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This particular question was asked my senior's quiz and I was unable to solve it. So, I am asking for help here.

Let $f:\mathbb{R}\to\mathbb{R}$ 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

By using definition of continuity along with boundedness definition, I can say that (B) is true.

A function can be defined on $x=1/2 + \mathbb{Z} $ such that it's $\infty $ at $1/2 + \mathbb{Z} $ and finite elsewhere, so (A) becomes false.

But I am unable to think about (C) and (D) .

Answer is :

B,D.

Kindly help.

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3 Answers 3

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The condition on $f$ is equivalent to saying that $f$ is a periodic function with period 1. Therefore we should really think of $f$ as a function on the unit interval $[0, 1]$ with $f(0) = f(1)$.

An arbitrary function on $[0, 1]$ has no nice properties like boundedness, so we can cross it off. If we're continuous, though, since $[0, 1]$ is compact, $f$ must be bounded. This gives us B.

For C, there are many continuous functions which are not differentiable (in fact, most continuous functions are not differentiable, in some sense).

For D, we use that continuity on a compact set implies uniform continuity.

The takeaway is that the condition $f(x) = f(x + 1)$ lets us just think of a function on $[0, 1]$ (with $f(0) = f(1)$) instead of the whole real line.

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A is false: $\cot(\pi x)$ if $x\notin \mathbb{Z}$ and $f(0)=0$

B is true: if it's continuous on $[0,1]$, it's bounded there by the extreme value theorem and periodicity gives boundedness on all of $\mathbb{R}$

C is false: consider, for instance, the Blancmange function

D is true: continuous on $[0,1]$ implies uniform continuity and again we use periodicity

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a) is false: take for example $f(x)$ to be the Conway base 13 function on $[0,1)$, and then simply demand $f(x+1) = f(x)$ for all $x$.

b) is true. If $f$ is continuous then the image of $[0,1]$ under $f$ is compact, hence the image of $\mathbb{R}$ under $f$, which is equal to the image of $[0,1]$ is also compact, hence bounded.

c) This is false. Take any continuous but not differentiable function $f$ on $[0,1]$ with $f(0) = f(1)$, and then extend periodically by setting $f(x+1) = f(x)$ for all $x$.

d) This is true. To show this we simply need to recall that any continuous function on a closed interval is automatically uniformly continuous.

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