Prove there exists $[a,b] \subset [0,1]$ such that $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx=\frac{ 1}n$ [duplicate]

Let $f(x)$ and $g(x)$ be two continuous functions on $[0,1]$ and $$\int_{0}^{1}f(x) dx= \int_{0}^{1}g(x)dx = 1$$

Show that there exist $[a,b]\subset [0,1]$, such that $$\int_{a}^{b}f(x) dx= \int_{a}^{b}g(x)dx = \frac{1}{2} $$

The question can be solved by considering the fundamental group of $S^1$, now I am wondering if we can solve it by real-analysis.

Here is the topological solution:

Assume that for any $[a,b]\subset[0,1]$, we always have $$(\int_{a}^{b}f(x) dx\neq \frac{1}{2}) \quad \vee \quad(\int_{a}^{b}g(x)dx \neq \frac{1}{2}) $$ Consider mapping $$\phi:D\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\},\,\,(x,y)\mapsto(\int_{y}^{x}f(t) dt,\int_{y}^{x}g(t) dt)$$ where $D=\{(x,y)|0\leq x\leq y\leq 1\}$.

Let $a$ be path from $(0,0)$ to $(0,1) $in $D$ and $b$ be path from $(0,1)$ to $(1,1) $in $D$, then $ab$ is a path from $(0,0)$ to $(1,1) $in $D$ and $$\phi\circ (ab):[0,1]\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\} $$ Notice that $$(\phi\circ (ab))(0)=(\phi\circ (ab))(1)=(0,0)$$

so $\phi\circ (ab)$ is a loop based on $(0,0)$ in $\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}$.

For any $t\in [0,\frac{1}{2}]$, it's not hard to get that $$(\phi\circ (ab))(t+\frac{1}{2})=(1,1)-(\phi\circ (ab))(t)$$ is equivalent to $$(\phi\circ (ab))(t+\frac{1}{2})-(\frac{1}{2},\frac{1}{2})=-((\phi\circ (ab))(t)-(\frac{1}{2},\frac{1}{2}))$$

Define retraction $$r:\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}\rightarrow S^1,\quad (x,y)\rightarrow\ \frac{(x,y)-(\frac{1}{2},\frac{1}{2})}{||(x,y)-(\frac{1}{2},\frac{1}{2})||}$$

Then $r\circ \phi\circ (ab)$ is a loop on $S^1$, such that $$(r\circ \phi\circ (ab))(t+\frac{1}{2})=-(r\circ \phi\circ (ab))(t),\quad\forall t\in [0,\frac{1}{2}]$$

So $<r\circ \phi\circ (ab)>$ is not trivial in $\pi_1(S^1)$.

However, $ab$ is path homotopic to $c$ in $D$, where $c$ is the path between $(0,0)$ and $(1,1)$ in $D$. In this case, $r\circ \phi\circ (ab)$ is a point-path in $S^1$ by $$(\phi\circ c)(t)=\phi(t,t)\equiv (0,0)$$

which leads to contradiction.

Solution 1:

Consider the following proposition.

Proposition 1: Let $T:=\{(a,b)\in[0,1]^2:a\leq b\}$ and let $f:T\to\mathbb{R}^2$ be continuous. Assume that $f(0,r)+f(r,1)=(1,1)$ holds for all $r\in[0,1]$ and that $f$ is $(0,0)$ on the edge $E:=\{(a,b)\in T:a=b\}$. Then there exists $v\in T$ such that $f(v)=(\frac12,\frac12)$.

In can easily be seen that your statement follows from Proposition 1 by considering $$F(a,b)=\left(\int_a^bf(x)\mbox{d}x,\int_a^bg(x)\mbox{d}x\right).$$ I think it is unlikely that you can prove your statement with a proof that can not be manipulated to also prove Proposition 1. However, Proposition 1 can be shown equivalent to a version of the Borsuk Ulam theorem, only using arguments from real analysis. Since I don't think anyone ever came up with a proof of the Borsuk Ulam theorem only using arguments from real analysis, I think it is unlikely that you can prove your statement only using arguments from real analysis. The version of the Borsuk Ulam theorem I am talking about is the following.

Borsuk Ulam theorem: Let $f:D^2\to\mathbb{R}^2$ be continuous such that $f(v)=-f(-v)$ for $v\in S^1$. Then there exists $v\in D^2$ such that $f(v)=(0,0)$.

Remarks: Here $D^2=\{v\in\mathbb{R}^2:\|v\|\leq1\}$ and $S^1=\{v\in\mathbb{R}^2:\|v\|=1\}$. A generalized form of this theorem is mentioned equivalent to the Borsuk Ulam theorem in the second bullet point of the first paragraph of the Wikipedia page. However, from now on, I will refer to this as the Borsuk Ulam theorem.

I think it is worth mentioning that, even though the Borsuk Ulam theorem is usually proven with algebraic topology, you can also prove the Borsuk Ulam theorem with just combinatorics and real analysis. At the end of this post, I give a direct proof of the Borsuk Ulam theorem, only using Tucker's lemma from combinatorics and the Bolzano Weierstrass theorem from real analysis.

The rest of this post will be showing the equivalence of Proposition 1 and the Borsuk Ulam theorem. The following proposition will be used in the proof of both implications.

