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

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.

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 $$\zeta\in[a,b].$$

• Thanks for the elaborate answer! :) – The Pointer Oct 17 '18 at 8:00
• You are welcome! :) – Arjun Banerjee Oct 17 '18 at 10:22

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 $$t\in[a,b]$$.

• Ahh, I understand now. Written slightly differently, $\frac{\int_a^b h(t)g(t)\,dt}{\int_a^b g(t)\,dt} \leq \frac{ M \int_a^b g(t)\,dt}{\int_a^b g(t)\,dt} = M = \frac{ \int_a^b M g(t)\,dt}{\int_a^b g(t)\,dt}$, which makes it clear. Thanks for the clarification, Ernie! – The Pointer Oct 15 '18 at 9:23