Continuous Function with Sequential Limits at Infinity has Limit at Infinity

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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:

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


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:

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

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:

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

Let:

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

From Union of Connected Sets with Common Point 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.

$\blacksquare$