# proof that there is $c \in [a,b]$ such that $f(c) = g(c)$

Let $$f,g: [a,b] \rightarrow \mathbb{R}$$ continuous functions such that $$\int_a^{b} f(x)dx = \int_a^{b}g(x)dx$$. Proof that there is $$c \in [a,b]$$ such that $$f(c)=g(c).$$

This questions has been asked before in this site, but the answer is wrong and I am studying real analysis so we must not to use the fundamental theorem of calculus, just upper and lower (Darboux) integrals.

My Attempt: I am trying this way: suppose that $$f(c) \neq g(c), \hspace{0.2cm} \ \forall c \in [a,b],$$ that is, suppose $$f(c) - g(c)>0$$ (whatever).

Since $$f$$ and $$g$$ are continuous so $$f-g$$ is continuous, that is, for every $$\epsilon >0$$, there is an $$\delta>0$$ such that $$x \in (c-\delta, c+ \delta)$$ implies $$f(x)-g(x)> 0.$$ (Call $$\epsilon= \frac{f(c) - g(c)}{2})$$

Considerer the partition $$P = \{x_0,x_1,...,x_n\}$$ of interval $$[a,b]$$ such that $$\{c - \delta, c+ \delta\} \subset P$$ then there is $$j\in \{1,2,..n-1\}$$ such that $$x_{j-1} = c-\delta$$ and $$x_j = c+\delta.$$

we have:

$$m_j(f-g) = inf \{f(x)-g(x) ; x \in [c-\delta,c+\delta] \}>0$$.

Then the lower Darboux sum of ƒ with respect to P is: $$L_{f,P} = \displaystyle \sum_{i=1}^{j-1}m_i(f-g)(x_i-x_{i-1}) + m_j(f-g)2\delta$$ + $$\displaystyle \sum_{i=j+1}^nm_i(f-g)(x_i-x_{i-1})$$

and my objective is to show that $$L_{f,P}$$ is positive so I can conclude that the $$\int_a^b f(x) - g(x)dx$$ is positive and this is a contradiction but the only thing that I know is $$m_j(f-g)2\delta>0.$$

• Do you know if $f-g$ attains its minimum and maximum values on $[a,b]$? – kimchi lover Mar 3 at 2:50
• Since $f,g$ are continuous. if $f(c) \neq g(c)$ for all $c \in [a,b]$, then $f(x) - g(x) > 0$ or $f(x) - g(x) < 0$ for all $x \in [a,b]$. What would this say about the two integrals? – Ethan Alwaise Mar 3 at 2:53
• did you use the intermediate value theorem, right? – user638057 Mar 3 at 3:07

If there is no $$c \in [a, b]$$ with

$$f(c) = g(c), \tag 1$$

then

$$\forall x \in [a, b], \; f(x) - g(x) \ne 0; \tag 2$$

this implies

$$f(x) - g(x) > 0, \; \forall x \in [a, b] \tag 3$$

or

$$f(x) - g(x) < 0, \; \forall x \in [a, b]; \tag 4$$

in the former case we have

$$\displaystyle \int_a^b f(x) \; dx - \int_a^b g(x) \; dx = \int_a^b (f(x) - g(x)) \; dx > 0; \tag 5$$

in the latter

$$\displaystyle \int_a^b f(x) \; dx - \int_a^b g(x) \; dx = \int_a^b (f(x) - g(x)) \; dx < 0; \tag 6$$

each of (5) and (6) imply

$$\displaystyle \int_a^b f(x) \; dx \ne \int_a^b g(x) \; dx; \tag 7$$

it follows then by contraposition that

$$\displaystyle \int_a^b f(x) \; dx = \int_a^b g(x) \; dx \Longrightarrow \exists c \in [a, b], \; f(c) = g(c). \tag 8$$

• Thank you, I should use the intermediate value theorem to concluse that (1) implies (3) and (4) but i didn't remember of him. This way it's easier than my attempt. – user638057 Mar 3 at 3:11
• @user638057: yes, the intermediate value theorem is ultimately what allows one to conclude that $f(x) \ne g(x), \forall x \in [a, b] \Longrightarrow$f(x) - g(x) <> 0\$; I was thinking of mentioning it explicitly but it seemed pretty obvious so . . . – Robert Lewis Mar 3 at 3:17