Continuous function satisfying $f(x+1) = f(x)$ 
Let $f : \mathbb{R} \to \mathbb{R} $ be a continuous function and $f(x+1) = f(x)$ for
  all $x \in \mathbb{R}$. Then
  
  
*
  
*$f$ is bounded above, but not bounded below.
  
*$f$ is bounded above and bounded below, but may not attain its bounds.
  
*$f$ is bounded above and bounded below, and attain its bounds.
  
*$f$ 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!
 A: 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 $[-1,1]$.

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

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

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