$\frac{\left(\int_a^b h(t) g(t) \ dt \right)}{\left( \int_a^b g(t) \ dt \right)}$ lies between $m = h(t_m)$ and $M = h(t_M)$?

Question

My vector calculus textbook states the following:

$$\int_a^b h(t) g(t) \ dt = h(c) \int_a^b g(t) \ dt,$$

provided $h$ and $g$ are continuous and $g \ge 0$ on $[a, b]$; here $c$ is some number between $a$ and $b$.$^3$

$^3$ Proof If $g = 0$, the result is clear, so we can suppose $g \not= 0$; thus, we can assume $\int_a^b g(t) \ dt > 0$. Let $M$ and $m$ be the maximum and minimum values of $h$, achieved at $t_M$ and $t_m$, respectively. Because $g(t) \ge 0,$

$$m \int_a^b g(t) \ dt \le \int_a^b h(t) g(t) \ dt \le M \int_a^b g(t) \ dt$$

Thus, $\dfrac{\left( \int_a^b h(t) g(t) \ dt \right)}{\left( \int_a^b g(t) \ dt \right)}$ lies between $m = h(t_m)$ and $M = h(t_M)$ and therefore, by the intermediate value theorem, equal $h(c)$ for some intermediate $c$.

Can someone please explain how $\dfrac{\left( \int_a^b h(t) g(t) \ dt \right)}{\left( \int_a^b g(t) \ dt \right)}$ lies between $m = h(t_m)$ and $M = h(t_M)$? This was simply asserted with no justification.

Thank you.

Answer 1

If $M(h(t))=\sup_{t\in[a,b]}h(t)$ and $m(h(t))=\inf_{t\in[a,b]}h(t)$,

$m(h(t))\le h(t)\le M(h(t))\forall t\in[a,b]\implies m(h(t))g(t)\le h(t)g(t)\le M(h(t))g(t)\forall t\in[a,b]$ if $g(t)\ge 0$.

$\because g(t),h(t)\in\mathfrak{R}[a,b]\implies g(t)h(t)\in\mathfrak{R}[a,b],$ where $\mathfrak{R}[a,b]$ is the set of all Riemann integrable functions defined on $[a,b]$

$\therefore \displaystyle\int_a^b m(h(t))g(t)\,dt\le \displaystyle\int_a^bh(t)g(t)\,dt\le \displaystyle\int_a^bM(h(t))g(t)\,dt$

$\implies m(h(t))\displaystyle\int_a^bg(t)\,dt\le \displaystyle\int_a^bh(t)g(t)\,dt\le M(h(t))\displaystyle\int_a^bg(t)\,dt$

$\implies m(h(t))\le \dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}\le M(h(t))$

Similarly, if $g(t)\le 0$, $M(h(t))\le \dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}\le m(h(t)).$

Thus $\dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}=\mu$ where $m(h(t))\le\mu\le m(h(t))$

If in case $h(t)\in\mathfrak{C}[a,b]$ then $h$ attains all values between its supremum and infimum. So $\exists\zeta\in[a,b]:\mu=h(\zeta).$

Hence $\dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}=h(\zeta)$ for some $\zeta\in[a,b].$

Answer 2

It follows from the inqualities right before. Since $M$ is the maximum, $h(t)\leq M$ for all $t\in[a,b]$. Hence $$ \frac{\int_a^b h(t)g(t)\,dt}{\int_a^b g(t)\,dt} \leq \frac{M \int_a^b g(t)\,dt}{\int_a^b g(t)\,dt} = M. $$ The other inequality follows similarly from $m\leq h(t)$ for all $t\in[a,b]$.