Show that $f_n(x)= \frac{\sin^2(n^{\alpha}x)}{nx}$ is not uniformly convergent for $\alpha \geq 1$ and $x \neq 0$ Can someone help me with proving that 
\begin{align}
f_n: \mathbb{R} \rightarrow \mathbb{R }, \hspace{0.5cm}  f_n(x)=
\begin{cases}
\frac{\sin^2(n^{\alpha}x)}{nx} & \text{ if } x \neq 0, \\ 0 & \text{ if } x = 0
\end{cases}
\end{align}
is not uniformly convergent if $\alpha \geq 1, \forall x \in \mathbb{R} $. 
I've unsucessfully tried to show that $ \underset{x \in \mathbb{R}} \sup |f_n(x)| \nrightarrow 0 $. Same with the definition of uniform convergence. 
 A: We first note that the pointwise limit of $f_n$ in $\mathbb{R}$ is the constant function zero. So $(f_n)_n$ converges uniformly in $\mathbb{R}$ if and only if $\sup_{x\in \mathbb{R}}|f_n(x)|\to 0$ as $n$ goes to infinity.
1) For $\alpha>1$, let $x_n=1/n^{r}$, with $r\in (\alpha,2\alpha-1)$. Then $n^\alpha x_n\to 0$ and
$$\lim_{n+\infty}f_n(x_n)=\lim_{n\to+\infty}\frac{\sin^2(n^{\alpha}x_n)}{nx_n}=\lim_{n\to+\infty}\frac{(n^{\alpha}x_n)^2}{nx_n}=\lim_{n+\infty}n^{2\alpha-1-r}=+\infty.$$
Hence $\sup_{x\in \mathbb{R}}|f_n(x)|\to +\infty$ and 
2) If $\alpha=1$ then after letting $nx=t$ we get
$$\sup_{x\in \mathbb{R}}|f_n(x)|=\sup_{t\in \mathbb{R}}\left|\frac{\sin^2(t)}{t}\right|\geq \frac{\sin^2(\pi/2)}{\pi/2}=\frac{2}{\pi}>0.$$
Therefore $\sup_{x\in \mathbb{R}}|f_n(x)|\not\to 0$.
3) Finally if $0\leq\alpha<1$, then after letting $n^{\alpha}x=t$ we get
$$\sup_{x\in \mathbb{R}}|f_n(x)|=n^{\alpha-1}\sup_{t\in \mathbb{R}}\left|\frac{\sin^2(t)}{t}\right|\leq n^{\alpha-1}\to 0$$
because $\sin^2(t)\leq |\sin(t)|\leq |t|$.
So we have uniform convergence in $\mathbb{R}$ for $0\leq\alpha<1$.
P.S. For any $\alpha$, $(f_n)_n$ converges uniformly to $0$ in $\mathbb{R}\setminus (-r,r)$ for all $r>0$:
$$\sup_{\mathbb{R}\setminus (-r,r)}|f_n(x)|\leq \frac{1}{nr}\to 0.$$
A: Use the logical negation of uniform convergence: exists some $\epsilon>0$ such that for any $N\in\Bbb N$ there is some $x_N\in\Bbb R$ such that exists $n_0\ge N$ and $$|f_{n_0}(x_N)-f(x_N)|\ge\epsilon$$
Because $f=0$ then for $\epsilon:=1/2$ and any $N\in\Bbb N_{>0}$ if we choose $x_N:=\pi/(2N^\alpha)$ then for $n_0:=N$ we find that $$|f_{n_0}(x_N)-f(x_N)|=|f_{n_0}(x_N)|=\frac{2N^{\alpha-1}}\pi\sin^2(\pi/2)=\frac{2N^{\alpha-1}}\pi\ge\frac2\pi>\frac12$$
when $\alpha\ge 1$.
