Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello

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.

Thanks

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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 $|t| < \epsilon$.

share|improve this answer
    
@John Jiang ... thank you, very much appreciated. –  aukm Mar 15 '11 at 5:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.