Show uniform convergence implies uniform convergence of squares/root Let $f_n ,\,f\colon [0,1]\rightarrow [0,\infty)$ be given.
I wanted to show that if $f_{n}{\underset  {n\ }{\rightrightarrows }}f,$ then $f^{1/2}_{n}{\underset  {n\ }{\rightrightarrows }}f^{1/2}$
Using $|\sqrt{a}-\sqrt{b}|\le \sqrt{|a-b|}$ we see that
$$\lim_{n\to \infty}\sup_{x\in[0,1]}\left | \sqrt{f_n(x)}-\sqrt{f(x)}\right | \le \lim_{n\to \infty}\sup_{x\in[0,1]}\sqrt{\left | f_n(x)-f(x)\right |}\le
\lim_{n\to \infty}\sup_{x\in[0,1]}\left | f_n(x)-f(x)\right |=0$$
and the claim follows.
Is this correct?
I however can not find an example where 
$f_{n}{\underset  {n\ }{\rightrightarrows }}f,$ then $f^{2}_{n}{\underset  {n\ }{\rightrightarrows }}f^{2}$
This cannot be true I think, can it?
 A: In the first part, the claim $f_{n}{\underset  {n\ }{\rightrightarrows }}f \implies f_{n}^{1/2}{\underset  {n\ }{\rightrightarrows }}f^{1/2} $ for $f_n,f :[0,1] \to [0,\infty)$ is true.  Your proof starts out correctly, but your last inequality is not valid.  You can complete the argument by stating that for all $\epsilon > 0$ there exists $N$ independent of $x$ such that when $n > N$ we have for all $x \in [0,1]$, 
$$|f_n(x) - f(x)| < \epsilon^2, $$
and, thus,
$$\left|\sqrt{f_n(x)} - \sqrt{f(x)}\right| \leqslant \sqrt{|f_n(x) - f(x)|} < \epsilon.$$
Second Part
Clearly, there are examples where both $f_{n}{\underset  {n\ }{\rightrightarrows }}f$ and $f^{2}_{n}{\underset  {n\ }{\rightrightarrows }}f^{2}$ hold, either by virtue of your first proposition or the trivial example $f_n(x) = 1/n, \, f(x) = 0$.
However, the implication$f_{n}{\underset  {n\ }{\rightrightarrows }}f\implies f^{2}_{n}{\underset  {n\ }{\rightrightarrows }}f^{2}$ is not true.
For a counterexample, take $f_n,f:(0,1] \to \mathbb{R}$ where 
$$f_n(x) = \log \frac{nx}{n+1}, \quad f(x) = \log x$$
We have $f_{n}{\underset  {n\ }{\rightrightarrows }}f$, since
$$|f_n(x) - f(x)| = \log \frac{n+1}{n} \to 0$$
However,
$$|f_n^2(x) - f^2(x)| =  \left|\log \frac{nx}{n+1 } - \log x\right|\left|\log \frac{nx}{n+1 } + \log x\right| = \log\frac{n+1}{n} \left|\log \frac{nx^2}{n+1 }\right|,$$
and by choosing $x_n = 1/(n e^n)$ we have
$$|f_n^2(x_n) - f^2(x_n)| =  \log\frac{n+1}{n} \log (n^2 + n) + 2n  \log\frac{n+1}{n} $$
Since the RHS does not converge to $0$ as $n \to \infty$, it follows that $f_{n}^2{\underset  {n\ }{\rightrightarrows }}f^2$ does not hold on $(0,1].$
This example is easily modified for the domain $[0,1]$ and range $[0,\infty)$ as $f_n(x) = -\log \frac{nx}{n+1} \mathbf{1}_{(0,1]}.$ 
