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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?

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    $\begingroup$ Your proof is correct . $x_n \to c$ follows directly from definition of the limit, hence you are done. $\endgroup$
    – Gono
    Sep 1, 2020 at 20:51
  • $\begingroup$ By density of rationnals the sets of point where f and g are equal is also dense. Then you can conclude by the sequential definition of continuity for real functions. $\endgroup$
    – nicomezi
    Sep 1, 2020 at 20:51
  • $\begingroup$ @Gono No, from the way OP has constructed $x_n$, it is possible that it converges to something strictly smaller than $c$. $\endgroup$
    – angryavian
    Sep 1, 2020 at 20:56

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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 $$f(x)=g(x)$$

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  • $\begingroup$ Nice one ! Good job. $\endgroup$
    – nicomezi
    Sep 1, 2020 at 20:53
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Your proof is needlessly complicated but correct. There is no need to invoke completeness of reals.

Assume on the contrary that there is a point $c$ with $h(c) \neq 0$. Then by definition of continuity there is a neighborhood $I$ of $c$ such that $h$ is non-zero on $I$ (has same sign as that of $h(c) $). Now take any two points in $I$, say $p$ and $q$, and by the hypotheses $h$ must vanish somewhere in $(p, q) $ and hence somewhere in $I$.

Contradiction! Done!!

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  • $\begingroup$ I hope all is well, Paramanand! We miss you in the chats, and on meta. You have been such an asset to this site! Stop in sometime to your favorite chat, to say "hi". We need you here! No problem if you've become terribly busy; but I don't want math.se to lose you! Be well! $\endgroup$
    – amWhy
    Oct 28, 2020 at 23:06
  • $\begingroup$ @amWhy: thanks a lot for concern! The long absence was not due to covid (thank god!) but other personal pressing reasons. However as a side benefit this experience has given me some hope that one can control the addiction of Math.SE $\endgroup$
    – Paramanand Singh
    Nov 8, 2020 at 1:51
  • $\begingroup$ @amWhy: will meet you all in chat. Btw congrats for the new President (don't know if you voted for or against)! $\endgroup$
    – Paramanand Singh
    Nov 8, 2020 at 1:53
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Your construction is on the right track, but you are right to be suspicious: from the way you have constructed the $x_n$, it is possible that the $x_n$ do not converge to $c$, since the construction only ensures $b < x_1 < x_2 < \cdots$; it is possible that the $x_n$ converge to some number strictly smaller than $c$. (This is why you were not able to verify $\sup_n x_n = c$.)

To remedy this, instead choose $x_n$ to lie in the interval $(c-1/n, c)$. The sequence you construct in this manner will not necessarily be increasing, but that is not important; what is important is that it is bounded from above by $c$ and converges to $c$.

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  • $\begingroup$ Ya, actually I had exactly the same doubt. $\endgroup$ Sep 1, 2020 at 20:57
  • $\begingroup$ Actually I can still construct a strictly increasing sequence within the interval $(c-1/n,c)$. And, I came up with the idea of a strictly increasing sequence for the clarity of definition of the sequence $(x_n)_{n\ge 1}$. $\endgroup$ Sep 1, 2020 at 21:03
  • $\begingroup$ @SanketBiswas Sure, that's pretty straightforward to do. $\endgroup$
    – angryavian
    Sep 1, 2020 at 21:06

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