prove/disprove: the functions $f_n(x)=\cos(x+n)+\ln\left(1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\right)$ are uniformly equicontinuous

Question

Prove/Disprove that the sequence of functions $(f_n)$ from $\Bbb R$ to $\Bbb R$ defined by $$f_n(x)=\cos(x+n)+\ln\left(1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\right)$$ are uniformly equicontinuous. That is, prove/disprove that for all $\epsilon>0$ there is a $\delta>0$ such that $$|x-t|<\delta, n\in\Bbb N\Rightarrow |f_n(x)-f_n(t)|<\epsilon.$$

Thoughts/Context. It is my gut feeling that $f_n$ are uniformly equicontinuous. I have a not-very-rigorous "proof" of this. For $n\in\Bbb N$ and $x\in\Bbb R$, $\sin^2(n^n x)$ is between $0$ and $1$, so $$0\le \frac{\sin^2(n^nx)}{\sqrt{n+2}}\le \frac{1}{\sqrt{n+2}}<1.$$ Then from the squeeze theorem $a_n(x)=\frac{\sin^2(n^nx)}{\sqrt{n+2}}\to 0$ as $n\to\infty$, regardless of the value of $x$. Hence $\ln(1+a_n(x))\to0$ as $n\to \infty$ as well. So for large $n$, $f_n$ is practically $\cos(x+n)$. I know that the set of functions $$\mathcal{E}=\left\{f\in C^1(\Bbb R,\Bbb R): |f'(x)|\le1\right\}$$ is uniformly equicontinuous. Thus the subset $$\{\cos(x+n):n\in\Bbb N\}\subset \mathcal{E}$$ is also uniformly equicontinuous. Of course this does not necessarily imply that functions that behave like $\cos(x+n)$ (in the aforementioned way that $f_n$ does) are also uniformly equicontinuous.

Unfortunately, $\{f_n(x):n\in\Bbb N\}$ does not seem to be a subset of $\mathcal{E}$, so I can't really take the easy way out here.

Is there some way to rigorize my argument, or is there a better way to go about things? Thanks.

Answer 1

So it is all about to show the uniform-equicontinuity of $g_{n}:x\rightarrow\log\left(1+\dfrac{\sin^{2}(n^{n}x)}{\sqrt{n+2}}\right)$.

Take a subsequence $(n_{k})$ of positive integers, which we will still denote as $(n)$ for simplicity, and consider the compact interval $[0,3\pi]$ in issue. We have \begin{align*} \log\left(1+\dfrac{\sin^{2}(n^{n}x)}{\sqrt{n+2}}\right)&\leq\dfrac{\sin^{2}(n^{n}x)}{\sqrt{n+2}}\\ &\leq\dfrac{1}{\sqrt{n+2}}, \end{align*} this shows that $g_{n}\rightarrow 0$ uniformly. So Arzela-Ascoli Theorem says that $g_{n}$ is uniform-equicontinuous on $[0,3\pi]$. Note that we must have a compact interval to use Arzela-Ascoli Theorem.

So now the issue is to extend to the whole $\mathbb{R}$, this uses the periodicity of $\sin$ function.

Now we write down the definition of uniform-equicontinuity of such functions on $[0,3\pi]$. Given $\epsilon>0$, there is a $\delta>0$ such that for every $n$ and $|x-t|<\delta$, $x,t\in[0,3\pi]$, then $|g_{n}(x)-g_{n}(t)|<\epsilon$.

Now for any $x,t\in\mathbb{R}$ with $|x-t|<\delta$, say, $x\leq t<x+\delta$, and $\delta<\pi$, for a fixed $n$, we notice that \begin{align*} \left\{\cdots,-\dfrac{4\pi}{n^{n}},-\dfrac{2\pi}{n^{n}},0,\dfrac{2\pi}{n^{n}},\dfrac{4\pi}{n^{n}},\cdots\right\} \end{align*} forms a partition on $\mathbb{R}$, so there is some $k$ such that \begin{align*} \dfrac{2k\pi}{n^{n}}\leq x\leq\dfrac{2(k+1)\pi}{n^{n}}, \end{align*} and hence \begin{align*} 0\leq x-\dfrac{2k\pi}{n^{n}}\leq\dfrac{2\pi}{n^{n}}\leq 2\pi<3\pi, \end{align*} we also have \begin{align*} 0\leq x-\dfrac{2k\pi}{n^{n}}\leq t-\dfrac{2k\pi}{n^{n}}<x+\delta-\dfrac{2k\pi}{n^{n}}\leq\delta+2\pi<3\pi. \end{align*} We now let $x'=x-\dfrac{2k\pi}{n^{n}}$ and $t'=t-\dfrac{2k\pi}{n^{n}}$, then $|x'-t'|<\delta$ and $x',t'\in[0,3\pi]$ so $|g_{n}(x')-g_{n}(t')|<\epsilon$.

