a_n|)^{1/n} & \le \limsup_{n} 2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} \\ & = \lim_{n}2\pi M\left(\int_{-M}^{M}|\hat{f}(\zeta)|d\zeta\right)^{1/n} = 2\pi M. \end{align} The only assumption needed for $\hat{f}$ is that it is absolutely integrable on $[-M,M]$.

For the converse, assume that $f(z)=\sum_{n=0}^{\infty}a_n z^n$, and assume that $$ \limsup_{n}(n!|a_n|)^{1/n} \le 2\pi M. $$ Let $\epsilon > 0$ be given. Then there exists $N$ such that $$ \sup_{n \ge N}(n!|a_n|)^{1/n} \le (2\pi M+\epsilon) \\ n!|a_n| \le (2\pi M+\epsilon)^{n} \;\;\; n \ge N \\ \sum_{n=0}^{\infty}|a_n||z|^{n} \le \sum_{n=0}^{N-1}\left(|a_n|-\frac{(2\pi M+\epsilon)^{n}}{n!}\right)|z|^{n}+\sum_{n=N}^{\infty}\frac{1}{n!}(2\pi M+\epsilon)^{n}|z|^{n} $$ A little juggling of constants to bound the first few terms by the full exponential series gives $$ |f(z)| \le C\sum_{n=0}^{\infty}\frac{1}{n!}(2\pi M+\epsilon)^{n}|z|^{n} = e^{(2\pi M+\epsilon)|z|}. $$

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  • $\begingroup$ So you didn't make use of the Paley-Wiener theorem right? $\endgroup$ – user162343 Mar 27 '16 at 1:44
  • $\begingroup$ How do you prove that $f$ is holomorphic in the complex plane? $\endgroup$ – user162343 Mar 27 '16 at 1:46
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    $\begingroup$ @user162343 : (a) No, we're proving the Paley-Wiener for this special case. (b) $f$ is holomorphic because $f(z) = \int_{-M}^{M}\hat{f}(\xi)e^{2\pi i\xi\,z}d\xi$ easily extends to $z \in \mathbb{C}$ and can be shown to be differentiable from this integral representation; you can also expand the exponential in a power series, interchange summation and integration and get a power series expansion in $z$ for $f$. $\endgroup$ – DisintegratingByParts Mar 27 '16 at 1:50
  • $\begingroup$ Right, so let me continue with the other one :) (math.stackexchange.com/questions/1710909/…) and I'll be connected :) $\endgroup$ – user162343 Mar 27 '16 at 1:52
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    $\begingroup$ @user162343 : Assume $f\in L^2$ or $f\in L^1$ so that the classical transform makes sense. In either case, because $\hat{f}$ is assumed to be compactly supported, then $\hat{f}\in L^1$. Then $e^{2\pi i\xi z}=\sum_{n=0}^{\infty} \frac{(2\pi i\xi z)^{n}}{n!}$ converges uniformly for bounded $z$ and $\xi$, which is enough to interchange summation of the series and integration to obtain $f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\int_{-M}^{M} \hat{f}(\xi)\xi^{n}d\xi\right)z^{n}$. The series automatically converges for all $z$, which proves an infinite radius of convergence. $\endgroup$ – DisintegratingByParts Mar 27 '16 at 2:00

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