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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with the property that $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x)$ exist and are equal. Prove that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$.

I applied Rolle's theorem on $\mathbb{R}$ since $\lim_{x \rightarrow \infty} f(x)= \lim_{x \rightarrow -\infty} f(x)$. So, there exists $c \in \mathbb{R}$ such that $f'(c) = 0$. This means that our function has a minimum or a maximum. Now, since the limits at $\infty$ and $-\infty$ are equal we can conclude that $\forall d > 0$ there exists $x_1, x_2 \in \mathbb{R}$ such that $x_1 - x_2 = d$ and $f(x_1) = f(x_2)$.

Is this a correct solution? If not, please help me find a good one.

Thank you!

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  • $\begingroup$ Are you sure that Rolle's theorem allows you to do this? $\endgroup$ – user384138 Nov 17 '16 at 17:26
  • $\begingroup$ It's not given that $f$ is differentiable, just continuous. $\endgroup$ – dxiv Nov 17 '16 at 17:27
  • $\begingroup$ Well, I haven't thought about the differentiability of $f$. $\endgroup$ – George R. Nov 17 '16 at 17:31
  • $\begingroup$ Hint: consider the continuous function $g(x)=f(x+d)-f(x)$ and prove that it must change sign. $\endgroup$ – dxiv Nov 17 '16 at 17:36
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    $\begingroup$ you may wish to compare with the en.wikipedia.org/wiki/Universal_chord_theorem $\endgroup$ – Mirko Nov 17 '16 at 18:49
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Since $f$ is not assumed differentiable you cannot use Rolle's theorem; but the IVT is available, and does the job.

After substracting a suitable constant from $f$ (which is irrelevant for the problem at hand) we may assume that $\lim_{x\to-\infty}f(x)= \lim_{x\to\infty}f(x)=0$, and that there is an $a\in{\mathbb R}$ with $f(a)>0$. Using a standard argument we then can infer that there is a $\xi\in{\mathbb R}$ with $$f(x)\leq f(\xi)\qquad(-\infty<x<\infty)\ .$$ Given a $d>0$ put $g(x):=f(x)-f(x+d)$. Then $$g(\xi-d)=f(\xi-d)-f(\xi)\leq0,\qquad g(\xi)=f(\xi)-f(\xi+d)\geq0\ .$$ As $g$ is continuous it follows that there is a point $x_1\in[\xi-d,\xi]$ with $g(x_1)=0$. Put $x_1+d=:x_2$; then $|x_2-x_1|=d$ and $$f(x_1)-f(x_2)=f(x_1)-f(x_1+d)=g(x_1)=0\ .$$

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  • $\begingroup$ I only have one question regarding this answer. Why can we assume what you wrote in the second paragraph? Let's say, for example that the limits are equal to $-10$. Wouldn't this change anything in the solution? $\endgroup$ – George R. Nov 18 '16 at 16:36
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    $\begingroup$ See my edit. Note that it could also be the case that $f$ is always below its $\pm\infty$ limiting value, but this is tacitly covered by symmetry. $\endgroup$ – Christian Blatter Nov 18 '16 at 16:50
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Fix $d>0$ and put $g(x) = f(x+d)-f(x)$. If for all $x \in \Bbb R$, $g(x) \neq 0$, then $g(\Bbb R) \subset (-\infty,0)\cup(0,\infty)$.

Let $A=g^{-1}((-\infty,0))$ and $B=g^{-1}((0,\infty))$. Assume that neither is empty. Note that $A\cup B = \Bbb R$, and $A$ and $B$ are both open (by continuity of $g$) and that they are disjoint. This contradicts the connectedness of $\Bbb R$. Therefore $A=\emptyset$ or $B = \emptyset$, i.e. $g(x) < 0$ for all $x\in \Bbb R$ or $g(x) > 0$ for all $x\in \Bbb R$.

Now assume $g(x)>0$ for all $x$, then $f(x+d)>f(x)>f(x-d)$ for all $x$. Then, for any fixed $x$,

  • $f(x)>f(x-d)>f((x-d)-d) = f(x-2d)>\cdots > f(x-nd)$, for all $n \in \Bbb N$

and

  • $f(x)<f(x+d)<f((x+d)+d) = f(x+2d) < \cdots < f(x+nd)$ for all $n \in \Bbb N$

i.e. $f(x+nd)>f(x)>f(x-nd)$ for all $n \in\Bbb N$. Taking $n\to \infty$, we get by continuity that $f(x) = L$. Since $x$ was arbitrary, we get $f(x) = L$ for all $x$. So $g(x) = 0$ for all $x$, contradiction.

Similarly, if $g(x)<0$ for all $x$, we get a contradiction.

We conclude that there exists $x_0$ such that $g(x_0)=0$, i.e. $f(x_0+d) = f(x_0)$, as desired.

Added later on

Looking at it now, there is no need for all of that juggling to prove that $g(x) > 0$ for all $x$ or $g(x) < 0$ for all $x$. Just argue that if at some two points $g(x_1)\le 0$ and $g(x_2)\ge 0$, then by IVT $g(x_0)=0$ somewhere.

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Let $m=\sup_{x\in \mathbb{R} } f(x)$ and assume that $m>\lim_{x\to \infty} f(x).$ Then for some $x_0 $ we have $f(x_0 )=m.$ Suppose that $f(x+d) - f(x) \neq 0 $ for all $x.$ Then we can assume that $f(x+d) - f(x) <0 $ for all $x .$ But this implies that $f(x_0 ) -f(x_0 -d ) <0$ but this is impossible.

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  • $\begingroup$ for some x_0 we have f(x_0)=m You don't know that $f$ attains its $\sup$, in other words that $x_0$ exists. Consider for example $f(x) = -e^{-|x|}$ where $\sup f(x) = 0$.. $\endgroup$ – dxiv Nov 17 '16 at 18:04
  • $\begingroup$ then you can take $\inf f(x)$ and go in analogous way $\endgroup$ – MotylaNogaTomkaMazura Nov 17 '16 at 18:15
  • $\begingroup$ No, the exact same argument applies to $\inf$. $\endgroup$ – dxiv Nov 17 '16 at 18:25
  • $\begingroup$ Analogous not exact $\endgroup$ – MotylaNogaTomkaMazura Nov 17 '16 at 18:28
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Since you can't use Rolle's theorem because your function is not continuous, maybe we can try something else. The idea is this: you take a point $x_0$ and calculate $x_0+d$. Now you are certain that the two points are at the right distance from each other.

Then calculate $f(x_0)$ and $f(x_0+d)$. Then define $g(x)=f(x)-f(x+d)$. If this function is zero, we are done. We have found our points $x_1$ and $x_2$.

So, start out with one point $x_0$ and calculate $g(x_0)$. It will either be positive or negative, say without loss of generality that it's positive. We have \begin{equation} 0<g(x_0)=f(x_0)-f(x_0+d). \end{equation} Now, we know that at some other point we must have \begin{equation} 0>g(x_{3})=f(x_{3})-f(x_{3}+d), \end{equation} otherwise we would have a monotonically decreasing sequence and then the limits at infinity can not be the same.

You know that g(x) is a continuous function, so you can apply the Intermediate Value Theorem, which is true for all continuous functions on a bounded interval.

The IVT gives you a point $x_1$ at which $g(x_1)=f(x_1)-f(x_1+d)=0$. This gives you your $x_1$ and $x_2$.

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