Riemann-Stieltjes Improper integral

Question

I'm trying to prove that $$\int_{-\infty}^{+\infty}\frac{1}{t-z}\, \mathrm{d}w(t)=0 \Longrightarrow \hbox{$w(t)$ is constant},$$ where $w\colon \mathbb{R}\to\mathbb{R}$ is of bounded variation and $z\in\mathbb{C}\setminus\mathbb{R}$. Can someone give me a hint? Thanks.

Answer 1

1. Solution

Plugging $z = \sigma + i\xi$ and considering imaginary parts only, it follows that

$$ \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\xi}{(\sigma - t)^{2} + \xi^{2}} \, dw(t) = 0. \tag{1} $$

Integrating both sides with respect to $\sigma$ on $[a, b]$, using Fubini's Theorem,

$$ \int_{-\infty}^{\infty} \frac{1}{\pi} \left\{ \arctan\left(\frac{b-t}{\xi}\right) - \arctan\left(\frac{a-t}{\xi}\right) \right\} \, dw(t) = 0. $$

Since $dw(t)$ has finite total variation and the integrand is uniformly bounded by $1$, we can take limit $\xi \to 0^{+}$ inside the integral and we have

$$ \int_{-\infty}^{\infty} \frac{\mathrm{sgn}(t - a) - \mathrm{sgn}(t - b)}{2} \, dw(t) = 0. $$

In particular, if both $a$ and $b$ are points of continuity of $w$, it reduces to

$$ 0 = \int_{a}^{b} dw(t) = w(b) - w(a). $$

We know that $w$ is continuous except for countably many points, so in particular $w(t) = c$ for some constant $c$ except for countably many $t$. Then the conclusion follows by the right-continuity of $w$.

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2. Motivation

As $\xi \to 0^{+}$, the Poisson kernel in $(1)$ converges to the Dirac delta in some sense:

$$\frac{1}{\pi} \frac{\xi}{(\sigma - t)^{2} + \xi^{2}} \xrightarrow[]{\xi \to 0^{+}} \delta(\sigma - t).$$

Thus at least intuitively, we have

$$ \int_{-\infty}^{\infty} \delta(\sigma - t) \ dw(t) = 0. $$

To get rid of the Dirac delta, we integrate both sides w.r.t. $\sigma$ on $[a, b]$ and we have

\begin{align*} 0 = \int_{-\infty}^{\infty} \{ u(b-t) - u(a-t) \} \ dw(t) = \int_{a}^{b} dw(t) = w(b) - w(a), \end{align*}

where $u$ is the unit-step function. These lines of heuristic argument suggest that $w$ is constant, which is indeed proved above by justifying this idea.