Fall 2016 - Problem 9

Solution.

1. Notice that

   $$\int_{\br{0,1}} \sum_{n=0}^\infty \frac{\abs{zt}^n}{n!} \,\diff\mu\p{t} \leq \sum_{n=0}^\infty \frac{\abs{z}^n}{n!} \int \,\diff\mu\p{t} = e^{\abs{z}} < \infty$$

   for any $z$. Thus, we may interchange the sum and integral in the following by Fubini's theorem:

   $$f\p{z} = \int_{\br{0,1}} \sum_{n=0}^\infty \frac{\p{izt}^n}{n!} \,\diff\mu\p{t} = \sum_{n=0}^\infty \p{\frac{i^n}{n!} \int_{\br{0,1}} t^n \,\diff\mu\p{t}} z^n,$$

   so $f$ is entire.

2. The assumption tells us that $f$ is a polynomial (e.g., by Cauchy estimates). Thus, we have

   $$f\p{z} = \sum_{k=0}^n a_k z_k = \sum_{k=0}^\infty \p{\frac{i^k}{k!} \int_{\br{0,1}} t^k \,\diff\mu\p{t}} z^k,$$

   so by uniqueness, $\int_{\br{0,1}} t^{n+1} \,\diff\mu\p{t} = 0$. Now suppose that $\mu$ is not the Dirac measure $\delta_0$, i.e., $\mu$ is supported on some $\br{a, b} \subseteq \br{0, 1}$ such that $a > 0$. But this means

   $$\int_{\br{0,1}} t^k \,\diff\mu\p{t} \geq \int_{\br{a,b}} t^k \,\diff\mu\p{t} \geq a\mu\p{\br{a,b}} > 0,$$

   a contradiction. Hence, $\mu = \delta_0$ to begin with.