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Let $g$ and $h$ measurable functions on $X $, and assume $g\in L^{1}(X)$ and $h\geq 0$. Show that if $$F(t)=\int g(x)e^{-th(x)}\,d\mu(x)$$ then for $t >0$, $$F(t)=\sum\limits_{n=0}^{\infty} \left[\int g(x)h^{n}(x)\,d\mu(x)\right] \frac{(-t)^{n}}{n!}$$

The question is, what convergence theorem can be used, as the monotone convergence theorem can not be used?

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1 Answer 1

So let $F_n(t)$ be the lower representation summed from $1$ to $n$, i.e. $$F_n(t) = \sum_{i=0}^n\int g(x)h^n(x)d\mu(x)\frac{(-t)^n}{n!}$$ When $n$ is finite we are able to rearrange the summation and integration to give an equivalent: $$F_n(t) = \int g(x) \sum_{i=o}^{n}\frac{(-th(x))^n}{n!}\,d\mu(x)$$ So to use Lebesgue dominated convergence we need to find some integrable function $l(x)$ such that $$\left|g(x)\right|\left| \sum_{i=o}^{n}\frac{(-th(x))^n}{n!}\right| \le l(x)$$ For all $n$. But we also have that $\exp(-x) \le 1$ for all $x \in \Bbb{R}^+$. And that for every $\epsilon > 0$ there exists an $N$ such that $$\sum_{n>N}\frac{(-th(x))^n}{n!}<\epsilon$$ for fixed $t$ and all $x$. We have $$\sum_{i=o}^{\infty}\frac{(-th(x))^n}{n!} \le 1$$ So for $n>N$ we have $$\sum_{i=o}^{n}\frac{(-th(x))^n}{n!} \le 1+\epsilon $$ Andthis gives us $l(x) = |g(x)|(1+\epsilon)$ as a bound if we reindex $\{F_n\}$ to $\{F_{n'}\}$ for $n > N$ this converges to $$g(x)\exp(-th(x))$$ And LDCT shows that the integral converges to our first definition of $F$ as well, giving the desired result.

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I had already tried to find a function that bounds the Fn but can not find it. –  Alexander Osorio Feb 4 '13 at 5:55
    
Sorry, by mistake, you're not trying to find a function that bounds $F_n$, you're trying to find a function that bounds the integrand. Make use to use absolute values where appropriate and think about convergence of the power series and what that implies. I'll add more to my answer. –  Sam DeHority Feb 4 '13 at 12:24
    
@AlexanderOsorio I have fleshed out my answer. I hope you find it satisfactory. –  Sam DeHority Feb 4 '13 at 20:34

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