$F(z)=e^{\int \frac{x+z}{x-z} \log h(x)dx}$ is analytic?

Question

I'm trying to show that $$F(z)=e^{\int _T \frac{x+z}{x-z} \log h(x)dx}$$ is analytic in the open disc if $\int_{T} \mid \log h(x) \mid dx< \infty$ and $h \ge 0 $.

I've tried to consider the radius of convergence of the exponent as an analytic function. But that leads one to needing

$$\lim \sup \mid \frac{(n+2) \int_{T}\frac{\log h(x)}{x^{n+2}}dx}{n!} \mid < 1$$

which I can't make sense of.

I also tried construction an harmonic function such that this is its real part, but all without success.

Answer 1

Clearly, $F$ will be analytic if we manage to prove that $f(z) = \int _T \frac {x+z} {x-z} \log h(x) \ \Bbb d x$ is analytic, where $T = \{ x \in \Bbb C \mid |x| = 1 \}$. Notice that the fraction inside $f$ is well defined because $|z| < 1 = |x|$ (by assumption).

Notice that

$$\frac 1 {x-z} = \frac 1 x \frac 1 {1 - \frac z x} = \frac 1 x \sum _{n \ge 0} \frac {z^n} {x^n}$$

where the geometric series converges because $\left| \frac z x \right| \le 1$.

This means that

$$\frac {x+z} {x-z} = 1 + 2z \frac 1 {x-z} = 1 + 2 \frac z x \sum _{n \ge 0} \frac {z^n} {x^n} = 1 + 2 \sum _{n \ge 1} \frac {z^n} {x^n}$$

so we get the (so far formal) series representation

$$f(z) = \int \limits _T \log h(x) \ \Bbb d x + 2 \sum _{n \ge 1} z^n \int \limits _T \frac {\log h(x)} {x^n} \ \Bbb d x $$

so

$$\begin{align} |f(z)| &\le \left| \int \limits _T \log h(x) \ \Bbb d x \right| + 2 \left| \sum _{n \ge 1} z^n \int \limits _T \frac {\log h(x)} {x^n} \ \Bbb d x \right| \\ &\le \int \limits _T |\log h(x)| \ \Bbb d x + 2 \sum _{n \ge 1} \left| \int \limits _T \frac {\log h(x)} {x^n} \ \Bbb d x \right| |z|^n \\ &\le \int \limits _T |\log h(x)| \ \Bbb d x + 2 \sum _{n \ge 1} \int \limits _T \left| \frac {\log h(x)} {x^n} \right| \Bbb d x \ |z|^n \\ &\le \int \limits _T |\log h(x)| \ \Bbb d x + 2 \sum _{n \ge 1} \int \limits _T | \log h(x) | \ \Bbb d x \ |z|^n \\ &= \int \limits _T | \log h(x) | \ \Bbb d x \left( 1 + 2 \sum _{n \ge 1} |z|^n \right) \\ &= \int \limits _T | \log h(x) | \ \Bbb d x \ \frac {1+|z|} {1-|z|} . \end{align}$$

If $\int _T | \log h(x) | \ \Bbb d x < \infty$, the last value is finite, so the formal series that represents $f$ in the open unit disk converges, therefore $f$ is analytic in the open unit disk. This, in turn, implies that $F = \Bbb e ^f$ is analytic there too.