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!
Thanks
P.S. Please let me know if you have a suggestion for a better title.
Best Answer
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.