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(This may look very silly to you but I don't understand the reason that was given to me, nor do I have the knowledge to find out by myself).

The domain for $z$ is the open unit disc $D=\{z\in\mathbb{C} | \vert z \vert < 1\}$.

Here $\sigma$ is a real-valued function of bounded variation on $[0,2\pi]$. $$f(z):=\int_0^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}d\sigma (\theta).$$

In any compact $K\subset D$, the modulus of the integrand is bounded by $(1+\lambda)/(1-\lambda)$, so $|f(z)|$ is bounded by $\int_0^{2\pi}\frac{1+\lambda}{1-\lambda}|d\sigma (\theta)| = \frac{1+\lambda}{1-\lambda}V_{[0,2\pi]}(\sigma) $ ($V_{[0,2\pi]}(\sigma)$ is the total variation of $\sigma$).

Why does this show that $f(z)$ is holomorphic in $D$? I was told "by Weirstrass' theorem" but that theorem deals with sequences of holomorphic functions.

Thank you.

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up vote 4 down vote accepted

Just to recall the Weierstrass Uniform Convergence Theorem: If $ f_n $ are holomorphic functions on an open set $ U \subset \mathbb C $ such that $ f_n $ converges to $ f $ uniformly on compact subsets of $ U $, then $ f $ is holomorphic.

Now we simply need to realize the integral as the limit of a sequence of holomorphic functions. For this we simply go back to the definition of the Riemann-Stieltjies integral (we don't need Lebesgue-Stieltjes since $ \sigma $ is of bounded variation). In particular recall that when it exists it is simply the limit of sums of the form

$$ f_n(z) = \sum_{i=0}^{n-1} k(z,\tilde\theta_i)(\sigma(\theta_{i+1})-\sigma(\theta_i)) $$

where in your case $ k(z,\theta) $ is the Schwarz kernel (or in general any function that is analytic in $ z $ and continuous in $ \theta $). By the details you've provided, these sums converge uniformly to the integral on compact subsets of $ U $. Furthermore they are analytic on $ D $ since they are finite sums of analytic functions. Hence by Weierstrass's theorem, $ f(z) $ is analytic on $ D $.

In general it is often fruitful to prove things about continuous objects by passing to statements about sequences.

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Clearly $f$ is a continuous function of $z$. Now, the bound you have for the integrand and the fact that the integrand is holomorphic on $D$ allows you to use Morera and Fubini to conclude.

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I think this solution is equally important to have seen. – R.R. Dec 1 '12 at 3:33

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