Let $f :\mathbb R\rightarrow\mathbb R$ be a function such that $f(x + 1) = f(x)$ for all x ∈ R. Which of the following statement(s) is/are true?

Question

The given options are: (A) f is bounded. (B) f is bounded if it is continuous. (C) f is differentiable if it is continuous. (D) f is uniformly continuous if it is continuous.

Any hints on how to approach this sum? I tried using Mean Value Theorem, that did not seem to work. One observation regarding the function is that it will have same value across all integral points but how do I analyze the non-integral points?

Answer 1

You might just try to explicitly create counterexamples for some of them. You should also realize that it's easy to "periodize" a function by chopping it off at $1$ and repeating. That is, given any function $f(x)$, you can "periodize it" by making the function

$$ F(x) = \begin{cases} f(x) & x \in [\,0, 1) \\ f(x-1) & x \in [\,1, 2) \\ ... \end{cases}$$

Stated slightly differently, you can precompose $f$ with the natural quotient map $\mathbb{R} \longrightarrow \mathbb{R}/\mathbb{Z}$.

Then you can take functions like $\lvert x - \tfrac{1}{2} \rvert$ and periodize it to get continuous, periodic, not-differentiable functions.

Finally, I encourage you to really try at this (and related) questions. This sort of question is very good at building intuition, which is very important.