Find a function not satisfying f(x) = f(x+a)

**nimon** · February 6th 2013, 04:34 AM

So I'm trying to find a solution to this question. In part (a), I have shown that for a function $f$ which is continuous on $[0,1]$ satisfying $f(0) = f(1)$ then for any number $n\in\mathbb{N}$ there is some $x\in[0,1-1/n]$ for which $f(x) = f(x+1/n).$ I did this by considering the function $g(x) = f(x) - f(x+1/n)$ and assuming $g(x) \neq 0$ for any $x.$ We can then show that either $f(1) > f(0)$ or $f(1) < f(0)$ by noting that, since $g$ is continuous we must have $g(x)<0$ or $g(x) > 0$ for all $x.$ So, in the latter case for example, we have $g(x) > 0 \quad \Rightarrow \quad f(0) > f(1/n) > f(2/n) > \ldots > f(1).$

This clearly relies on the fact that $n \cdot \frac{1}{n} = 1$ with $n\in\mathbb{N}.$

Now for part (b), which has me totally stumped. For $0<a<1$ with $a \neq 1/n$ for any $n\in\mathbb{N},$ find a function $f$ continuous on $[0,1]$ satisfying $f(1) = f(0)$ for which $f(x) \neq f(x + a)$ for any $x \in [0,1-a].$ I just can't see how that could work. Imagine any function, and connect the points $(x,f(x))$ and $(x+a,f(x+a))$ with a straight line - surely if we have $f(0) = f(1)$ this line must be horizontal at some point!

How do I find such an $f$ ?

**Asharma** · February 6th 2013, 04:34 PM

What about a constant function?
If f(x)=c , where c is any constant then we can't find any a for which f(x)=f(x+a)

**nimon** · February 7th 2013, 02:18 AM

On the contrary, we would then have $f(x) = f(x+a) = c$ for all $x\in[0,1-a].$

**nimon** · February 7th 2013, 09:55 AM

Ok, I found the solution online. The graph of $f$ can be a saw function starting at $f(0)$ and with $f(a)>0$ which gradually climbs. Where $1/(n+1) < a < 1/n$ it has local minima of $-kf(a)$ at $1-ka$ for $k = 0 \ldots n$ and local maxima of $f(1-ka)+(n+1)f(a)$ at $ka$ . Now you can just connect the minima and maxima with straight lines. If you sketch it, you'll see that $f(x+a) -f(a) = f(a) > 0$ for all appropriate $x$ and that $f(1) = 0$ .

Now I've seen a solution, I think this was probably easier than I was making out. Did get me stuck for a bit though

**Asharma** · February 7th 2013, 03:41 PM

It's perfect

**johng** · February 7th 2013, 06:13 PM

I found this to be an interesting challenging problem. Anyway, I thought up a solution similar to the one you describe, but thought I'd post it anyway. The first image is of such a function f (a is .3) and the dotted red graph is f(x+a):

Find a function not satisfying f(x) = f(x+a)-mhfcalc.png

The entire description is in two image files (I find it much easier to use my own math editor and paste images):

Find a function not satisfying f(x) = f(x+a)-mhfcalc1.png

Find a function not satisfying f(x) = f(x+a)-mhfcalc2.png

**nimon** · February 8th 2013, 12:29 AM

Thank you for that johng,

I'm really impressed (and a bit jealous!) that you achieved this on your own! Apologies for not putting a spoiler alert on the thread, I didn't realise people were working on it. My own image is drawn below with a bit of labelling in case anyone is confused by my sloppy description above!

The function from my previous post.

In case you're interested, the problem is question 19.(b) of chapter 7 from Calculus by Spivak. This book has lots of problems that make you think like this and I highly recommend it.

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