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 as a consequence integrable which would make the riemann-lebesgue lemma applicable. So the limit above may be computed. But I dont see this. Could anyone help me with this step of the proof.



The uniform boundedness suggested by the OP is not true, if $\phi$ is say the following function: it equals $t^2$ for $t \in [k,k + 1/k^3]$ for all $k \in \mathbb{Z}$ and $0$ elsewhere. Verify for yourself that $\phi(t)$ is still integrable, hence in $L^1$. But for the sake of Fourier inversion, all you need is that $\underset{t}{\Delta} [\frac{\phi(t) \exp(-itx)}{it}]$ be bounded by a constant $K$ for all $t$ with $|t| < \epsilon$ for some $\epsilon > 0$. The reason this suffices is that $\phi(t)/t$ will be integrable away from $t=0$. Now using the fact that an $L^1$ function $f$ can be written as $f_1 +f_2$, where $f_1$ is uniformly bounded, and $f_2$ has arbitrarily small $L^1$ norm, we can show that $\lim_{h \to \infty} \int_{-\infty}^\infty f(t) \cos(ht) dt = 0$ using Riemann Lebesgue lemma. So now let's show $\underset{t}{\Delta} [\frac{\phi(t) \exp(-itx)}{it}]$ is uniformly bounded in a neighborhood of $0$ in $t$.

Define $g(t) = \phi(t) e^{-itx}$. Then we have $$\displaystyle \underset{t}{\Delta} [\frac{\phi(t) \exp(-itx)}{it}] = \frac{g(t)}{t} + \frac{g(-t)}{-t}$$ $$\displaystyle = \frac{g(t)-g(0)}{t} + \frac{g(0) - g(-t)}{t}$$ The latter is the sum of two difference quotient, whose limit as $t \to 0$ goes to $2g'(0)$. Since $\phi(t)$ is differentiable at $t=0$, so is $g(t)$ (an easy exercise). Differentiability means $\lim_{t \to 0} |\frac{g(t)-g(0)}{t} - g'(0)| \to 0$, which means for any $\delta > 0$, there is an $\epsilon > 0$ such that for all $t$ with $|t| < \epsilon$, the left hand side $|\frac{g(t)-g(0)}{t} - g'(0)| < \delta$. Thus by triangle inequality, $\frac{g(t)}{t} + \frac{g(-t)}{-t} < 2g'(0) + 2\delta$, for all $t \in [-\epsilon, \epsilon]$, which gives uniform boundedness in $|t| < \epsilon$.

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  • $\begingroup$ @John Jiang ... thank you, very much appreciated. $\endgroup$ – mark Mar 15 '11 at 5:51

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