Riemann-Lebesgue Lemma, Jordan's, and Dini's Theorem Review

We will now review some of the recent material regarding the Riemann-Lebesgue Lemma, Jordan's Theorem, and Dini's Theorem.

• We began on the Lebesgue Integrable Functions with Arbitrarily Small Integrable Terms page by looking at a very nice theorem which said that if $f$ is Lebesgue integrable on an interval $I$ then for all $\epsilon > 0$ there exists upper functions $u^*$ and $v^*$ on $I$ for which $f = u^* - v^*$, $v^*$ is nonnegative almost everywhere on $I$, and:

(1)

\begin{align} \quad \int_I v^*(x) \: dx < \epsilon \end{align}

• We also saw that if $f$ is Lebesgue integrable on an interval $I$ then for all $\epsilon > 0$ there exists a Lebesgue integrable function $g$ on $I$, and a step function $s$ on $I$ for which $f = g + s$ and:

(2)

\begin{align} \quad \int_I \mid g(x) \mid \: dx < \epsilon \end{align}

• To summarize these two very important results, if $f \in L(I)$ then we can firstly find an upper function representation $f = u^* - v*$ of $f$ whose difference contains a upper function $v^*$ that is nonnegative almost everyone on $I$ and whose Lebesgue integral over $I$ can be made arbitrarily small. We can also find a Lebesgue integrable function $g$ and a step function $s$ where $f = g + s$ and such that the Lebesgue integral of $g$ over $I$ can be made arbitrarily small.

• On The Riemann-Lebesgue Lemma page we used the result above to prove the very important Riemann-Lebesgue lemma which says that if $f \in L(I)$ and $\beta \in \mathbb{R}$ then:

(3)

\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0 \end{align}

• We proved this by first showing that it is true for step functions and then wrote $f = g + s$ where $g \in L(I)$, $s \in S(I)$, and where $\displaystyle{\int_I \mid g(x) \mid \: dx}$ was made arbitrarily small from the previous result.

• As a simple corollary to the Riemann-Lebesgue lemma we obtained the following identities by setting $\beta = 0$ and $\displaystyle{\beta = \frac{\pi}{2}}$ respectively for any $f \in L(I)$:

(4)

\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \sin \alpha t \: dt = 0 \end{align}

(5)

\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \cos \alpha t \: dt = 0 \end{align}

• On the Bonnet's Theorem page we looked at an important result. It said that if $f$ is an increasing function on $[a, b]$, $g$ is continuous on $[a, b]$, and $A, B \in \mathbb{R}$ are such that $A \leq f(a+) \leq f(b-) \leq B$ then there exists an $x_0 \in [a, b]$ for which:

(6)

\begin{align} \quad \int_a^b f(x) g(x) \: dx = A \int_a^{x_0} g(x) \: dx + B \int_{x_0}^b g(x) \: dx \end{align}

• We then proved Bonnet's theorem as a consequence for when $f$ is an increasing function on $[a, b]$, $g$ is continuous on $[a, b]$, $B \in \mathbb{R}$ is such that $f(b-) \leq B$, and $f(x) \geq 0$ for all $x \in [a, b]$, which says that there then exists an $x_0 \in [a, b]$ for which:

(7)

\begin{align} \quad \int_a^b f(x) g(x) \: dx = B \int_{x_0}^b g(x) \: dx \end{align}

• Here we applied the previous theorem by setting $A = 0 \leq f(a+)$ (since $f(x) \geq 0$ for all $x \in [a, b]$).

• On the Dirichlet Integrals page we looked at a special type of integral. We said that a Dirichlet Integral is an integral of the following form (for $g \in L([0, b])$ and $b > 0$):

(8)

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

• Assuming that $g(0+) = \lim_{t \to 0+} g(t)$ exists, we began to investigate the nature of the limit $\displaystyle{\lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt}$. We saw that if $g$ is a constant function on $[0, b]$ then:

(9)

\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = \frac{\pi}{2} \cdot g(0+) \quad \Leftrightarrow \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+) \quad (*) \end{align}

• We then began to look at sufficient conditions which allowed for the property $(*)$ to hold. On the Jordan's Theorem for Dirichlet Integrals page we proved that if $g$ is of bounded variation on $[0, b]$ for some $b > 0$ then $\displaystyle{\lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+)}$.

• On the Dini's Theorem for Dirichlet Integrals page we looked at another sufficient condition which said that if $g \in L([0, b])$ for some $b > 0$, $g(0+)$ exists, and the Lebesgue integral $\displaystyle{\int_0^b \frac{g(t) - g(0+)}{t} \: dt}$ exists, then $\displaystyle{\lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+)}$.