Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a complex-valued continuous function on $\mathbb{R}_+$ with compact support and let $g, h$ be two complex-valued continuous functions on $\mathbb{R}_+$ such that $g$ is bounded and $|h(t)|$=1. Let $0 \leq s < t < \infty$.

Find a limit as $n \to \infty$ of

$\prod \limits_{j=k+1}^{r} \left[ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+f(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\right] \times \prod \limits_{j\notin \{k+1, \ldots, r \}}\left(1+\frac{1}{n}f(\frac{j}{n})\right) $

where $\frac{k}{n}< s \leq \frac{k+1}{n} \leq \frac{r}{n} \leq t < \frac{r+1}{n}$.

It should be of the form $\exp\left[\int_{0}^{\infty}f(x)dx + \int_{s}^{t} \left(-\frac{1}{2}|g(x)|^2 + f(x)(h(x)-1)\right) dx\right]$.

Hints are mean value theorems and Plancherel theorem, but I have no idea how to apply it.

Thank you for any hints and answers.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

There is a $\frac1n$ missing before the term $f(\frac{j}{n})h(\frac{j}{n})$.

Taking logarithms turns this product into a Riemann sum plus some terms which are negligible in the limit. For example, observe that, since $f$ has compact support, we can assume that $f(x)$ vanishes for $x> L$, for some fixed $L>t$. Then the product over $j\notin\{k+1,\ldots,r\}$ can be cut off at $Ln$, so \begin{eqnarray*} &&\log \prod_{j\notin\{k+1,\ldots,r\}} (1+\frac1n f(\frac jn))\\ &=&\sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} \log (1+\frac1n f(\frac jn))\\ &=&\sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} (\frac1n f(\frac jn) + O(\frac{1}{n^2})),\\ \end{eqnarray*} where, since $|f|$ is bounded on ${\Bbb R}_+$, the $O()$ term is uniform in $j$ $$ =O(\frac1n) + \sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} \frac1n f(\frac jn). $$ The first term vanishes in the limit and the second is a Riemann sum whose limit is $$\int_0^s f(x) \, dx+\int_t^L f(x) \, dx.$$

The term in the other part of the product, $$ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+\frac1nf(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2},$$ factors as $$ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\left(1+\frac1nf(\frac{j}{n})h(\frac{j}{n})\right).$$ You can then deal with in it the same way, giving a logarithmic limit of $$ \int_s^t -\frac12 |g(x)|^2 + f(x)h(x) \, dx. $$ Adding and exponentiating then gives the desired result.

Re the comments below, suppose that you replaced the term in the first product by $$ T_{jn}:=\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+\frac1n \left(h(\frac{j}{n})+f(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\right).\ \ (1)$$ Then, use the Taylor expansion $$ \sqrt{1-\frac{1}{n} |g(\frac{j}{n})|^2}=1-\frac{1}{2n} |g(\frac{j}{n})|^2+O(\frac{1}{n^2}).\qquad (2) $$ Since $g$ is bounded, the $O()$ term is uniform in $j$. Therefore, substituting $(2)$ into $(1)$, $$ T_{jn}=1-\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn) + O(\frac{1}{n^2}), $$ and so $$ \log T_{jn}=-\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn) +O(\frac{1}{n^2}).\qquad (3) $$ Since $f$, $g$ and $h$ are all bounded on the interval being considered, the $O()$ term is still uniform in $j$ in both of these expressions. Then \begin{eqnarray*} &&\log \prod_{k+1\le j\le r} T_{jn}\\ &=&\sum_{k+1\le j\le r} \log T_{jn}\\ &=&O(\frac 1n)+\sum_{k+1\le j\le r} -\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn),\\ \end{eqnarray*} using $(3)$, and as before the sum on the right-hand side is a Riemann sum.

share|improve this answer
    
What if there will be an extra term like $\prod \limits_{j=k+1}^{r} \left[ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+\frac{1}{n}\left(h(\frac{j}{n})+f(\frac{j}{n})h(\frac{j}{n}) \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\right)\right]$ ? Is there any way to decompose it nicely? –  Frank Tessla Feb 20 '13 at 10:02
1  
It still works out in the same way. The extra term gives an extra $\int h(x) \, dx$ in the logarithmic limit. –  David Moews Feb 20 '13 at 17:57
    
I can't see how, in the first situation due to the logarithm properties $\ln \left[\left(\sqrt{1- \frac{1}{n}|g(\frac{j}{n})|^2} \right)\left(1+ f(\frac{j}{n})h(\frac{j}{n})\right)\right]= \frac{1}{2}\ln \left( 1- \frac{1}{n}|g(\frac{j}{n})|^2 \right)+ \ln\left( 1+ f(\frac{j}{n})h(\frac{j}{n})\right)$ and now we can use the trick with Taylor expansion. If there is an extra term $h(\frac{j}{n})$ we cannot factorize it in a nice way. –  Frank Tessla Feb 20 '13 at 18:38
1  
It's nicer to factorize the term you're taking the product of, but it's not necessary. You can write it as $1+(1/n) \Phi(j/n) +O(1/n^2)$ for some function $\Phi$. Then take logarithms. –  David Moews Feb 20 '13 at 18:50
1  
I added a working of this case in the main body of the answer above. –  David Moews Feb 20 '13 at 20:04

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.