Question: Let $f,g:\mathbb{R}\to\mathbb{R}$ be continuous functions such that given any two points $x_1<x_2$, there exists a point $x_3$ between $x_1$ and $x_2$ such that $f(x_3)=g(x_3).$ Show that $f(x)=g(x)$ for every $x\in\mathbb{R}$.
Solution: Let $h:\mathbb{R}\to\mathbb{R}$ be such that $$h(x)=f(x)-g(x), \forall x\in\mathbb{R}.$$ Note that $h$ is continuous on $\mathbb{R}$.
Now observe that the property described in the statement of the problem can be restated to the following: given any two points $x_1<x_2$, there exists a point $x_3$ between $x_1$ and $x_2$ such that $h(x_3)=0$. Let this restated property be denoted by $(*)$.
Now fix any point $c\in\mathbb{R}$. Next select any point $b>c$. Thus, by $(*)$, there exists a point $x_1\in (b,c)$ such that $h(x_1)=0$. Again, by $(*)$, there exists a point $x_2\in (x_1,c)$ such that $h(x_2)=0$. Continuing this procedure we would end up with a sequence $(x_n)_{n\ge 1}$ such that $x_1\in (b,c)$ and $x_{n+1}\in (x_n,c), \forall n\in\mathbb{N}$ and $h(x_n)=0, \forall n\in\mathbb{N}.$
Also note that $(x_n)_{n\ge 1}$ is a strictly increasing sequence and $x_n<c, \forall n\in\mathbb{N}$, that is, $(x_n)_{n\ge 1}$ is bounded above. This implies that $(x_n)_{n\ge 1}$ is convergent and since $$\sup\{x_n:n\in\mathbb{N}\}=c,$$ implies that $$\lim_{n\to\infty}x_n=c.$$
Now since $h$ is continuous on $\mathbb{R}$, implies that $h$ is continuous at $c$. Thus, by the sequential definition of limit, since $\lim_{n\to\infty}x_n=c$, implies that $(h(x_n))_{n\ge 1}$ converges to $h(c)$. But, since $h(x_n)=0, \forall n\in\mathbb{N}$, implies that $$\lim_{n\to\infty}h(x_n)=0.$$ This implies that $h(c)=0$. Now since $c\in\mathbb{R}$ is arbitrary, implies that $h(x)=0, \forall x\in\mathbb{R},$ i.e, $f(x)=g(x), \forall x\in\mathbb{R}$. Hence, we are done.
Is this solution correct and rigorous enough, except that I haven't found a way to prove that $$\sup\{x_n:n\in\mathbb{N}\}=c,$$ or more specifically that $$\lim_{n\to\infty}x_n=c?$$ Also is there any alternative approach to solve the problem?
Best Answer
Let $x \in \mathbb{R}$. For all $n \geq 1$, apply the hypothesis on the interval $[x, x + \frac{1}{n}]$ : there exists $x_n$ in this interval such that $$f(x_n)=g(x_n) \quad \quad (1)$$
Now because $x \leq x_n \leq x + \frac{1}{n}$, you have that $(x_n)$ tends to $x$. Let $n$ tend to $+\infty$ in the equality $(1)$ : by continuity of $f$ and $g$, you get $$f(x)=g(x)$$