Home > Analysis, Topology > Asymptotic characterization in terms of sequence limits

Asymptotic characterization in terms of sequence limits


Suppose {f:(0,\infty) \rightarrow \Bbb{R}} is a continuous function such that for every {x>0} we have

\displaystyle \lim_{n \rightarrow \infty} f(nx)=0.

Prove that {\lim\limits_{x \rightarrow \infty} f(x)=0}.

Solution: In problems like this Baire’s theorem almost suggests itself. For some {\varepsilon>0} define the following sets

\displaystyle A_n = \bigcap_{m \geq n}^\infty \{x : |f(mx)| \leq \varepsilon \}

and note that the hypothesis implies that for every {x >0} there exists {n_x \in \Bbb{N}} such that {x \in A_{n_x}}. This proves that {\bigcup_{n=1}^\infty A_n =(0,\infty)}. Moreover, from the definition we have {A_n \subset A_p} for {n\leq p}. The continuity of {f} assures us that the sets {A_n} are closed.

The Baire category theorem implies that at least one {A_n} has non-void interior and that means that there exists a non-trivial interval {I} such that

\displaystyle \forall x \in I, \forall m \geq n \Rightarrow |f(mx)|\leq \varepsilon.

The main idea now is to realize that {\bigcup_{m \geq n} mI} contains every real number which is sufficiently large. To do this, suppose {I=(a,b)} with {a<b} and note that the inequality {nb<(n+1)a} can only hold for a finite number of values for {n}.

Thus we have proved that for {\varepsilon>0} there exists {K_\varepsilon} such that {|f(x)|\leq \varepsilon} for every {x \geq K_\varepsilon}. This is exactly the definition of the fact that the limit of {f} at {\infty} is zero.

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