@PeterTamaroff Do you have available the fact that if $0<x$, there is a positive integer $n$ such that $\frac1n<x$, the Archimedean property?

@BrianM.Scott Yes.

@BrianM.Scott I have proven it.

Then you’re in business. If $f(x)\ne f(y)$, choose $n$ so that $\frac1n<\frac{|f(x)-f(y)|}{(y-x)^2}$, and get a contradiction.

What is there to prove? That's the supremum property or am I a reeetard now?

@JonasTeuwen You are a reeetard independently of the lub property.

Good.

Not extra deluxe now?

Gotta go to the Post Office. BBL

@BrianM.Scott Done. The proof looks pretty!

@PeterTamaroff It really is a rather nice one, yes.

@BrianM.Scott This is really valuable for me. Thanks.

No problem!

But I think that it is easy for differentiable onces, hence...

Which holds a.e. at least for this thing.

time for sleep - night all

Night.

@JonasTeuwen I have a new one. This is a $**$ in Spiavk's, meaning it is extra hard. I know you like when stuff is extra hard.

Suppose $f$ is continuous in $[0,1]$ and $f(0)=f(1)$. Let $n$ be any natural number. Prove there exists an $x$ such that $f(x)=f(x+1/n)$

I think I need to exploit the fact that for continuous $f$ $$\lim_{\epsilon \to0}f(x+\epsilon)=f(x)$$

If $\forall n\forall x$ $$g(x)=f(x)-f(x+1/n)$$ is $\neq0$, $f$ would not be continuous.

This stems for $\forall \epsilon >0 \exists n\in\Bbb N$ such that $$\frac 1 n <\epsilon$$.

@BrianM.Scott I'm on the right track, yes?

@PeterTamaroff Not sure: I don’t immediately see how this one’s going to go.

@BrianM.Scott Well, fix some $n$, and consider $g(x)=f(x)-f(x+1/n)$. The hypothesis is that $f$ is continuous. I want to prove that for any natural $n$, there is some $x$ with $f\left( x \right) = f\left( {x + \frac{1}{n}} \right)$ assuming $f(0)=f(1)$.

I'm failing to apply that last condition, namely $f(0)=f(1)$.

Without loss of generality I can assume $f(0)=f(1)=0$

Do you have available anything about connected (intermediate value theorem) yet?

@BrianM.Scott Yes. Not proven, but stated as a theorem.

The hint is "Consider $g(x)=f(x)-f(x+1/n)$: What would happen if $g(x)\neq 0$ for all $x$?"

You can use that to prove the $n=2$ case. My guess is that if you can find a way to prove the $n=3$ case, you’ll be able to prove the general result.

@BrianM.Scott Why does it matter if it is $n=2$ or $n=3$?

Try it and see.

@BrianM.Scott But wouldn't the negation consists of $\forall n\forall x$, $g(x)=f(x)-f(x+1/n)\neq0$?? Then, I can use the definition of limit and the Archimedean property to prove that $\lim_{\epsilon \to 0} f(x+\epsilon)=f(x)$ is false, can't I?

However, I think that the hint gives an easier approach. Suppose that $g(0)>0$ and that for $g(x)\ne 0$ for $x\in[0,1-1/n]$. Then by continuity $g(x)>0$ for all $x\in[0,1-1/n]$. Then $g(0)>g(1/n)>g(2/n)>\ldots>g(1)$.

@BrianM.Scott Why does $g(x)>0$ by continuity?

@PeterTamaroff Intermediate value theorem.

@PeterTamaroff No: if you could, you could prove that $f(x)=x$ wasn’t continuous!

@BrianM.Scott Oh. GAWD. How?!

It satisfies the same hypothesis: $f(x)-f(x+1/n)\ne 0$ for every $x$ and $n$.

Summation of g gives a neat proof: math.stackexchange.com/questions/16374/universal-chord-theorem

Rather, it ends up being equivalent to the one above :)