Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 7

The continuous function $|h|$ attains a finite maximum $M$ on $[a,b]$ by the extreme value theorem. Additivity and the triangle inequality for the Riemann integral imply, for any $x,y\in [a,b]$,For reversed endpoints, reverse the orientation of the integral before taking its absolute value. Given $\epsilon >0$, choose $\delta =\epsilon /(M+1)$. Then $|x-y|<\delta$ implies $|H(x)-H(y)|<\epsilon$. Thus $H$ is continuous on $[a,b]$, including one-sided continuity at the endpoints; it is even Lipschitz continuous. If $a=b$, the function is defined on one point and the conclusion is immediate.

Write $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. By the extreme value theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$. Because $g\ge 0$, the monotonicity of the Riemann integral givesIf $G=0$, this forces $I=0$, and every $c\in [a,b]$ satisfies the required identity. This includes $a=b$.

If $G>0$, divide to obtain $m\le I/G\le M$. The intermediate value theorem, applied between points where the continuous function $f$ attains $m$ and $M$, supplies $c\in [a,b]$ with $f(c)=I/G$. Hence the weighted mean value theorem for integrals gives

Again let $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. Since $f$ is a nondecreasing monotonic function, $f(a)\le f(t)\le f(b)$ on $[a,b]$. The condition $g\ge 0$ and the monotonicity of the Riemann integral giveSet $H(x)={\int }_a^{x}g(t)dt$. By part (a), $H$ is continuous. Therefore the functionis continuous on $[a,b]$. Its endpoint values are $K(a)=f(b)G$ and $K(b)=f(a)G$, which enclose $I$. The intermediate value theorem provides $d\in [a,b]$ with $K(d)=I$, soIf the two endpoint values coincide, the preceding bounds force $I$ to equal that common value and either endpoint works. Thus the proof also covers $G=0$, $f(a)=f(b)$ and $a=b$, without dividing by any possibly zero quantity.