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Let $h:[0,\infty)$ be an integrable function. Prove that the function $$g(z)=\int_0^\infty h(t)e^{tz}\,dt$$ is analytic on $\{z=x+yi:x<0,y\in\mathbb{R}\}$.

How do I start for this question?

I know Cauchy's integral theorem, but that does not seem relevant here?

Also, I am not sure if this is related to my previous question (Lebesgue integral question (double integral)). $$\int_{-\infty}^\infty g(x)^2h(x)\,dx=\int_0^\infty\int_{\{t\in\mathbb{R}:g(t)>x\}}2h(t)x\,dtdx.$$

Thanks for any help.

Community
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yoyostein
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    Try to show that you can differentiate under the integral. – PhoemueX Jun 30 '16 at 11:00
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    if $Re(z) < 0$ and $Re(z_0) < 0$ then $\int_0^\infty h(t) e^{tz} dt = \int_0^\infty h(t) e^{tz_0} e^{t (z-z_0)} dt =\int_0^\infty h(t) e^{tz_0} \sum_{k=0}^\infty \frac{t^k (z-z_0)^k}{k!} dt$ $ =\sum_{k=0}^\infty \frac{(z-z_0)^k}{k!}\int_0^\infty h(t) e^{tz_0} t^k dt$ i.e. it is analytic (and interchanging $\sum$ and $\int$ is justified by absolute convergence) – reuns Jul 01 '16 at 02:20
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    of course you can use instead that holomorphic $\implies$ analytic and show that $g(z)$ is complex differentiable on $Re(z) < 0$, but it requires a little more complex analysis knowledge – reuns Jul 01 '16 at 02:23
  • @user1952009 Thanks. How do we know $\int_0^\infty h(t)e^{tz_0}t^k,dt$ exists? – yoyostein Jul 01 '16 at 03:07
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    because $Re(z_0) < 0 \implies |e^{t z_0} t^k|$ is bounded (and exponentially decreasing) on $t \ge 0$. and sorry I said something dumb, you need that $|z_0-z| < |Re(z)|$ for being sure that everything converges and that interchanging $\sum$ and $\int$ is allowed – reuns Jul 01 '16 at 03:24
  • @user1952009 Nice. I wonder if using Fubini to change the sum and integral is possible? By considering sum as counting measure? – yoyostein Jul 01 '16 at 03:35

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To show this function is analytic, you need to show that the Cauchy Riemann equations are valid; equivalently, the function satisfies $$ \frac{\partial g}{\partial \bar z} = \frac{1}{2}\bigg( \frac{\partial g}{\partial x} - \frac{1}{i}\frac{\partial g}{\partial y} \bigg) = 0.$$ If you are unfamiliar with this very useful formulation of the Cauchy Riemann equations, it's a nice, quick exercise to verify they are equivalent. We'll use it to finish the problem.

Now, we want to apply $\partial/ \partial \bar z$ to $g(z)$, and we want to show this is $0.$ To accomplish this, we want to switch the order of differentiation and integration for both $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$. For example, the integrand satisfies the hypothesis of Theorem 2.27 in Folland's Real Analysis. To apply this, just differentiate the integrand and use the fact that $h(t)$ is integrable and $te^{tz}$ decays as $t$ goes to infinity because of the domain's restriction to $x<0$ to show the derivatives are integrable. The upshot is we can exchange the order of $\frac{\partial}{\partial \bar z}$ and integration. Now, since $e^{tz}$ is analytic, it satisfies the Cauchy Riemann equations, so $$\frac{\partial}{\partial \bar z}e^{tz} =0.$$ Therefore, the integral is 0, and $g$ is analytic, as desired.

Silvia Ghinassi
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Jack Burkart
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  • Thanks. Why do we differentiate with respect to $\bar z$ instead of $z$? – yoyostein Jul 01 '16 at 02:00
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    @yoyostein : $\partial f / \partial \overline{z} = 0$ is a weird way to say that $f$ fulfills the Cauchy Riemann equations – reuns Jul 01 '16 at 02:24
  • I think that the following answer to this stack exchange question will help both of you! http://math.stackexchange.com/questions/314863/what-is-the-intuition-behind-the-wirtinger-derivatives – Jack Burkart Jul 01 '16 at 02:48
  • I made a mistake about applying the switching differentiation and integration theorem. But it still works, I outlined what you can do above – Jack Burkart Jul 01 '16 at 03:46
  • Thanks so much. Just one last question: How do we know $g$ is differentiable as a function $g:\mathbb{R}^2\to\mathbb{R}^2$, which is necessary for Cauchy-Riemann to imply analytic? – yoyostein Jul 01 '16 at 14:10
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    By showing you can exchange differentiating with respect to $\bar z$, you've shown that the partial derivatives with respect to $x$ and $y$ exist - and since the result is a continuous function (it's the integral of an integrable function), $g$ is continuously differentiable – Jack Burkart Jul 01 '16 at 14:34