The problem asks you to prove that if a function $f : \mathbb{R} \to \mathbb{R}$ is periodic and continuous, then it attains both a maximum and a minimum on $\mathbb{R}$. To do this, we can use the Weierstrass Extreme Value Theorem.

📚 Weierstrass Extreme Value Theorem

The theorem states:

• If a function $f$ is continuous on a closed and bounded interval $[a, b]$, then it attains both a maximum and a minimum value on that interval.

✍️ Proof Outline

1. Periodicity and Reduction to a Closed Interval:

   • Since $f$ is periodic, there exists a period $T > 0$ such that: $f(x + T) = f(x) \quad \forall x \in \mathbb{R}$
   • It suffices to examine the behavior of $f$ on one period, say the interval $[0, T]$.

2. Application of Weierstrass Theorem:

   • The interval $[0, T]$ is closed and bounded.
   • The function $f$ is continuous on $[0, T]$.
   • By the Weierstrass Extreme Value Theorem, $f$ attains both a maximum and a minimum on $[0, T]$.

3. Extending to $\mathbb{R}$:

   • Due to periodicity, the maximum and minimum on $[0, T]$ repeat on all intervals $[nT, (n+1)T]$, where $n$ is any integer.
   • Therefore, $f$ attains its maximum and minimum on all of $\mathbb{R}$.

✅ Conclusion

If $f$ is periodic and continuous, it attains both a maximum and a minimum on $\mathbb{R}$.

Would you like a more detailed explanation of any step or further examples?

🔢 Related Questions:

1. What happens if the function is periodic but not continuous?
2. Can you prove the Weierstrass Theorem directly?
3. How does periodicity simplify the analysis of functions?
4. What are some real-world examples of periodic continuous functions?
5. How would the proof change if the domain were restricted to a finite interval?

💡 Tip:

When dealing with periodic functions, always reduce the analysis to one period. It simplifies the problem significantly.