I'd appreciate if somebody could check if my proofs below are correct and also give me some hints on the equicontinuity of $(k_n)$ (part c)). $\mathbb{R}$ represents the real line with the standard topology.
Consider the following sequences of functions in $\mathcal{C}(\mathbb{R}, \mathbb{R})$, (the set of continuous functions $f : \mathbb{R} \to \mathbb{R}$):
a) $f_n (x) = n + \sin(x)$
b) $h_n (x) = |x|^{1/n}$
c) $k_n (x) = n\sin(x/n)$
where $n\ge 1$.
Question
Which sequences are equicontinuous / pointwise bounded?
Answer
a) $(f_n)$ is not pointwise bounded, since given any $x, |n + \sin(x)| \to \infty$. For equicontinuity at a point $x$, we need to show that given any $\epsilon > 0$, there is an open neighborhood $U$ of $x$ such that
$$ \forall n\ge 1, \forall x^* \in U: |f_n(x^*) – f_n(x)| < \epsilon $$
Since $x \mapsto \sin(x)$ is continuous, let $U$ be an open set containing $x$ such that
$$x^* \in U \implies |\sin(x^*) – \sin(x)| < \epsilon$$
Since $|f_n(x^*) – f_n(x)| = |\sin(x^*) – \sin(x)|$ for all $n$, we see that $U$ satisfies the required condition. Since $x$ was arbitrary, we conclude that the sequence $(f_n)$ is equicontinuous.
b) $(h_n)$ is indeed pointwise bounded, since for any given $x$, $|x|^{1/n} \to 1$. I claim that it is not equicontinuous, since it is not equicontinuous at $x=0$: Assume it is equi. cont. at $x=0$. Pick $\epsilon = 1/2$. Then there exists $(-\delta, \delta)$ for some $\delta > 0$ such that
$$ \forall x^* \in (-\delta, \delta) \land \forall n\ge 1: |x^*|^{1/n} < 1/2$$
But this is not possible, since $|x^*|^{1/n} \to 1$.
c) $(k_n)$ is clearly not pointwise bounded. For equicontinuity, I'm a bit lost. Any tips?
Best Answer
I agree with your answers to 1) and 2).
Regarding 3), $k_n$ is pointwise bounded as for any $x \in \mathbb R$ you have
$$\left\lvert k_n(x) \right\rvert = \left\lvert n \sin \frac{x}{n} \right\rvert \le n \left\lvert \frac{x}{n} \right\rvert = \lvert x \rvert.$$ Moreover, $k_n$ is equicontinuous as for any $x,y \in \mathbb R$ you have the inequality (that can be proven using mean value theorem)
$$\lvert \sin x - \sin y \rvert \le \lvert x - y \rvert$$ and therefore $$\lvert k_n(x) - k_n(y) \rvert \le \lvert x- y \rvert.$$ So, $k_n$ is even uniformly equicontinuous.