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

(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.

It is a periodic function with period 1. I don't know how to proceed further...

Any Hints will be appreciated...

Mirunalini_UML
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  • For A try for example $f(x) = \frac{1}{\sin(2\pi x)}$ with $f(x)=0$ for points where $f$ is not defined. In general: to disprove a statement you only need (and the simplest way if often) to construct a single counter-example. – Winther Apr 20 '15 at 15:42

3 Answers3

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A. It doesn't have to be bounded: Does a periodic function have to be bounded?

B. If it is continuous, then the image $f(\mathbb R) = f([0,1])$ is compact as $[0,1]$ is compact and hence must be bounded. If results about compact sets is not familiar to you, use the extreme value theorem.

What are your thoughts on C and D?

Community
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Simon S
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For C, consider $f(x) = |x - 1/2|$ for $x \in [0,1]$, and then extend to all $x$ by using $f(x+1) = f(x)$. For D, note that a continuous function on a closed interval is uniformly continuous, because the closed interval is compact. Can you see how to apply that to your case?

user2566092
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if $S$ denotes the unit circle then there is a continuous map $\varphi:\mathbb{R} \to S$ which sends $r \mapsto e^{2\pi r i}$ any function satisfying $f(x)=f(x+1)$ can be written as $\hat f \circ \varphi$ where $\hat f:S \to \mathbb{R}$. so you only need think of the properties of $\hat f$, which has a compact domain

David Holden
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