Show that $\omega \mapsto \int_A \lambda^{i\omega}d\mu(\lambda)$ is analytic

Question

Let $\alpha\in \mathbb{R}$ and $D_\alpha$ be the open horizontal strip in the complex plane $\mathbb{C}$ bounded by $\mathbb{R}$ and $\mathbb{R}-i\alpha$. Let $\mu$ be a complex measure on $[0, \infty[$ and consider the function $$D_\alpha \ni \omega \mapsto \int_{1/k}^k \lambda^{i\omega}d\mu(\lambda)$$ where $k\in \mathbb{N}$ is fixed. I want to show that this function is analytic.

Since every complex measure is a linear combination of finite positive measures, we may assume that $\mu$ is a finite positive measure.

Consider a sequence $\{\omega_n\}\subseteq D_\alpha$ and $\omega\in D_\alpha$ with $\omega_n \ne \omega$ for all $n$ and $\omega_n\to \omega$. It suffices to show that $$\lim_{n \to \infty} \int_{1/k}^k \frac{\lambda^{i\omega_n}-\lambda^{i\omega}}{\omega_n - \omega}d\mu(\lambda) = \int_{1/k}^k i \operatorname{Log}(\lambda) \lambda^{i\omega}d \mu(\lambda)$$ so basically we need to justify why we can interchange limit and integral. For this, I wanted to apply the dominated convergence theorem. However, then I have to argue that there exists an integrable function $g$ on $[1/k,k]$ such that $$\left|\frac{\lambda^{i\omega_n}-\lambda^{i\omega}}{\omega_n - \omega}\right|\le g(\lambda).$$ How do I find this function? Thanks in advance for your help/comments!

Answer 1

I have another suggestion: Please tell me if I made a mistake: Fix $\lambda \in [\tfrac{1}{k}, k]$. Then we can find the $k$-th derivative of $$ g_\lambda:\omega \mapsto \lambda^{i\omega} $$ by $$ g_\lambda^{(k)}(\omega) := \log(\lambda)^k i^k \lambda^{i\omega}. $$ Here $\log$ denotes the usual real (!) logarithm. So for any given $\omega \in D_\alpha$ we can write $$ g_\lambda(z) = \sum_{\nu=0}^\infty \frac{\log(\lambda)^\nu i^\nu \lambda^{i\omega}}{\nu!}(z - \omega)^\nu $$ where $z$ is in a small, open neighborhood with radius $\varepsilon$ of $z$ such that the power series converges. Then observe that for fixed $\omega$ and $z$ (as above) we have $$ \int^k_{1/k} \lambda^{iz}~\mathrm{d}\mu(\lambda) = \int^k_{1/k} \lim_{n \rightarrow \infty} \sum_{\nu=0}^n \frac{\log(\lambda)^\nu i^\nu \lambda^{i\omega}}{\nu!}(z - \omega)^\nu~\mathrm{d}\mu(\lambda). $$ Then estimate for arbitrary $n$: $$ \left \lvert \sum_{\nu=0}^n \frac{\log(\lambda)^\nu i^\nu \lambda^{i\omega}}{\nu!}(z - \omega)^\nu \right \rvert \leq \sum_{\nu=0}^\infty \frac{\lvert \log(\lambda) \rvert^\nu \lvert \lambda^{i\omega} \rvert \varepsilon^\nu}{\nu!} = \lambda^\varepsilon \lvert \lambda^{i\omega} \rvert $$ I am not very well versed with complex measures, but the RHS looks $\mu$-integrable to me. So we can apply dominated convergence to find $$ \int^k_{1/k} \lim_{n \rightarrow \infty} \sum_{\nu=0}^n \frac{\log(\lambda)^\nu i^\nu \lambda^{i\omega}}{\nu!}(z - \omega)^\nu~\mathrm{d}\mu(\lambda) = \sum_{\nu=0}^\infty \int^k_{1/k} \frac{\log(\lambda)^\nu i^\nu \lambda^{i\omega}}{\nu!}\mathrm{d}\mu(\lambda) (z - \omega)^\nu~ $$ So around every $\omega \in D_\alpha$ we were able to write your function as a power series. So it is holomorphic.

Answer 2

Even more is true, and there is nothing special about your domain $D_{\alpha}$, or the endpoints $1/k$ and $k$.

Let $U\subset\Bbb{C}$ be any open set, $0<a<b<\infty$ given, and $\mu$ any positive measure on $[0,\infty)$ which assigns finite values to compact sets (like $[a,b]$). Then, the mapping $\omega\mapsto\int_{[a,b]}\lambda^{i\omega}\,d\mu(\lambda)$ is analytic as a function $U\to\Bbb{C}$ (it doesn’t matter if the integration domain is any variant of $[a,b]$ with/without endpoints).

And even this is not the most general, but let’s just stick to this. The first proof uses the holomorphic version of Leibniz’s rule, as explained in The Gamma function is holomorphic on V. For this, fix a point $\omega_0\in U$, and a small disk $\Delta_r$, say of radius $r$, which contains $\omega_0$ and is contained in $U$ (actually any bounded open set works… nothing special about a disk). Then, we have that for any $(\lambda,\omega)\in [a,b]\times \Delta_r$, \begin{align} |\lambda^{i\omega}|&=|e^{i\omega\log\lambda}|\leq e^{|\omega\log\lambda|}\leq e^{(|\omega_0|+r)\sup\limits_{\lambda\in [a,b]}|\log\lambda|}<\infty. \end{align} In words, $\lambda^{i\omega}$ is uniformly bounded on $[a,b]\times \Delta_r$, and since $\mu$ assigns finite values to compact sets, we have found an integrable upper bound. Thus, by the Leibniz integral rule stated in the link, the claim follows. Also, note that there are no issues with branch cuts; here I’m just using the usual logarithm on the positive real axis.

Let’s say we didn’t know the Cauchy inequalities from complex analysis. Then, what we’d have to do is show that for each point $\omega_0\in U$, there is an open neighborhood $\Delta$ of $\omega_0$ in $U$, such that the partial derivative $\frac{\partial}{\partial \omega}(\lambda^{i\omega})=i(\log\lambda) \lambda^{i\omega}$ has an upper bound $g(\lambda)$, which is independent of $\omega\in \Delta$, and such that $g$ is integrable on $[a,b]$. Well, once again, by taking $\Delta$ to be a bounded open set, and by noting that $\lambda$ varies in an interval $[a,b]$ strictly away from the origin and infinity, we see that we can bound it above uniformly by a constant $C$ (depending only on the bounded set $\Delta$, and $a$ and $b$), and since constants are $\mu$-integrable, by the Wikipedia version, we can apply Leibniz’s rule safely.

The key thing you’re missing is that the difference quotient can be bounded by the partial derivative using the mean-value theorem (see Wikipedia for some details). Once you take this into account, you’ll arrive at the argument of the previous paragraph. But like I said, in the holomorphic case, Cauchy’s inequalities are wonderful and they allow us to seemingly weaken the hypothesis further, so we have fewer things to verify.