Continuous function satisfying $f(x+1) = f(x)$

Question

Let $f : \mathbb{R} \to \mathbb{R} $ be a continuous function and $f(x+1) = f(x)$ for all $x \in \mathbb{R}$. Then

1. $f$ is bounded above, but not bounded below.

2. $f$ is bounded above and bounded below, but may not attain its bounds.

3. $f$ is bounded above and bounded below, and attain its bounds.

4. $f$ is uniformly continuous.

I have answered that 3 and 4 are correct.

Reason: We can see the image of $f$ depends on the image of $[0,1]$ under $f$.

Now , since $f$ is continuous and $[0,1]$ is a closed and bounded interval therefore $f$ is uniformly continuous on it and hence uniformly continuous on the whole of $\mathbb{R}$.

Also $f$ being continuous will attain its bound on the closed and bounded interval $[0,1]$ and the since the image of $f$ is in some sense same as image of $[0,1]$ under $f$ , therefore the bounds of $f$ on $\mathbb{R}$ will be same as that in $[0,1]$

Is this reasoning correct?

Thanks in advance!

Answer 1

I think that 3 and 4 are correct. If we look at $f$ restricted on $[0,1]$, which is compact, then it is uniformly continuous by Heine-Cantor theorem and bounded from above (by $\sup_{x\in [0,1]}f(x)$) and from below (by $\inf_{x\in [0,1]}f(x)$). The bounds are attained because it is a continuous map on a compact space.

All this properties remains true also for $f$ from $\mathbb{R}$ to $\mathbb{R}$. It is maybe useful to note the following.

For uniform continuity consider an arbitrary $\epsilon>0$ choose $\hat{\delta}>0$ such that for all $x,\,y\in[0,1]$ with $|x-y|<\hat{\delta}$ we have $|f(x)-f(y)|<\epsilon$. Such a $\hat{\delta}$ exists by above discussion. Now, by periodicity, if we take $x,\,y\in[n,n+1]$ ($n\in\mathbb{N}$) there is no problem. Indeed, for $\alpha,\,\beta\in(0,1)$, $|\alpha-\beta|<\hat{\delta}$ and $x=n+\alpha$, $y=n+\beta$, we have \begin{equation} |f(x)-f(y)|=|f(\alpha)-f(\beta)|<\epsilon. \end{equation} This argument, however, does not consider the case $x\in[n,n+1)$, $y\in(n+1,n+2]$ with $|x-y|<\hat{\delta}$ and $n\in\mathbb{N}$. Indeed, using just periodicity it seems hard to motivate $|f(x)-f(y)|<\epsilon$. However, $f$ is continuous, in particular continuous in $n+1$. Thus there exists a $\delta_{n+1}>0$, for which \begin{equation} |x-(n+1)|<\delta_{n+1} \Rightarrow |f(x)- f(n+1)| < \frac{\epsilon}{2}. \end{equation} By periodicity, such $\delta_m$ is the same for all $m\in\mathbb{Z}$. Taking $\delta = \min\{\hat{\delta},\delta_m\}$ we are done.

Answer 2

A proof for 4.

Let $[x]$ denote the integer closest to $x$.

Since $f$ has period $1$, $\forall p\in \mathbb Z, f(x+p) = f(x)$.

Since $f$ is continuous over $[-1,1]$, it is uniformly continuous over $[-1,1]$.

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Let $\epsilon >0$.

By uniform continuity of $f$ over $[-1,1]$, there is some $\delta>0$ such that $$|x|\leq 1, |y|\leq 1, |x-y|\leq \delta \implies |f(x)-f(y)|\leq \epsilon$$

Let $\gamma =\min(\frac 12, \delta)$.

Consider $x$ and $y$ two real numbers such that $|x-y|\leq \gamma$. Without loss of generality assume that $x<y$.

Note that $|f(x)-f(y)| = |f(x-[x])-f(y-[x])|$

By definition of $[x]$, $$|x-[x]|\leq \frac 12 \leq 1$$

and $$|y-[x]|\leq|y-x| + |x-[x]|\leq \gamma + \frac 12 \leq \frac 12 + \frac 12 = 1$$

Moreover, $$|(x-[x]) - (y-[x])| = |x-y| \leq \gamma \leq \delta$$

Using uniform continuity of $f$ over $[-1,1]$ with $(x-[x])$ and $(y-[x])$ yields $$|f(x-[x])-f(y-[x])|\leq \epsilon$$

Hence $|f(x)-f(y)|\leq \epsilon$.

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Putting everything together, $$\forall \epsilon > 0, \exists \gamma >0, \forall x,y, |x-y|\leq \gamma \implies |f(x)-f(y)|\leq \epsilon$$

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3. is true as well, quite simple to prove formally.