If $f:[0,1] \rightarrow \mathbb{R}$ is continuous function such that $f(0)=f(1)$ then there exists $ x\in [0,1]$ such that $f(x) = f(x+\frac{1}{n})$.

Question

If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.

Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.

Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$. If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or $[c_2,c_2-\frac{1}{n}]$ respectively. If $c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.

Is this proof correct? Thanks.

$c_1,c_2$ are definitely in $[0,1].$ Do you mean "If $c_1,c_2$ both do not belong to $\left[0,1-\frac{1}{n}\right],\cdots$ — Thomas Andrews, Oct 11 '18 at 15:56
You don't need to appeal to the maximum value theorem here. Just the IVT does it. — Ted Shifrin, Oct 11 '18 at 17:02
@Ted Shifrin How only IVT is sufficient? Can you please show your proof? Or give me some hint. — ramanujan, Oct 11 '18 at 17:04
@Ted Shifrin If we apply IVT to $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$, we get $g(0)=f(\frac{1}{n})-f(0)$ and $g(1-\frac{1}{n})=f(0)-f(1-\frac{1}{n})$. Now how to proceed? — ramanujan, Oct 11 '18 at 17:10
If $g\ne 0$, $g$ must always have the same sign, and then $f(0)=f(1)$ is impossible. — Ted Shifrin, Oct 11 '18 at 17:18

score 5 · Accepted Answer · answered Oct 11 '18 at 17:12

5

I don't see what your last paragraph leads to. And you must use the fact that $n\in \Bbb N$ because if $r\in (0,1)\setminus \{1/n: n\in \Bbb N\}$ then there is a continuous $f:[0,1]\to \Bbb R$ with $f(0)=f(1),$ such that $f(x)\ne f(x+r)$ for all $x\in [0,1-r].$

I saw this in American Mathematical Monthly. The title was (approximately?) The Theorem Of The Horizontal Chord. The idea is that $$\sum_{j=0}^{n-1}g(j/n)=\sum_{j=0}^{n-1}f((j+1)/n)-f(j/n)=$$ $$=f(1)-f(0)=0 $$ so the members of $\{g(j/n):j=0,...,n-1\}$ cannot be all positive or all negative, so by the IVT, $g(x)$ must be $0$ for some $x\in [0,1-1/n].$

answered Oct 11 '18 at 17:12

DanielWainfleet

57,985

Thanks. Very nice and elegant answer. By the way, in my last paragraph after applying IVT on $g$, $g$ will be negative at one end point and positive at other. So $g(x)=0$ somewhere between. – ramanujan Oct 11 '18 at 17:22
Yep,and this makes clear why other $\epsilon>0$ not of the form $\frac{1}{n}$ do not work. – Thomas Andrews Oct 11 '18 at 19:40
To the proposer: I still can't follow your last paragraph and I don't see what would differ in it if you replaced $1/n$ with an arbitrary $r\in (0,1).$ – DanielWainfleet Oct 14 '18 at 09:33
1

@DanielWainfleet sorry. Actually my proof is incorrect. – ramanujan Oct 21 '18 at 13:44

If $f:[0,1] \rightarrow \mathbb{R}$ is continuous function such that $f(0)=f(1)$ then there exists $ x\in [0,1]$ such that $f(x) = f(x+\frac{1}{n})$.

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