# Show that for any positive integer $n$ there is $x \in [0,1-\frac{1}{n}]$ for which $f(x) = f(x+\frac{1}{n})$

Suppose $f$ is a continuous function over $[0,1]$ such that $f(0) = f(1).$ Show that for any positive integer $n$ there is $x \in [0,1-\frac{1}{n}]$ for which $f(x) = f(x+\frac{1}{n})$.

We seem to be saying that $f(x+\frac{1}{n})$ is periodic with respect to any positive integer $n$. Also this seems to make sense since we start and end at the same spot there have to be values of $x$ with the same $f(x)$ as other $x$'s. But I am struggling to see how to prove this specific statement.

• It's much less than being periodic. Being periodic would be "for all $x\in \bigl[0,1-\frac{1}{n}\bigr]$", we only have "there exists an $x\in \bigl[0,1-\frac{1}{n}\bigr]$". Commented Feb 23, 2016 at 15:54
• @DanielFishcher That's why I said with respect to $n$. Commented Feb 23, 2016 at 15:59

Consider $g(x) = f(x) - f(x+\frac{1}{n})$. We want to prove that $g(x) = 0$ for some $x\in [0,1-\frac{1}{n}]$. Consider the values $g(0), g(1/n), g(2/n), \ldots, g(1-1/n)$. At least one of these values must be negative and at least one must be positive (if one is zero, we're done). Otherwise, they are all positive or all negative. But the first case would mean $f(0) > f(1/n) > f(2/n) > \cdots > f(1-1/n) > f(1)$ which is false, and the second case would mean $f(0) < f(1)$. Thus, one of the values is positive and one is negative, and therefore by continuity $g(x) = 0$ for some $x$ in the given interval.