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.

• Proof: It suffices to show that Theorem 1 holds for increasing functions since from the Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions we saw that every function of bounded variation can be written as the difference of two increasing functions.

• 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}