Continuous Function with Sequential Limits at Infinity has Limit at Infinity

Theorem
Let $f : \openint 0 \infty \to \R$ be a continuous function such that:


 * for each $x \in \openint 0 \infty$, the sequence $\sequence {\map f {n x} }$ converges to $0$.

Then:


 * $\ds \lim_{x \to \infty} \map f x = 0$

Proof
Fix $\epsilon > 0$.

For each $n \in \N$, define $g_n : \openint 0 \infty \to \R$ by:


 * $\map {g_n} x = \map f {n x}$

From Composite of Continuous Mappings is Continuous, we have:


 * $g_n$ is continuous for each $n$.

For each $m \in \N$, define the set $X_m$ by:


 * $X_m = \map { {g_m}^{-1} } {\closedint {-\epsilon} \epsilon} = \set {x \in \openint 0 \infty : \size {\map f {m x} } \le \epsilon}$

From Continuity Defined from Closed Sets, we have:


 * $X_m$ is closed for each $m$.

Now, for each $n \in \N$, define:


 * $\ds K_n = \bigcap_{i \mathop = n}^\infty X_i = \set {x \in \openint 0 \infty : \size {\map f {m x} } \le \epsilon \text { for all } m \ge n}$

From Intersection of Closed Sets is Closed, we have:


 * $K_n$ is closed.

We now prove that:


 * $\ds \openint 0 \infty = \bigcup_{n \mathop = 1}^\infty K_n$

Let:


 * $x \in \openint 0 \infty$

By hypothesis, we have that:


 * the sequence $\sequence {\map f {m x} }$ converges.

That is, there exists some $N \in \N$ such that:


 * $\size {\map f {m x} } \le \epsilon$

for $m \ge N$.

That is:


 * $x \in K_N$

so:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty K_n$

Since by construction we have:


 * $\ds \bigcup_{n \mathop = 1}^\infty K_n \subseteq \openint 0 \infty$

We have, by the definition of set equality:


 * $\ds \openint 0 \infty = \bigcup_{n \mathop = 1}^\infty K_n$

From Space of Positive Real Numbers in Non-Meager:


 * $\openint 0 \infty$ is not meager.

That is:


 * for some $n$, $K_n$ is not nowhere dense.

Fix this $n$, then we have:


 * $\paren {\overline {K_n} }^\circ$ is non-empty.

That is, there exists some non-empty open interval $\openint c d$ such that:


 * $\openint c d \subseteq \paren {\overline {K_n} }^\circ$

Since $\openint c d$ is non-empty, we have:


 * $\closedint a b \subseteq \openint c d$

for some $a, b \in \R$.

For $x \in \closedint a b$, we have:


 * $\size {\map f {m x} } \le \epsilon$

for $m \ge n$.

That is:


 * $\size {\map f {\paren {n + j} x} } \le \epsilon$

for all integers $j \ge 0$ and $x \in \closedint a b$.

Now, let:


 * $\ds K = \bigcup_{j \mathop = 0}^\infty \closedint {\paren {n + j} a} {\paren {n + j} b}$

For every $x \in K$, we have:


 * $\size {\map f x} \le \epsilon$

It remains to show that $K$ contains a closed interval of the form $\hointr N \infty$.

Then:


 * $\size {\map f x} \le \epsilon$

for $x \ge N$, so we would obtain the result.

Note that if:


 * $j \ge \dfrac a {b - a}$

we have:


 * $\paren {n + j} b \ge \paren {n + j + 1} a$

So, for integer $j$ with:


 * $j \ge \dfrac a {b - a}$

we have:


 * $\closedint {\paren {n + j} a} {\paren {n + j} b} \cap \closedint {\paren {n + j + 1} a} {\paren {n + j + 1} b} \ne \emptyset$

Let:


 * $\ds S = \bigcup_{j \in \N : j \ge \frac a {b - a} } \closedint {\paren {n + j} a} {\paren {n + j} b} \subseteq K$

From Countable Union of Overlapping Connected Sets is Connected:


 * $S$ is connected.

From Subset of Real Numbers is Interval iff Connected:


 * $S$ is an interval.

Note that $S$ is unbounded above and contains its infimum.

So $S$ has the form:


 * $S = \hointr N \infty$

for some $N \in \openint 0 \infty$, as required.