Bonnet's Theorem

# Bonnet's Theorem

Theorem 1: Let $f$ be an increasing function on $[a, b]$, $g$ be continuous on $[a, b]$, and let $A, B \in \mathbb{R}$ be such that $A \leq f(a+) \leq f(b-) \leq B$. Then there exists an $x_0 \in [a, b]$ such that $\displaystyle{\int_a^b f(x)g(x) \: dx = A \int_a^{x_0} g(x) \: dx + B \int_{x_0}^b g(x) \: dx}$. |

**Proof:**Define the function $\alpha$ by:

\begin{align} \quad \alpha (x) = \int_a^x g(t) \: dt \end{align}

- This function is continuous since $g$ is continuous and moreover, $\alpha'(x) = g(x)$. Since $f$ is increasing on $[a, b]$ and $\alpha$ is continuous on $[a, b]$ we have by The Second Mean-Value Theorem for Riemann-Stieltjes Integrals that there exists a point $x_0 \in [a, b]$ such that:

\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = f(a) \int_a^{x_0} \: d \alpha (x) + f(b) \int_{x_0}^b \: d \alpha (x) \end{align}

- Using the theorem on the Reducing Riemann-Stieltjes Integrals to Riemann Integrals page we see that:

\begin{align} \quad \int_a^b f(x) \: d \alpha (x) &= f(a) \int_a^{x_0} \alpha'(x) \: dx + f(b) \int_{x_0}^b \alpha'(x) \: dx \\ \quad \int_a^b f(x) g(x) \: dx &= f(a) \int_a^{x_0} g(x) \: dx + f(b) \int_{x_0}^b g(x) \: dx \end{align}

- So if $A = f(a)$ and $B = f(b)$ the theorem is proved. If not, redefine $f$ at the end points of $a$ and $b$. Then the integral of $f$ keeps its value since changing the value of $f$ at a finite number of points does not change the value of the integral. However, if we ensure that $A, B \in \mathbb{R}$ are such such that $A \leq f(a+) \leq f(b-) \leq B$ then $f$ is still an increasing function and so:

\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 \quad \blacksquare \end{align}

Theorem 2 (Bonnet's Theorem): Let $f$ be an increasing function on $[a, b]$, $g$ be continuous on $[a, b]$, $B \in \mathbb{R}$ be such that $f(b-) \leq B$, and $f(x) \geq 0$ for all $x \in [a, b]$. Then there exists an $x_0 \in [a, b]$ such that $\displaystyle{\int_a^b f(x)g(x) \: dx = B \int_{x_0}^b g(x) \: dx}$. |

**Proof:**Since $f(x) \geq 0$ for all $x \in [a, b]$ we have that $f(a+) = \lim_{x \to a+} f(x) \geq 0$. So let $A = 0$ and $B \in \mathbb{R}$ be such that $f(b-) \leq B$. By Theorem 1 there exists an $x_0 \in [a, b]$ such that:

\begin{align} \quad \int_a^b f(x)g(x) \: dx &= (0) \int_a^{x_0} g(x) \: dx + B \int_{x_0}^b g(x) \: dx \\ &= B \int_{x_0}^b g(x) \: dx \quad \blacksquare \end{align}