Uniformly continuous function on a set that has sequence with accumulation point cannot have $f(x_n)\rightarrow \infty$

Question

Need help in proving the following statement;

Suppose that $f$ is uniformly continuous on a set S, that $\{x_n\}_{n=1}^{\infty}$ is a sequence in S, and that $\{x_n\}_{n=1}^{\infty}$ has an accumulation point $x\in\mathbb{R}$. Prove that it is impossible to have $f(x_n) \rightarrow \infty$ as $n\rightarrow \infty$.

I'm unable to think systematically about this. This is what we know from the statement;

$f$ is uniformly continuous on $S$, this means that for every $\epsilon > 0$ there exists a $\delta >0$ so that if $x,y \in S$ and $|x-y|<\delta$ then $|f(x) -f(y)|<\epsilon$.

Also since $\{x_n\}_{n=1}^{\infty}$ has a partial limit in $S$, this means that there exist $\delta >0$ such that the sequence $\{x_n\}_{n=1}^{\infty}$visits $(x-\delta,x+\delta)$ infinitely often.

I'm not sure how to use the above two definitions to prove that $f(x_n)$ cannot approach $\infty$ as $n\rightarrow\infty$.

Will appreciate any help/hints.

Answer 1

Suppose that $f(x_n)\rightarrow \infty$ as $n\rightarrow \infty.$

First, since $x$ is an accumulation point of $(x_n)$, there is a subseq $(x_{n_{k}})$ such that $x_{n_k}\rightarrow x$, and then $f(x_{n_k})\rightarrow f(x)$ since $f$ is uniformly continuous. (here we use the property that uniformly continuous function can be extented to its boundary)

Now, since $f(x_{n_k})\rightarrow f(x)$, there is $k_0\in \mathbb{N}$ such that for all $k\geq k_0$, we have $$ |f(x_{n_k})|\leq |f(x)|+1. $$

On the other hand, since $f(x_n)\rightarrow \infty$, there is $N\in\mathbb{N}$ such that for all $n\geq N$, we have $$ |f(x_n)|>|f(x)|+1. $$ However, $n_k\rightarrow \infty$, and hence there is a $k'$ such that $n_{k'}\geq N$, and $k'\geq k_0$, we obtain $$ |f(x_{n_{k'}})|\leq |f(x)|+1,\ \ \mbox{and also}\ |f(x_{n_{k'}})|>|f(x)|+1. $$

Answer 2

We recall that $f (x_n)\to \infty$ if for all $M\in \mathbb {R}$, there is an $N\in \mathbb {N} $ such that $f (x_n) > M $ for all $n\geq N $ (without loss of generality let $x_n\to x $, as otherwise, we can just let $x_{n_k}\to x $ and define $y_k = x_{n_k}$, and then the rest of the argument holds with $y_k $ instead of $x_n $). For some $\epsilon > 0$, we choose $m$ such that $\| x-x_n\| < \frac {\delta}{2} $ for all $n\geq m $, where $\delta$ is chosen such that $\|x-y\| < \delta $ implies $\lvert f (x)-f (y)\rvert < \epsilon $ for $x $, $y\in S $. Then, for $M > f (x_m)+\epsilon $, if $n\geq m $, then $$\| x_n-x_m\|\leq \| x-x_m\|+\| x-x_n\| < \delta $$ and $$\lvert f (x_n)-f (x_m)\rvert < \epsilon \Rightarrow f (x_n) < M $$ so there is no $N $ such that $f (x_n) > M $ for all $n\geq N $.

Answer 3

The key here is this neat little lemma:

Let $X$ and $Y$ be complete metric spaces and suppose $f:S\rightarrow Y$ is a uniformly continuous function on a dense set $S\subset X$. Then there exists a unique uniformly continuous extension, i.e. a uniformly continuous function $F:X\rightarrow Y$ on $X$ such that $F\big|_S=f$.

The proof is elementary so I'll just provide the outline.

• Show that if $\{x_n\}$ is Cauchy in $S$ then $\{f(x_n)\}$ is Cauchy in $Y$.
• Show that if $\{y_n\}$ is any other Cauchy sequence in $S$ with $d_X(x_n,y_n)\to0$, then $d_Y(f(x_n),f(y_n))\to0$.
• Define $F(x):=\lim_{n\to\infty}f(x_n)$, where $\{x_n\}$ is any sequence converging to $x$. Use the previous parts to argue this uniquely defines a function $F$.
• Prove that this function satisfies the requirements of the lemma.

In this case, $S$ is dense in $\overline S\subseteq\mathbb R$ which is complete. If $F$ is the uniformly continuous extension of $f$, and $x_{n_k}\to x$, then obviously $F(x_{n_k})\to F(x)\in\mathbb R$. In particular $f(x_n)=F(x_n)\not\to\infty$.

This lemma also has many applications in functional analysis, where it allows us to define bounded linear operators only along a dense subset. We can use this method to define the Ito isometry and Brownian motion, for example.