# Construct a continuous function $f$ over $[0,1]$ satisfying $f(0) = f(1)$ but $f(x) \neq f(x+a)$

Suppose $0 < a < 1$ is not of the form $\dfrac{1}{n}$ for positive integer $n$. Construct a continuous function $f$ over $[0,1]$ satisfying $f(0) = f(1)$ but $f(x) \neq f(x+a)$ for all $x \in [0,1-a]$.

This is a follow up question to this. I am wondering how it is possible to construct such a function. I would start by saying $g(x) = f(x)-f(x+a) \neq 0$ for all $x$ in the domain and then doing casework on the values of $g(x)$. But this unlike the last question doesn't have a nice casework for the values of $g(x)$ so I am stuck.

• Hint: try a function of the form $\cos(2\pi x/a) + mx$ for a well-chosen $m$. Commented Feb 24, 2016 at 17:43
• We must have $1 = \cos{2\pi/a}+m$. Commented Feb 24, 2016 at 17:45

Let $n$ be the largest integer such that $na < 1$.

Let $g$ be any continuous function on $[0, a]$ such that

$$g(0) = 0$$ $$g(1-na) = -n$$ $$g(a) = 1$$

Then choose $f(ka+x) = g(x) + k$ for $k \in \mathbb{N}, x \in [0,a)$

Edit: I drew a picture of $f$:

Basically, $f$ is found by first setting $f(x+a) = f(x) + 1$, with $f(0) = f(1) = 0$. This gives you all the points in the above drawing. Then choose a continuous $g$ on the first 3 points, copy it and translate it by $(a,1)$ a bunch of times to obtain $f$.

• How does $f(x) \neq f(x+a)$ here? Commented Feb 24, 2016 at 18:57
• @user19405892 $f(x + a) = f(x) + 1$ Commented Feb 24, 2016 at 19:02
• What was the point of saying $g(1-na) = -n$? Commented Feb 24, 2016 at 19:20
• @user19405892 This gives $f(1) = f(1-na) + n = g(1-na) + n = -n + n = 0$ Commented Feb 24, 2016 at 19:21
• Why did you define $n$ to be the largest integer such that $na < 1$? Was that really necessary? Commented Feb 24, 2016 at 21:01