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Let $f,g:[0,1] \rightarrow \Bbb R$ be continuous such that $\int_0^1 f(x)\,dx=\int_0^1 g(x)\,dx=1$. Let $n \in \Bbb N$. I want to prove that there exists $[a,b] \subset [0,1]$ such that $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx=\frac{ 1}n$.

I tried everything I can for a few hours but couldn't get anything. could somone please help on this? We can use everything involving 1-variable real analysis - including results from topology if needed.

Michael Hardy
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D.N
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  • A partial result: Let $h(x)=f(x)-g(x)$. Since $h$ is continuous and $\int _0^1 h(x),dx=0$, we must have $h(c)=0$ for some $c\in(0,1)$. Let $Z$ be the set of zeroes of $h$ in $(0,1)$. If $h(x)$ is not identically zero, $h$ must change sign at at least one $c\in Z$, else we'd contradict $\int _0^1 h(x),dx=0$; then $c$ isn't a local extremum of $h$. Under the further assumption $c$ is not an accumulation point of $Z$, Tong and Braza essentially showed that given continuity on an interval $[a,b]$ there is a subinterval with $ h(c) = \frac{1}{b_1-a_1}\int _{a_1}^{b_1} h(x),dx$. – Integrand May 08 '20 at 00:02
  • In other words: if $c$ isn't an accumulation point, we can come up with arbitrarily small intervals $I_n$ such that $\int _{I_n}h(x),dx=0$. If we also had $f,g$ positive, we might be able to make $\int _{I_n}f,dx=1/n$ as well... – Integrand May 08 '20 at 00:02
  • I think I have seen this earlier on this website. If someone is comfortable with approach0 then do try to find out. – Paramanand Singh May 08 '20 at 13:43
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    See https://math.stackexchange.com/q/3036340/72031 and also have a look at Universal chord theorem. – Paramanand Singh May 08 '20 at 13:48

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