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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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    @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](http://ora.ouls.ox.ac.uk/objects/uuid%3Aa4c3ad11-74fe-458c-8d58-6f74511a476c/datastreams/ATTACHMENT01)2011-03-14

2 Answers 2

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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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    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.2011-03-15
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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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    @aukie *Which can be expanded*... but should not be, in any circumstances, under a penalty of divergent-at-zero integrals.2011-03-14