- Suppose that I'm dealing with a family of complex functions analytic in the right-half plane and that each $f$ has a representation:
$$f(z) = \int^1_{-1} \frac{2z}{(1+ z^2) + t(1 – z^2)} \, d\mu(t),$$
where $d\mu(t)$ is some probability measure. How do I show that such a family is compact?
- Similarly, let each $f(z_1, z_2)$ be analytic in $\mathbb{C}_+ \times \mathbb{C}_+$.
$$f(z_1, z_2) = \int^1_{-1}\int^1_{-1} \frac{2z_1}{(1+ z_1^2) + t_1(1 – z_1^2)} \frac{2z_2}{(1+ z_2^2) + t_2(1 – z_2^2)} \, d\mu_1(t)\, d\mu_2(t).$$
Do those functions form a compact family?
Best Answer
These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $t\in [-1,1]$, define $$K(z,t) = \frac{2z}{(1+z^2)+t(1-z^2)}.$$ It is a routine exercise that the map $\mu\mapsto f(z)=\int_{[-1,1]} K(z,t) \mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $\mu_n\to\mu$ weak* then the corresponding $f_n\to f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $C\times[-1,1]$. Hence the collection of restrictions $K_t: z\mapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $\omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $\delta$-balls, and bound $\|f_n-f\|_C=\sup\{|f_n(z)-f(z)|: z\in C\}$ with a finite sum of differences of integrals, one per ball center, plus a term $\omega(\delta)$. The weak* convergence of the $\mu_n$ to $\mu$ shows the differences of the integrals converges to $0$; choice of $\delta$ small enough then forces $\|f_n-f\|_C<\epsilon$ for any desired $\epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact. Alternatively, if $f_n$ are elements of your set, corresponding to measure $\mu_n$, there is a weak* convergent subsequence of the $\mu_n$ with limit $\mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $\mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.