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I want to prove that

$\lim_{h\rightarrow\infty}\left(\int_{0}^{\infty}\left(\cos ht-1\right)\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt\right)=-\int_{0}^{\infty}\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt$

where $\underset{t}{\triangle}\eta(t)=\eta(t)+\eta(-t)$ and $\phi$ is an integrable function (in the lebesgue sense), to be precise it is the fourier transform of an integrable density function and thus continuous. Also $\phi$ is differentiable at $0$.

According to the authors of this paper (see proof of theorem 3), this can be achieved by showing $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is integrable and the result will follow from the Riemann Lebesgue lemma.

They do this by showing that $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded. And this is the part of the proof I am stuck on. Can anyone show me how to prove $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded and integrable?

Thanks

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2  
Did you try using the Riemann Lebesgue lemma? –  Prometheus Mar 13 '11 at 19:40
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If $\phi(t)/t$ is integrable, then the result would seem to follow from the Riemann--Lebesgue lemma. As it's stated, I think it might be false. Consider $\phi(t) = i e^{i t x} \chi_{[0,1]}(t)$ where $\chi_{A}(t)$ is the (set) characteristic function. Then $\int_0^1 |\cos(ht)/t| = \infty$ for each $h$ and so the integrand is not Lebesgue integrable. Have I screwed something up? (Likely.) –  cardinal Mar 13 '11 at 19:43
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I agree with cardinal, you probably need to give more information about $\phi$. –  Willie Wong Mar 14 '11 at 0:32
    
@cardinal , @Willie, @Prometheus ... The only extra information i have on $\phi$ is that it is the fourier transform of an integrable density function and hence continuous. I think what needs to be shown is that $\frac{\phi(t)\exp\left(-itx\right)}{it}$ is lebesgue integrable and then the result would follow from the Riemann-Lebesgue lemma. But I couldn't show this is an integrable function. Any ideas? Incidentally the above result is required to prove theorem 3 of the following paper –  aukie Mar 14 '11 at 9:21

2 Answers 2

up vote 1 down vote accepted

I hope I'm not misunderstanding the question, but here's what I think (I apologize ahead of time as it is late!)

Write $\eta(t) = \displaystyle\mathop{\Delta}_{t} \left[\frac{\phi(t) e^{-itx}}{it} \right] = \frac{\phi(t) e^{-itx} - \phi(-t) e^{itx}}{it}$.

Then,

$$ \begin{array} \\ \int_0^{\infty} (cos(ht) - 1) \eta(t) dt & = \left.\left( \frac{1}{h} \sin(ht) - t\right) \eta(t)\right|_0^{\infty} + \int_0^{\infty} \left(1 - \frac{\sin(ht)}{ht}\right)t \eta'(t) dt \\ &= \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta'(t) dt \end{array} $$ by parts (which is justified as the author assumes $\phi$ is differentiable at the origin). The Dominated Convergence Theorem implies that $$ \lim_{h \to \infty} \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta'(t) dt = \int_0^{\infty} t \eta'(t) dt. $$ Again, by parts, we have $$ \int_0^{\infty} t \eta'(t) dt = \left. t \eta(t)\right|_0^{\infty} - \int_0^{\infty} \eta(t) dt = - \int_0^{\infty} \eta(t) dt. $$

Putting it altogether gives:

$$ \int_0^{\infty} \left( \cos(ht) - 1\right) \eta(t) dt = - \int_0^{\infty} \eta(t) dt. $$

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impressive! ... This is the trail of thought I was following. If $\eta(t)$ is lebesgue integrable. Then by the Riemann lebesgue lemma the fourier transform $\widehat{\eta}(h) \rightarrow 0$ as $h \rightarrow \infty$, in particular $\int_{0}^{\infty}\cos(ht)\eta(t)dt \rightarrow 0$. But the crux is trying to show $\eta(t)$ is lebesgue integrable. If this could be done it would certainly make the proof shorter. I've almost given up at this stage. But if you see this please let me know. –  aukie Mar 14 '11 at 9:31
    
Infact I'm sure now there is a shorter method. See my edits to the question above. –  aukie Mar 14 '11 at 10:02
    
It was merely a "follow the logic train" exercise. I have no doubt that it can be improved, if it is even correct (I did not justify each step rigorously). I think the crux of the matter is actually showing that $\eta(t)$ is Lebesgue integrable at $0$, as the exponential seems to demand convergence at $0$. Be careful, however, about splitting up the integral into $\int \left( \cos(ht) - 1 \right) \eta(t) dt = \int\cos(ht) \eta(t) - \int \eta(t)$, as Didier points out that each integral need not converge separately. –  JavaMan Mar 14 '11 at 11:49
    
why though? if $\eta(t)$ is integrable, as the proof of the theorem claims (which I think I now know why) then both the integrals on the right hand side should be finite. –  aukie Mar 15 '11 at 5:59

The proof of Theorem 3 is discussing $\int_0^\infty exp(-itx) \phi(t) \frac{\cos(ht)-1}{t}\, dt$. That -1 makes a big difference.

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@Robert Israel, I dont agree, the integral the author claims to tend to $0$ as $h \rightarrow \infty$ (i.e. the $\cos ht$ part of the integrand, see line after equation A.2) is precisely, $\int_{0}^{\infty}\cos\left(ht\right)\frac{\phi(t)\exp\left(-itx\right)+\phi\lef‌​t(-t\right)\exp\left(itx\right)}{it}dt$, which equals $\int_{0}^{\infty}\cos\left(ht\right)\frac{\phi(t)\exp\left(-itx\right)}{it}dt+\‌​int_{0}^{\infty}\cos\left(ht\right)\frac{\phi\left(-t\right)\exp\left(itx\right)}‌​{it}dt$. The first term is the one I asked about. –  aukie Mar 14 '11 at 5:53
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@aukie That's funny: (A.2) does discuss the integral @Robert says, with the $(\cos(ht)-1)$ term (and certainly not $\cos(ht)$ alone, for good reasons), but you don't agree... Oh well. –  Did Mar 14 '11 at 6:40
    
@Didier, @Robert, hmmm, seems like im being silly. Equation A.2 reads $\frac{2}{2\pi}\int_{0}^{\infty}(\cos ht-1)\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]‌​dt$. Which can be expanded as $\frac{2}{2\pi}\int_{0}^{\infty}\cos ht\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt-‌​\frac{2}{2\pi}\int_{0}^{\infty}\underset{t}{\triangle}[\frac{\phi(t)\exp(-itx)}{i‌​t}]dt$. From my understanding the author claims the first term in this sum goes to zero (which i wonder how?). And the second term goes to $2F(x)-1$. Are you saying this is not right?? –  aukie Mar 14 '11 at 7:15
    
@aukie Which can be expanded... but should not be, in any circumstances, under a penalty of divergent-at-zero integrals. –  Did Mar 14 '11 at 7:55

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