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$\Omega$ is a domain in the complex plane and $F(z,t)$ is a continuous function on $\Omega\times I$ where $I=[0,1]$ is the unit interval in $\mathbb{R}$. Suppose further that $F(z,t)$ is analytic in $z$ on $\Omega$ for each fixed $\in I$. Prove that

$$g(z)=\int_0^1 F(z,t) dt$$

Is analytic on $\Omega$. What can be said if $F(z,t)$ is only assume to be analytic on $z\in\Omega$ For all rational values of $t$ (when held fixed in $I$)?

Prove: $$\lim_{h\rightarrow0} \frac{g(z+h)-g(z)}{h}=\lim_{h\rightarrow 0} \int _0^1 \frac{F(z+h,t)-F(z,t)}{h}$$

Here I want to switch the integral but I do not know which theorem can I use?

Edit 1 Another approach, Using Morera's theorem, $\int_{\gamma}\int_0^1F(z,t)dtdz=0$ for any closed curve $\gamma$. Since $F(z,t)$ is uniformly continuous, we can switch the integral and get the result. Is this right? Can we switch the integral?

Edit 2 Second part: Let $t\in [0,1]$ and let $\{t_n\}$ be a sequence in $\mathbb{Q}$ such that $t_n\rightarrow t$ uniformly.

$lim_{n\rightarrow\infty}F(z,t_n)=F(z,t)$ uniformly since $F$ is uniformly continuous in $t\in[0,1]$

Then we have: $\lim_{n\rightarrow \infty} \int_{\gamma}\int_0^1F(z,t_n)dtdz=\int_{\gamma}\int_0^1lim_{n\rightarrow\infty}F(z,t_n)dtdz=\int_{\gamma}\int_0^1F(z,t)dtdz=0$ by first part.

claire
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    Do you know other characterisations of analyticity? In the case where $F(z,t)$ is assumed analytic in $z$ for all $t$, you could also use some standard theorems about differentiation under the integral, but for the second part, that would become incredibly ugly if at all possible, as far as I can see. There is a very useful characterisation of analyticity that makes the exercise a breeze. – Daniel Fischer Jun 21 '14 at 20:58
  • I only know when $f_n$ uniformly convergent we can switch the integral and limit. But it seems not apply here. All the other theorem can be used as I know is DCT, MCT, fubini toneli theorem. They are all in real analysis... – claire Jun 21 '14 at 21:17
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    Do you have to switch integral and limit? As I said, if you're doing it in a certain different way, it's much easier. The dominated convergence theorem, rightly applied, would get you the first part (but, one must first establish a situation where one can prove it applicable). – Daniel Fischer Jun 21 '14 at 21:21
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    I double checked the notes and found the morera's theorem. Is that the right approach? – claire Jun 21 '14 at 22:08
  • Yes, Morera's theorem is exactly what I was getting at. Do you see how to use it for the second part, where analyticity is only assumed for rational $t$? – Daniel Fischer Jun 21 '14 at 22:13
  • Use the density of rational numbers? – claire Jun 21 '14 at 22:17
  • Yes. But what else? – Daniel Fischer Jun 21 '14 at 22:19
  • I wrote edit 2. But I am a bit unclear what theorem to use to switch the integrals and limit in both parts. Could you give me some reference? – claire Jun 21 '14 at 22:30
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    Just a suggestion, to get $\rightarrow$ in your $\LaTeX$ code, type \rightarrow instead of $->$. – Cameron Williams Jun 21 '14 at 22:30
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    @CameronWilliams \to is even shorter for $\to$. – Viktor Vaughn Jun 21 '14 at 22:46
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    @SpamIAm Woah! Neat. I didn't know that one though I guess it makes sense.. Thanks for that. – Cameron Williams Jun 21 '14 at 22:48

1 Answers1

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For the case where $F(z,t)$ is analytic in $z$ for all $t\in I$, if you already knew that $\frac{\partial F}{\partial z}(z,t)$ is continuous - or, if you can use Lebesgue integration theory, that it is (locally with respect to $z$) dominated by an integrable [over $I$] function - you could use a standard theorem about differentiation under the integral sign, e.g. the dominated convergence theorem.

But seeing that one can apply the dominated convergence theorem is not trivial.

One trick to make the legitimacy of the interchange of limit and integration easier to see is to write $F(z,t)$ as a Cauchy integral, so we get

$$g(z) = \int_0^1 \int_{\lvert \zeta-z\rvert = \rho} \frac{F(\zeta,t)}{\zeta-z}\,d\zeta\,dt.$$

Now, if we're looking at $\frac{g(z+h)-g(z)}{h}$ only for $\lvert h\rvert < \rho/2$, since the integrand is a continuous function of $(\zeta,z,t)$, it is easy to see that the convergence of the integrand of the difference quotient $\frac{g(z+h)-g(z)}{h}$ is uniform on the product of the interval and the circle, so interchanging limit and integral is allowed.

This does not work well, if at all, if the analyticity of $F$ in $z$ is only assumed for rational $t\in I$.

Both situations, however, are nicely dealt with by Morera's theorem. By that, $g$, which is known to be continuous, is holomorphic if and only if

$$\int_{\partial\Delta} g(z)\,dz = 0$$

for all triangles $\Delta\subset \Omega$. Inserting the definition of $g$, we find

$$\int_{\partial \Delta} g(z)\,dz = \int_{\partial\Delta} \int_0^1 F(z,t)\,dt\,dz = \int_0^1 \int_{\partial\Delta} F(z,t)\,dz\,dt,$$

where changing the order of integration is allowed since the integrand is continuous and the domains of integration are compact. In the first case, since the inner integral is $0$ for all $t$, Morera's theorem directly asserts the analyticity of $g$.

In the second case, where the analyticity of $F$ in $z$ is only assumed for rational $t\in I$, we note that

$$h(t) = \int_{\partial\Delta} F(z,t)\,dz$$

is a continuous function of $t$, since the integrand is continuous (and the domain of integration compact). Since $h$ vanishes in all rational points, and the rational points are dense in $I$, it follows that $h \equiv 0$ (so in fact, the analyticity of $F$ in $z$ follows for all $t\in I$), and once again, Morera's theorem establishes the analyticity of $g$.

Daniel Fischer
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