Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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.
