Sequence $(f_n)$ equibounded, equicontinuous but with not uniformly convergent subsequence

Question

Let $f_n:[0,\infty)\rightarrow R$, given by $f_n(x) = \sin(\sqrt{x+4\pi^2n^2})$. Prove that $(f_n)$ is equicontinuous and equibounded, but has no uniformly convergent subsequence.

I'm having trouble to prove that this sequence is equicontinuous and has no uniformly convergent subsequence.

Clearly, $|f_n(x)\rvert= \lvert\sin\sqrt{x+4\pi^2n^2}|\leq1$ for every $x$ and every $n\in N$. So $(f_n)$ is equibounded.

Now in order to prove that it is equicontinuous, what I have noted so far using the mean value theorem is that:

$$|f_n(x)-f_n(y)|=$$

$$|\sin\sqrt{x+4\pi^2n^2}-\sin\sqrt{y+4\pi^2n^2}|\leq\bigg|\frac{\cos\sqrt{c_n+4\pi^2n^2}}{2\sqrt{c_n+4\pi^2n^2}}\bigg||x-y|\leq\bigg|\frac{1}{2\sqrt{c_n+4\pi^2n²}}\bigg||x-y|$$

but I can't control the term $\bigg|\frac{1}{2\sqrt{c_n+4\pi^2n²}}\bigg|$ properly, since it is dependent of $n$.

Also, any help on how to prove that there is no uniformly convergent subsequence?

Thanks in advance!

Answer 1

Let's prove that this sequence of function is equicontinuous. Indeed fix $\epsilon > 0$, then consider $\delta > 0$ for a $\delta$ to be chosen shortly. Now consider $|f_n(x) - f_n(y)|$ = $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$. Indeed, now (assume $x < y$) by MVT we see there exists a $\xi \in (x,y)$ such that $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$ = $\frac{\cos(\sqrt{4 \pi^2 n^2 + \xi}}{2 \sqrt{4 \pi^2n^2 + \xi}}$$|x-y|$ $\leq$ $\frac{1}{2 \sqrt{4 \pi^2n^2 + \xi}}|x-y|$. Notice that $x \in [0, \infty)$, and as $\xi \geq 0$, we have $2 \sqrt{4 \pi^2n^2 + \xi}$ $\geq$ $2\sqrt{4\pi^2n^2}$. In particular, we also have $2\sqrt{4\pi^2n^2} \geq 2\sqrt{4 \pi^2}$ for any $n \in \mathbb{N}$. Hence, we have the following inequality $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$ $\leq$ $\frac{1}{2 \sqrt{4\pi^2}}|x-y|$ < $\frac{1}{2 \sqrt{4\pi^2}}\delta$. Now choose $\delta = 2\epsilon\sqrt{4 \pi^2} $ and we have equicontinuity.

Sorry, I didn't originally see the no uniformly convergent subsequence. Let's assume for contradiction that there exists a sequence $\{f_{n_k}(x)\} \rightarrow f$ uniformly. In particular, $\sup_{x \in [0,\infty))} |f_{n_k}(x) - f(x)| \rightarrow 0$. As the space of continuous bounded functions is complete, this is possible if and only if the sequence is cauchy. E.g. $\sup_{x \in [0,\infty))}|f_{n_k}(x) - f_{m_k}(x)| \rightarrow 0$. Hence, there exists an $N \in \mathbb{N}$ such that for any $n,m \geq N$,$\sup_{x \in [0,\infty))}|f_{n_k}(x) - f_{m_k}(x)| < \frac{1}{2}$. In particular, $|\sin(\sqrt{x + 4 \pi^2n_k^2}) - \sin(\sqrt{x + 4 \pi^2m_k^2})| < \frac{1}{2} $ for every $x \in [0,\infty)$. Let $x = 12 \pi^2n_k^2 $, then $\sin(\sqrt{12 \pi^2n_k^2 + 4 \pi^2n_k^2}) = \sin(\sqrt{16 \pi^2n_k^2}) = 0 $. So we have $|\sin(\sqrt{12\pi^2n_k^2 + 4 \pi^2m_k^2})| < \frac{1}{2} \Rightarrow |\sin(2\pi\sqrt{3n_k^2 + m_k^2})| < \frac{1}{2}$ for any $m_k > N$, which is clearly a false statement due to the periodicity of our function with period 1. We can choose a $m_k$ to make this a false statement.

I hope this helps.

Answer 2

You were very close on the equicontinuity part. Note that for $n=1,2,\dots,$

$$\sqrt{c_n+4\pi^2n^2} \ge \sqrt{0+4\pi^2\cdot1^2} = 2\pi.$$

Thus $|f_n(x)-f_n(y)| \le |x-y|/4\pi$ for all $n$ and all $x,y\in [0,\infty).$

To show there is no uniformly convergent subsequence, one approach would be to first show that $f_n\to 0$ pointwise everywhere on $[0,\infty).$ That's a nice little exercise that I'll leave to you. So if a subsequence of $f_n$ converged uniformly, it would converge uniformly to $0.$ But that can't happen, because for each $n,$ $f_n([0,\infty)) = [-1,1].$