Question

Exercise 25: Let f : [a, b] → R be continuous and ϕ : [a, b] → R be integrable and
non-negative, i.e. ϕ ≥ 0. Then there exists c ∈ (a, b) such that
the integral of f(x)ϕ(x)dx from a to b = f(c) the integral of ϕ(x)dx from a to b.

Answer 1

Let $m=\min\limits_{x\in[a,b]}f(x)$ and $M=\max\limits_{x\in[a,b]}f(x)$.

Then $\varphi(x)(f(x)-m)\ge 0$ on $[a,b]$ , so $\int\limits_a^b \varphi(x)(f(x)-m)dx\ge 0$, that is $m\int\limits_a^b\varphi(x)dx\le\int\limits_a^b \varphi(x)f(x)dx$.

Similarly we can obtain $\int\limits_a^b \varphi(x)f(x)dx\le M\int\limits_a^b \varphi(x)dx$

So we have 2 cases:

1)If $\int\limits_a^b \varphi(x)dx=0$, then $\int\limits_a^b \varphi(x)f(x)dx=0=f(c)\int\limits_a^b \varphi(x)dx$ for every $c\in(a,b)$

2)If $\int\limits_a^b \varphi(x)dx\neq 0$, then $m\le\frac{\int\limits_a^b \varphi(x)f(x)dx}{\int\limits_a^b \varphi(x)dx}\le M$. Then by the intermediate value theorem there is $c\in[a,b]$ such that $\frac{\int\limits_a^b \varphi(x)f(x)dx}{\int\limits_a^b \varphi(x)dx}=f(c)$, that is $\int\limits_a^b \varphi(x)f(x)dx=f(c)\int\limits_a^b \varphi(x)dx$.

Prove that if $c\in\{a,b\}$, then there is $c'\in(a,b)$ such that $\int\limits_a^b \varphi(x)f(x)dx=f(c')\int\limits_a^b \varphi(x)dx$. We need to consider 2 subcases:

2.1)$c=a$. Then we have $\int\limits_a^b \varphi(x)(f(x)-f(a))dx=0$.

If there is $c'\in(a,b)$ such that $f(a)=f(c')$, then $\int\limits_a^b \varphi(x)f(x)dx=f(c')\int\limits_a^b \varphi(x)dx$

Suppose that there is not $x\in(a,b)$ such that $f(x)=f(a)$, then by the intermediate value theorem $f(x)-f(a)$ does not change the sign on $(a,b)$. Without loss of generality we can assume that $f(x)>f(a)$ for every $x\in(a,b)$.

So we have that $\varphi(x)(f(x)-f(a))\ge 0$ on $(a,b)$. Then since $\int\limits_a^b \varphi(x)(f(x)-f(a))dx=0$, we have $\varphi(x)(f(x)-f(a))=0$ almost everywhere on $(a,b)$. Since $f(x)-f(a)>0$ on $(a,b)$, we have that $\varphi(x)=0$ almost everywhere on $(a,b)$, that is $\int\limits_a^b \varphi(x)dx=0$.

By the point 1) we obtain that $\int\limits_a^b \varphi(x)f(x)dx=f(c')\int\limits_a^b \varphi(x)dx$ for every $c'\in(a,b)$

So we obtain that there is $c'\in(a,b)$ such that $\int\limits_a^b \varphi(x)f(x)dx=f(c')\int\limits_a^b \varphi(x)dx$

2.2)$c=b$. This case is similar to 2.1.

So for the case 2 we obtain that $\int\limits_a^b \varphi(x)f(x)dx=f(c')\int\limits_a^b \varphi(x)dx$ for some $c'\in(a,b)$.

Therefore in every case we obtain that $\int\limits_a^b \varphi(x)f(x)dx=f(c)\int\limits_a^b \varphi(x)dx$ for some $c\in(a,b)$