It remains to observe that $g_{n}(x)=g_{n}(x')$ and $g_{n}(t)=g_{n}(t')$ and hence $|g_{n}(x)-g_{n}(t)|=|g_{n}(x')-g_{n}(t')|<\epsilon$, this shows the uniform-equicontinuity of $g_{n}$ on the whole $\mathbb{R}$.

Answer 2

$0\leq ln (1+\frac {sin^{2}(n^{n}x)} {\sqrt {n+2}} )\leq ln (1+\frac 1 {\sqrt n+2})$ which proves that $ln (1+\frac {sin^{2}(n^{n}x)} {\sqrt {n+2}}) \to 0$ uniformly. If a sequence of continuous functions converges uniformly to $0$ then it is uniformly equi-continuous.

Answer 3

Legit approach? If not, flaw in logic/conceptual understanding pointer would be appreciated. Think my notation is fine, but first post.

Let $\epsilon\gt0$ be given and fix $n\in\mathbb{N}$. Let $s,t\in\mathbb{R}$. First, note that for $$f_n(x)=\cos(n+x)+\log\biggl(1+\frac{1}{\sqrt{n+2}}\sin^2(n^nx)\biggl),$$ the fact that (1) the logarithm is nonnegative for all relevant $n,x$ coupled with (2) the fact that $0\leq\sin^2(z)\leq1$ for any real input $z$ and (3) number sense regarding $\frac{1}{\sqrt{n+2}}$ with $n\in\mathbb{N}$ allows us to write $$\cos(n+x)\leq f_n(x)\leq \cos(n+x)+\log(1+\frac{1}{\sqrt{3}}).$$ Note that this inequality holds for all $n\in\mathbb{N}$ and $x\in\mathbb{R}$. Thus, write $$\cos(n+s)\leq f_n(s)\leq \cos(n+s)+\log(1+\frac{1}{\sqrt{3}})$$ and $$\cos(n+t)\leq f_n(t)\leq \cos(n+t)+\log(1+\frac{1}{\sqrt{3}}).$$ Subtraction and simplification yields $$\cos(n+s)-\cos(n+t)\leq f_n(s)-f_n(t)\leq \cos(n+s)-\cos(n+t).$$ No longer regarding the LHS, we note that $$\left\lvert f_n(s)-f_n(t)\right\rvert\leq \left\lvert\cos(n+s)-\cos(n+t)\right\rvert.$$ In what follows, I will make use of angle sum formula, reorganization of terms, triangle inequality, absolute value properties, and the Mean Value Theorem, in that order: $$\begin{equation}\begin{aligned} \left\lvert f_n(s)-f_n(t)\right\rvert &\leq \left\lvert\cos(n+s)-\cos(n+t)\right\rvert \\ &= \left\lvert\cos(n)\cos(s)-\sin(n)\sin(s)-(\cos(n)cos(t)-\sin(n)\sin(t))\right\rvert \\ &= \left\lvert\cos(n)(\cos(s)-\cos(t))-\sin(n)(\sin(s)-\sin(t))\right\rvert \\ &\leq \left\lvert\cos(n)(\cos(s)-\cos(t))\right\rvert+\left\lvert-\sin(n)(\sin(s)-\sin(t))\right\rvert \\ &\leq \left\lvert\cos(s)-\cos(t)\right\rvert+\left\lvert\sin(s)-\sin(t)\right\rvert \\ &\leq \left\lvert s-t \right\rvert+\left\lvert s-t \right\rvert \\ \end{aligned}\end{equation} $$ So let $\delta=\frac{\epsilon}{2}$. Then from the work above we see that if $\left\lvert s-t \right\rvert\lt\frac{\epsilon}{2},$ we have: $$f_n(s)-f_n(t)\leq\left\lvert f_n(s)-f_n(t)\right\rvert \leq \left\lvert s-t \right\rvert+\left\lvert s-t \right\rvert \lt \epsilon. \blacksquare$$

If there's a big error it's like 0205 so I'm tired.