Given a continuous, unbounded function ( f: \mathbb{R} \to \mathbb{R} ), we need to determine if there exists an ( x \in \mathbb{R} ) such that the set ( { |f(nx)| \, | \, n \in \mathbb{Z} } ) is unbounded.
- Key Considerations:
- The function ( f ) is continuous and unbounded.
- We need to check if there exists an ( x ) such that ( |f(nx)| ) is unbounded as ( n ) varies over the integers.
- Examples and Counterexamples:
- For simple functions like ( f(y) = y ) or ( f(y) = y^2 ), every non-zero ( x ) works.
- For oscillating functions like ( f(y) = y \sin(y) ), there exist ( x ) values (e.g., ( x = \pi/2 )) where ( |f(nx)| ) is unbounded.
- Constructed examples with sparse peaks (e.g., at ( 2^{2^k} )) were considered, but the argument using the Baire Category Theorem showed that such ( x ) must exist.
- Baire Category Theorem Argument:
- For each ( M ), define ( A_M = { x \in \mathbb{R} \, | \, \exists n \in \mathbb{Z} \text{ with } |f(nx)| > M } ).
- Each ( A_M ) is open and dense because ( f ) is continuous and unbounded.
- By the Baire Category Theorem, the intersection ( \bigcap_{M=1}^\infty A_M ) is dense, hence non-empty. This intersection consists of ( x ) such that ( { |f(nx)| } ) is unbounded.
- Conclusion:
- Despite initial counterexamples with sparse peaks, the Baire Category Theorem ensures the existence of such ( x ). The continuity of ( f ) and the density arguments guarantee that even in such cases, there exist ( x ) values where ( |f(nx)| ) is unbounded.
Thus, the answer is (\boxed{Yes}).
