I have the following homework problem that is quite confusing.

Let $I$ be an open subinterval of $\mathbb{R}$, and let $f:\mathbb{R} \to \mathbb{R}$ be a Borel measurable function such that $x \mapsto \exp^{tx}\,f(x)$ is Lebesgue integrable for each $t \in I$. Define $h:I \to \mathbb{R}$ by

$h(t) = \int_{\mathbb{R}}\,\exp^{tx}\,f(x)\,\lambda(dx).$

Show that $h$ is differentiable, with derivative given by

$h'(t) = \int_{\mathbb{R}}\,x\,\exp^{tx}\,f(x)\,\lambda(dx),$ at each $t \in I$.

Use Maclaurin expansion of $\exp^{u}$ to show that $|\exp^u - 1| \leq |u|\,\exp^{|u|}$ holds for each $u \in \mathbb{R}$, and apply a suitable modified form of:

$\int\;f\;d\mu = \lim_{t\to{}^+\infty}\;\int\;f_t\;d\mu$ holds for measure space $(X,\mathcal{A},\mu)$ with $g$ a $[0,{}^+\infty]$-valued integrable function on $X$ and $f$ and $f_t$ (for $t \in [0,{}^+\infty]$) as real-valued $\mathcal{A}$-measurable functions on $X$ s.t. $f(x) = \lim_{t\to {}^+\infty}\;f_t(x)$ and $|f_t(x)| \leq g(x)$ for $t\in [0,{}^+\infty]$ holding at $x \in X$ a.e.

I seem to have an idea of a Laplace transform for one thing. I know that

$|\exp^u-1| \leq |u|\;\exp^{|u|} \Longleftrightarrow \frac{1}{2!}|u|^2 + \frac{1}{3!}|u|^3 + \cdots \leq |u|^2 + \frac{1}{2!}|u|^3 + \cdots$ and so validates that claim.

So that is where I am at and I appreciate any help or suggestions!


P.S. Please let me know if you have a suggestion for a better title.

  • 2
    $\begingroup$ It might help to think about the problem more generally, that you have a function $h(t)=\int F(x,t)dx$, and you want to show that you can calculate the derivative by moving $d/dt$ inside the integral sign. You can do this by noting that the derivative is the limit of the difference quotient, and then move the limit inside the integral, but only if you know that the difference quotients are (absolutely) bounded by an $L^1$ function. Finding the bounding function is where the specifics of the problem enter in. $\endgroup$ – Aaron Nov 22 '11 at 2:39
  • $\begingroup$ I cleaned up the question and believe I have an answer for it. To wrap this question up and make it usable I can provide it or not. Any suggestions? $\endgroup$ – nate Nov 26 '11 at 1:41
  • 1
    $\begingroup$ Yes, if you have a good solution, you can write it up and accept it. That way, it isn't left as an unanswered question, and other people can gain from your knowledge. $\endgroup$ – Aaron Nov 26 '11 at 7:15

Let $g(x,t):=\exp(tx)f(x)$. We have to show that we can differentiate $g$ under the integral. To see that, we have to show that for all $t_0\in I$, we can find $\delta_t$ such that $|\partial_t g(x,u)|\leq G(x)$ for all $u\in (t-\delta_t,t+\delta_t)$ and $x\in\Bbb R$, where $G$ is integrable. We have $\partial_t g(x,u)=x\exp(xu)f(x)$. Let $r$ such that $(t-r;t+r)\subset I$. Then $$|\partial_t g(x,u)|=|x|\exp(x(u+r))e^{-rx}|f(x)|\\ \leq C_r \exp(x(u+r))|f(x)|\leq C_r\exp(x(t_0+2r))|f(x)|$$ for $x\geq 0$, and for $x<0$, $$|\partial_tg(x,u)|=|x|\exp(x(u-r))e^{rx}|f(x)|\leq C_r\exp(x(t_0-2r))|f(x)|.$$ As $x\mapsto \chi_{\Bbb R_+}(x)\exp(x(t_0+2r))|f(x)|$ and $x\mapsto \chi_{\Bbb R_-}\exp(x(t_0-2r))|f(x)|$ are integrable, we get the wanted result.


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