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Edit: I very much appreciate alternate solutions to the problem, but I would also like to know if there are any problems or suggestions regarding the way I solved it.

This is a problem I'm attempting from a set of practice Putnam questions (it's also very similar to a problem in Spivak's Calculus. I was wondering if there were any problems with my solution.

Problem: Let $h(t)$ be a continuous function on the interval $[0, 1]$ such that $h(0) = h(1) = 0$. Show that there exists a real number $x \in [0, \frac{2012}{2013}]$ such that $h(x) = h(x + \frac{1}{2013})$.

My attempt at a solution:

I'm going to generalize it with the case $h(x) = h(x + \frac{1}{n})$ where $n \in \mathbb{N}$ (assuming that won't cause any difficulties).

Let $$g(x) = h(x) - h(x + \frac{1}{n}) \text{ on } [0, \frac{n-1}{n}]$$

We now attempt to prove the problem by contradiction. Assume $g(x) \neq 0$ for all $x$ in $[0, \frac{n-1}{n}]$.Then $h(x) \neq h(x + \frac{1}{n})$ for all $x$ in $[0, \frac{n-1}{n}]$.Certainly $g$ is continuous on the interval, since $h$ is continuous on the interval, so by the intermediate value theorem, it cannot be both positive and negative on the interval (for if it were, then we would have $g(a) > 0$ and $g(b) < 0$ (or vice versa) on some interval, which would mean that there is some $c \in [a, b]$ such that $g(c) = 0$, which we are assuming is not true). First we assume that $g(x) > 0$ on $[0, \frac{n-1}{n}]$. If this were the case, then $h(x) > h(x + \frac{1}{n})$ on $[0, \frac{n-1}{n}]$. This implies that $h$ is always decreasing on $[0, \frac{n-1}{n}]$, which is not possible, since $h(0) = h(1)$. If we assume the opposite case, that $g(x) < 0$ on $[0,\frac{n-1}{n}]$, then $h(x + \frac{1}{n}) > h(x)$ on $[0, \frac{n-1}{n}]$, which is also not possible, because it would mean $h$ is always increasing on $[0,\frac{n-1}{n}]$. Since neither of these cases is possible, then $g(x)$ must equal $0$ at some point in $[0, \frac{n-1}{n}]$, so by extension, $h(x) = h(x + \frac{1}{n})$ for some $x \in [0, \frac{n-1}{n}]$. To solve the given problem, we simply let $n = 2013$.

JRN
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user28375028
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    Since $n$ is given, $h(x) > h(x + \frac{1}{n})$ doesn't mean $h$ is decreasing. – mfl Aug 01 '14 at 01:17
  • Universal Chord Theorem – Brad Aug 01 '14 at 02:06
  • As @mfl says, the following conclusion is not valid: "This implies that $h$ is always decreasing...". – 2'5 9'2 Aug 01 '14 at 02:18
  • But if $g(x)>0$ on $[0,(n-1)/n]$, then $g(0)$ is positive, so $0=h(0)>h(1/n)$. And $g((n-1)/n)$ is positive, so $h((n-1)/n)>h(1)=0$. So $h$ is negative at $1/n$ and positive at $(n-1)/n$. Now $g(1/n)$ is positive, so $0>h(1/n)>h(2/n)$. And $g((n-2)/n)$ is positive, so $h((n-2)/n)>h((n-1)/n)>0$. So now $h$ is negative at $2/n$ and positive at $(n-2)/n$. See where this is going? – 2'5 9'2 Aug 01 '14 at 02:26
  • Yes, I see my faulty logic now. I'm assuming I'd have to deal with the possibility that h can increase between the natural numbers. As stated in the answers, it's clear to me now that such a condition isn't fully necessary. Perhaps more true to the original idea of the proof, I'm assuming it suffices to say: $g(0) > g(1/n) > ...> g(1)$, which is a clear contradiction. – user28375028 Aug 01 '14 at 02:32
  • My mistake. Those are supposed to be h, not g. – user28375028 Aug 01 '14 at 02:39
  • You say g is decreasing from 0 to n-1/n. Okay, but you say that's impossible because g (1)=g (0). Well g can be decreasing from 0 to n-1/n but then jump right be to g(0) in the remaining n-1/n to 1. – fleablood Dec 18 '15 at 07:11

3 Answers3

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By contradiction, assume that:

$$ \forall x \in \left[0, \frac{n-1}{n} \right] \; h(x) > h \left(x + \frac{1}{n}\right) $$

Then

$$ h(0) > h \left( \frac{1}{n} \right) > h \left( \frac{2}{n} \right) > \dots > h(1) $$

But $h(0) > h(1)$ is obviously false since $h(0) = h(1) = 0$, hence our assumption must be false, that is:

$$ \exists x \in \left[0, \frac{n-1}{n} \right] : h(x) \leq h \left(x + \frac{1}{n}\right) $$

And with an analogous argument we can prove that:

$$ \exists x \in \left[0, \frac{n-1}{n} \right] : h(x) \geq h \left(x + \frac{1}{n}\right) $$

This proves that the continuous function

$$ g(x) = h(x) - h \left( x + \frac{1}{n} \right) $$

assumes both positive and negative values on the interval (or is constant), and hence it must have a zero.

sbabbi
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$$g(x)=h(x)-h\left(x+\frac{1}{n}\right)$$

$$g(0)=-h\left(\frac{1}{n}\right)$$ $$g\left(1-\frac{1}{n}\right)=h\left(\frac{1}{n}\right)$$

Because $g(0)$ and $g\left(1-\frac{1}{n}\right)$ have opposite signs...

rlartiga
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Let: $$g:x\in[0,1-\dfrac{1}{n}]\longmapsto h\left(x+\dfrac{1}{n}\right)-h(x)$$ Evaluate $g$ in $n$ different points: $$\left\lbrace\begin{array}{rcl} g(0)&=&{\color{green}{h\left(\dfrac{1}{n}\right)}}-0 \\ g\left(\dfrac{1}{n}\right)&=&{\color{blue}{h\left(\dfrac{2}{n}\right)}}-{\color{green}{h\left(\dfrac{1}{n}\right)}}\\ g\left(\dfrac{2}{n}\right)&=&h\left(\dfrac{3}{n}\right)-{\color{blue}{h\left(\dfrac{2}{n}\right)}}\\ \dots & = & \dots \\ g\left(\dfrac{n-1}{n}\right) & = & 0-h\left(1-\dfrac{1}{n}\right) \\ \end{array}\right.$$

Let's sum: $$\sum_{k=0}^{n-1} g\left(\dfrac{k}{n}\right)=0$$

So $g$ is either the zero function, or has strictly positive and strictly negative values. Conclude with continuity of $g$ and the intermediate value theorem!

anderstood
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