Proposition 2: There exists a homeomorphism $\phi:T\setminus E\to D^2\setminus\{(1,0)\}$ such that $\phi(0,r)=e^{ri\pi}$ for all $r\in(0,1]$ and $\phi(r,1)=-e^{ri\pi}$ for all $r\in[0,1)$.

Remark: For the definitions of $T$, $E$ and $D^2$, refer to Proposition 1 and the Borsuk Ulam theorem.

If you want, you can ask in the comments for an explicit proof of this proposition. However, the proof will be very tedious and not very enlightening at all. The reader is advised to draw pictures of the situation and convince themselves that such a homeomorphism really exists.

Borsuk Ulam $\implies$ Proposition 1

Let $f$ be given as in Proposition 1. Let $\phi$ be as in Proposition 2. We can define $g:D^2\to\mathbb{R}^2$ by \begin{equation} g(v)= \begin{cases} (0,0) & \mbox{if $v=(1,0)$} \\ f(\phi^{-1}(v)) & \mbox{otherwise} \end{cases}. \end{equation} Combining the properties of $f$ and $\phi$, we know that $g$ is continuous, and $g(v)+g(-v)=(1,1)$ holds for all $v\in S^1$. We find that $g-(\frac12,\frac12)$ obeys all conditions of the Borsuk Ulam theorem. Hence, there exists $v\in D^2$ such that $g(v)=f(\phi^{-1}(v))=(\frac12,\frac12)$.

Proposition 1 $\implies$ Borsuk Ulam

Let $f:D^2\to\mathbb{R}^2$ be as in the Borsuk Ulam theorem. The goal is to find $v\in D^2$ such that $f(v)=(0,0)$. It is a simple application of the intermediate value theorem to show there exists $v\in S^1$ such that $f(v)$ has the same $x$ and $y$ coordinate. Note that this means either $f(v)=(0,0)$, or one of $f(v)$ and $f(-v)$ has equal negative coordinates. In the first case, we are done. In the second case, we can assume without loss of generality (by scaling and rotating) that $f(1,0)=(-\frac12,-\frac12)$.

We can define $g:T\to\mathbb{R}^2$ by \begin{equation} g(v)= \begin{cases} (0,0) & \mbox{if $v\in E$} \\ f(\phi(v))+(\frac12,\frac12) & \mbox{otherwise} \end{cases}. \end{equation} Combining the properties of $f$ and $\phi$, we know that $g$ obeys all conditions of Proposition 1. Hence, there exists $v\in T$ such that $g(v)=(\frac12,\frac12)$, so $f(\phi(v))=(0,0)$.

Combinatorial proof of the Borsuk Ulam theorem

Since $D^2$ is compact, $f$ is uniformly continuous. So for all $\varepsilon>0$ there exists $\delta>0$ such that for all $v,w\in D^2$ with $\|v-w\|<\delta$ we have $\|f(v)-f(w)\|<\varepsilon$.

Let $T$ be a triangulation of $D^2$ such that all edges have length smaller than $\delta$. We then give every vertex a color from the set $\{1,-1,2,-2\}$. The color of vertex $v$ is purely based on $f(v)$. If $f(v)$ has an absolutely larger first coordinate than second coordinate, $v$ gets color $1$ or $-1$, depending on the sign of the first coordinate of $f(v)$. Otherwise, $v$ gets color $2$ or $-2$, depending on the sign of the second coordinate of $f(v)$.

By the given properties of $f$ we can apply Tucker's lemma. We find adjacent vertices $v,w\in D^2$ with opposite colors. Because of our choice in triangulation, we find $\|f(v)-f(w)\|<\varepsilon$. Then because of our way of coloring the vertices, we see that both coordinates of both $f(v)$ and $f(w)$ must be absolutely smaller than $\varepsilon$, so in particular $\|f(v)\|<2\varepsilon$.

We find for each $\varepsilon>0$ some $v\in D^2$ such that $\|f(v)\|<2\varepsilon$. Hence, we can find a sequence $\{v_i\}$ in $D^2$ such that $f(v_i)\to(0,0)$. Because $D^2$ is compact, by the Bolzano Weierstrass theorem there is a convergent subsequence of $\{v_i\}$ with limit $v\in D^2$. By the continuity of $f$, we find $f(v)=(0,0)$.

Solution 2:

Here is a proof for the case $f > 0$.

Let $A$ satisfy $$ \int_0^A f(x) \, dx = \frac{1}{2}. $$ For every $a \leq A$, there exists a minimal point $\beta(a) > a$ such that $$ \int_a^{\beta(a)} f(x) \, dx = \frac{1}{2}. $$ Note that $\beta(0) = A$ and $\beta(A) = 1$.

Define $$ G(a) = \int_a^{\beta(a)} g(x) \, dx, $$ and notice that $$ G(0) + G(A) = \int_0^A g(x) \, dx + \int_A^1 g(x) \, dx = \int_0^1 g(x) \, dx = 1. $$ Therefore either $G(0) \geq 1/2 \geq G(A)$ or $G(0) \leq 1/2 \leq G(A)$. Either way, since $G$ is continuous, there must exist a point $a \leq A$ such that $G(a) = 1/2$. Taking $b = \beta(a)$, we obtain $$ \int_a^b f(x) \, dx = \int_a^b g(x) \, dx = \frac{1}{2}. $$