The question asks you to prove that if $f : \mathbb{R} \to \mathbb{R}$ is periodic and continuous, then $f$ attains a maximum and a minimum value on $\mathbb{R}$.

Hint for the proof:

1. Periodic Function Definition:

   • If $f$ is periodic, there exists a period $T > 0$ such that $f(x + T) = f(x)$ for all $x \in \mathbb{R}$.

2. Focus on One Period:

   • Since the function repeats every $T$, it suffices to show that $f$ attains a maximum and a minimum in the interval $[0, T]$.

3. Continuity and Compactness:

   • The interval $[0, T]$ is closed and bounded, so it is compact.
   • By the Extreme Value Theorem, a continuous function on a compact interval must attain both a maximum and a minimum on that interval.

4. Extend the Result:

   • Because $f$ is periodic, the maximum and minimum on $[0, T]$ will apply to the whole domain $\mathbb{R}$.

Suggested Steps:

• Prove that $f$ is bounded on $[0, T]$ and achieves its bounds (using the Extreme Value Theorem).
• Use the periodicity of $f$ to argue that the bounds repeat for all $x \in \mathbb{R}$.

Let me know if you'd like a full solution or clarification!

Relative Questions:

1. What does the Extreme Value Theorem state, and how is it applied here?
2. Why does periodicity allow us to limit our analysis to one period?
3. Can $f$ have different maximum/minimum values in other periods?
4. What role does continuity play in ensuring $f$ is bounded?
5. What happens if $f$ is not continuous or not periodic?

Tip:

Always verify compactness (closed and bounded) before applying the Extreme Value Theorem.