Jordan's Theorem for Dirichlet Integrals

# Jordan's Theorem for Dirichlet Integrals

Recall from the Dirichlet Integrals page that if $g$ is a function defined on $[0, b]$, $b > 0$, and $\alpha \in \mathbb{R}$ then a Dirichlet integral is of the following form:

(1)
\begin{align} \quad \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt \end{align}

We will now investigate two theorems known as Jordan's Theorem (below) and Dini's Theorem for Dirichlet Integrals which give sufficient conditions for the following equality to hold (provided $g(0+)$ exists of course):

(2)
\begin{align} \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+) \end{align}
 Theorem 1 (Jordan's Theorem): If $g$ is of bounded variation on $[0, b]$ for some $b > 0$ then $\displaystyle{\lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^{b} g(t) \frac{\sin \alpha t}{t} \: dt = g(0+)}$

Recall that $\displaystyle{g(0+) = \lim_{t \to 0+} g(t)}$ denotes the limit of $g$ as $t$ approaches $0$ from the right.

• So, let $g$ be an increasing function on $[0, b]$ for some $b > 0$ and let $h$ be such that $0 < h < b$. Then we have that:
(3)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt &= \lim_{\alpha \to \infty} \left ( \int_0^h g(t) \frac{\sin \alpha t}{t} \: dt + \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt \right ) \\ &= \lim_{\alpha \to \infty} \left ( \int_0^h [g(t) - g(0+)] \frac{\sin \alpha t}{t} \: dt + \int_0^h g(0+) \frac{\sin \alpha t}{t} \: dt + \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt \right ) \\ &= \underbrace{\lim_{\alpha \to \infty} \int_0^h [g(t) - g(0+)] \frac{\sin \alpha t}{t} \: dt}_{(*)} + \underbrace{\lim_{\alpha \to \infty} \int_0^h g(0+) \frac{\sin \alpha t}{t} \: dt}_{(**)} + \underbrace{\lim_{\alpha \to \infty} \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt}_{(***)} \quad (\dagger) \end{align}
• We will now look each of the three limits in $(\dagger)$ separately. For $(**)$ we see that:
(4)
\begin{align} \quad (**): \quad \lim_{\alpha \to \infty} \int_0^h g(0+) \frac{\sin \alpha t}{t} \: dt &= g(0+) \lim_{\alpha \to \infty} \int_0^h \frac{\sin \alpha t}{t} \: dt \\ &= g(0+) \lim_{\alpha \to \infty} \int_0^{\alpha h} \frac{\sin t}{t} \: dt \\ &= g(0+) \int_0^{\infty} \frac{\sin t}{t} \: dt \\ &= g(0+) \cdot \frac{\pi}{2} \end{align}
• For $(***)$ we see that since $g$ is increasing on $[a, b]$ that $g \in L([h, b])$ (since an increasing function can have at most countably many discontinuities which implies $g$ is Riemann integrable on $[h, b]$ which implies $g$ is Lebesgue integrable on $[h, b]$). Moreover, $\frac{1}{t}$ is continuous and is of bounded variation on $[h, b]$ (since $0 < h$), so $\displaystyle{\frac{g(t)}{t} \in L([h, b])}$. So by The Riemann-Lebesgue Lemma we have that:
(5)
\begin{align} \quad (***): \quad \lim_{\alpha \to \infty} \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt = 0 \end{align}
• All that remains to show is that the limit $(*)$ equals $0$. Let $\epsilon > 0$ be given. Then choose $M \in \mathbb{R}$, $M > 0$ such that for all $0 < x_1 < x_2$ we have that for $\alpha$ sufficiently large that:
(6)
\begin{align} \quad \biggr \lvert \int_{x_1}^{x_2} \frac{\sin \alpha t}{t} \: dt \biggr \rvert \leq M \quad (****) \end{align}
• Also, since $\lim_{t \to 0+} g(t) = g(0+)$ we have that for $\displaystyle{\epsilon_1 = \frac{\epsilon}{M} > 0}$ that for $t$ sufficiently close to $0$ from the right we have that:
(7)
\begin{align} \quad \mid g(t) - g(0+) \mid < \epsilon_1 = \frac{\epsilon}{M} \quad (*****) \end{align}
• Now, since $g$ is increasing on $[0, b]$ then clearly $g$ is increasing on $[0, h] \subset [0, b]$. So then for all $t \in [0, h]$ we have that $\displaystyle{g(0+) = \lim_{t \to 0+} g(t) \leq g(t)}$ and so $g(t) - g(0+) \geq 0$ for all $t \in [0, h]$. Also note that $\frac{\sin \alpha t}{t}$ is continuous on $[0, h]$ when we define this function to be equal to $0$ at $t = 0$. So we can apply Bonnet's Theorem, $(****)$, and $(*****)$ we have that there exists an $x_0 \in [0, h]$ such that:
(8)
\begin{align} \quad \biggr \lvert \int_0^h [g(t) - g(0+)] \frac{\sin \alpha t}{t} \: dt \biggr \rvert = \mid g(t) - g(0+) \mid \biggr \lvert \int_{x_0}^h \frac{\sin \alpha t}{t} \: dt \biggr \rvert < \epsilon_1 \cdot M < \frac{\epsilon}{M} \cdot M = \epsilon \end{align}
• Therefore:
(9)
\begin{align} \quad (*) : \quad \lim_{\alpha \to \infty} \int_0^h [g(t) - g(0+)] \frac{\sin \alpha t}{t} \: dt = 0 \end{align}
• So by combining $(*)$, $(**)$, and $(***)$ into $(\dagger)$ we see that:
(10)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt & = g(0+) \cdot \frac{\pi}{2} \\ \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt & = g(0+) \quad \blacksquare \end{align}