Differentiability of a function on $\mathbb R$ such that $f(x+1)=f(x)$.

Question

Let, $f:\mathbb R \to \mathbb R$ be a function such that $f(x+1)=f(x)$ for all $x\in \mathbb R$. Then which of the followings are correct?

(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.

Here, we find that $f$ is a periodic function with period $1$. So, it is bounded. So, option (a) is true & (b) is false. Am I right?

But what about the continuity & differentiability?

Answer 1

First: Logically (a) implies (b), so it cannot be that (a) is true and (b) is false. In fact $$\mbox{bounded $\Rightarrow$ (continuous $\Rightarrow$ bounded )}$$

Now, one can see that if $f$ is continuous, then it is bounded on $[0,1]$ by the Weierstrass theorem. By periodicity, it is bounded on the whole $\mathbb{R}$. So (b) is true. Moreover $f$ is also uniformly continuous using hte same argument (and Heine - Cantor theorem). So (d) is true.

A counterexample for (a) s $f(x) = \tan (\pi x) $ extended by $f(\frac{1}{2}) = f(1) = \dots = 0$, which is unbounded and periodic. So (a) is false.

Finally, (c) is clearly false. A counterexample can be constructed just extending by periodicity a continuous non differentiable function on $[0,1]$ having $f(0) = f(1)$.

Answer 2

a) is false, periodic doesn't imply boundedness, say $f(x)=\tan(\pi x)$ for $x\in\mathbb{R}\setminus\mathbb{Z}$, and $f(x)=0$ for $x=n\in \mathbb{Z}$ .

b) and d) is correct. Since it's periodic, we only need to consider $[0,1]$. If $f(X)$ is continuous, it's continuous on $[0,1]$ hence uniformly continuous, then use periodic, we have uniformly continuous and bounded on $\mathbb{R}$.

c) is false, just pick a uniformly continuous but not differentible function, say $f(x)=|x| ,x\in [-\frac{1}{2},\frac{1}{2}]$ .

Answer 3

Yes, you are right, that $f$ is periodic with period $1$. So it is sufficient to know what values $f$ has at $[0,1)$. To generate all such $f$, take any! function $g:[0,1)\to\mathbb R$. Then $f$ is given by $f(x)=g(x-\lfloor x\rfloor)$ and $f=g$ on $[0,1)$.

a) false: Take $g(x)=\frac{1}{1-x}$. Then $g$ is unbounded, and so is $f$.

b) true: If $f$ is countinous, then $f$ is bounded on $[0,1)$. But $f=g$ on $[0,1)$, so $g$ is bounded. Thus, $f$ is bounded on all of $\mathbb R$.

c) false: If $g$ is continuous with $\lim_{x\to1}g(x)=g(0)$, then $f$ is continuous. But if $g$ is not differentiable, then so $f$ cannot be differentialbe. Thus, we take $g(x)=|x-1/2|$, which is not differentiable at $x=\frac{1}{2}$, which makes $f$ not differentiable at each $x\in\mathbb Z+\frac{1}{2}$. And since $\lim_{x\to1}g(x)=\frac{1}{2}=g(0)$, $f$ is continuous. (In fact, $f$ is not differentiable even at each $x\in\frac{1}{2}\mathbb Z$.)

d) true: Since $f$ is continuous, $g$ must be continuous as well and $\lim_{x\to1}g(x)=g(0)$, so we can assume that $g$ is defined on $[0,1]$ and is continuous there with $g(0)=g(1)$. Thus, $g$ is uniformly continuous, since $[0,1]$ is compact. But then $f$ is uniformly continuous as well.