Continuous and periodic function from $\Bbb R$ to $\Bbb R$

Question

Let $f : \mathbb R \to \mathbb R$ be a continuous function and $f(x + 1) = f(x)$, $\forall x \in \Bbb R$ . Then

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

2. $f$ is bounded above and below but may not attain it's bounds.

3. $f$ is bounded above and below and $f$ attains its bounds.

4. $f$ is uniformly continuous.

My Attempt

If I take constant function then options 1,2 discarded. Help me.

Answer 1

Hint: Instead of thinking of a periodic function, you might find it easier to think of an equivalent construct: take $f$ to be a function over $[0,1]$ for which $f(0) = f(1)$.

You have probably already seen some useful statements that apply to functions over the closed interval $[a,b]$; all of those apply here.

Answer 2

Hints: on $[0,1]$ the function is bounded and attains its bounds. Since $f(x+n)=f(x)$ for any integer $n$ and any real number can be written as $y+n$ with $y \in [0,1]$ it follows that 3) is true.

For 4) use the fact that $f$ is periodic uniformly continuous on $[-1,2]$. [Let $\delta <1$. Then $|x-y| <\delta$ implies that $x=n+u$ and $y=n+v$ for some $n$ and some $u, v \in [-1,2]$ with $|u-v| <\delta$]. So 4) is also true and this implies that 1) and 2) are false.