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

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

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

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

How do I find such an $\displaystyle f$?

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

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)

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

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

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

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

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 (Wondering)

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

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

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):

Attachment 26896

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

Attachment 26898

Attachment 26897

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

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.