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Let $\phi: [0,1] \rightarrow \mathbb{C}$ be a continuous function. For all $z \in \mathbb{C} \setminus [0,1] $ define $f(z) = \int_0^1\frac{\phi(t)}{t-z} \ dt$. Prove that f is holomorphic on $\mathbb{C} \setminus [0,1]$.

I can't express $f$ as a composition of holomorphic functions, is there another way to prove $f$ is holomorphic? Thanks in advance!

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I believe it is true that, in general, $\int_a^b F(t,z) dt = h(z)$ is always holomorphic whenever $F$ is continuous in $t$ and holomorphic in $z$. The function $F(t,z) = \frac{\phi(t)}{t-z}$ is certainly holomorphic in $z$ for $z \in \mathbb{C}\setminus [0,,1]$. Can you think of how to prove this? –  snarski Aug 22 '12 at 20:08
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3 Answers

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By standard theorems on differentiation of parameter dependent integrals, for $z\notin [0,1]$, (Edit: to make it a bit clearer: $f$ is differentiable in $z$ and)$$\frac{\partial}{\partial \bar{z}}f(z)=\int_0^1\left(\frac{\partial}{\partial \bar{z} }\frac{\phi(t)}{t-z}\right)dt = 0$$

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By Morera's theorem: Let $\triangle$ be any triangle not containing values in $[0,1]$. Then $\int_{\triangle} f(z) = \int_{\triangle}\int_0^1 \frac{\phi(t)}{t-z}dtdz = \int_0^1\phi(t)\left(\int_\triangle\frac{dz}{t-z}\right)dt = 0$. As the integral is $0$ over all triangles in the domain, $f$ is analytic.

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Here is a more analytical proof.

First we prove that the function is continuous on $\mathbb{C}\setminus[0,1]$. Since $\phi$ is contiuous and $[0,1]$ is compact, $|\phi|$ attains some maximum, say $M$. Also $\mathbb{C}\setminus[0,1]$ is open. Fix some $z_0$ such that $D(z_0, 2r) \subseteq \mathbb{C}\setminus[0,1]$. Then $$|f(z)-f(z_0)|=\left|\int_0^1\frac{\phi(t)}{t-z} - \frac{\phi(t)}{t-z_0}\ dt\right|$$

$$=\left|\int_0^1\phi(t)\cdot\frac{z-z_0}{(t-z)(t-z_0)}\ dt\right|$$ from the $LM$ estimate, we then have as $\epsilon \rightarrow 0$ $$\le |z-z_0|\cdot \max_{t\in[0,1]}\left|\frac{\phi(t)}{(t-z)(t-z_0)}\right|$$ Taking $|z-z_0|<\epsilon<r$ we have $$\le\frac{M}{2r^2}\epsilon \rightarrow 0$$ With continuity at hand, consider then the differential quotient $$\frac{f(z)-f(z_0)}{z-z_0}=\int_0^1\frac{\phi(t)}{(t-z)(t-z_0)}\ dt$$ now the function $\varphi(t) \equiv \frac{\phi(t)}{t-z_0}$ is also continuous. So that as $z\rightarrow z_0$ $$\int_0^1\frac{\varphi(t)}{(t-z)}\ dt\rightarrow \int_0^1\frac{\varphi(t)}{(t-z_0)}\ dt$$ which shows that $$f'(z) = \int_0^1\frac{\phi(t)}{(t-z_0)^2}\ dt$$

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