Showing uniform convergence I am struggling to show that $f_n$ converges uniformly to $|f|$, where $\displaystyle{f_n=\frac{|f|^2}{|f|+\frac{1}{n}}}$. 
I have that 
$\displaystyle{|f_n-f|=|\frac{|f|^2}{|f|+\frac{1}{n}}-|f||=|f|*|\frac{|f|}{|f|+1/n}-1| \leq|f|*|\frac{|f|}{|f|+1/n}| \leq |f|*1 =|f|}$, but I am not sure if that is sufficient. Any help would be great.
 A: $$\left| \frac{|f|^2}{|f| +\frac{1}{n}} -|f|\right| =\left| \frac{|f|^2 -\frac{1}{n^2}}{|f| +\frac{1}{n}} +\frac{\frac{1}{n^2}}{|f| +\frac{1}{n}}-|f|\right| =\left|-\frac{1}{n} +\frac{\frac{1}{n^2}}{|f| +\frac{1}{n}}\right|\leq \frac{1}{n} +\frac{1}{n} =\frac{2}{n}$$
A: $$\displaystyle{|f_n-|f||=\left|\frac{|f|^2}{|f|+\frac{1}{n}}-|f|\right|=|f|\cdot \left|\frac{|f|}{|f|+1/n}-1 \right| \leqslant |f| \cdot \left|\frac{-\frac{1}{n}}{|f|+1/n}\right| \leqslant \frac{1}{n}}$$
A: For all $x>0$, $\frac{x^2}{x+\frac{1}{n}}-x=\frac{nx^2}{nx+1}-\frac{x(nx+1)}{nx+1}=-\frac{x}{nx+1}$.
So, $|\frac{x^2}{x+\frac{1}{n}}-x|=|\frac{x}{nx+1}|<\frac{x}{nx}=\frac{1}{n}$.
Then, let $x=|f(y)|$, we have $|f_n(y)-|f(y)||<\frac{1}{n}$, if $f(y)\neq 0$.
If $f(y)=0$, $f_n(y)=0=f(y)$.
In every case,$|f_n(y)-|f(y)||<\frac{1}{n}$, for all $y$.
A: Your mistake is in the third step.  What you should have is
$$
|f| \left| \frac{|f|}{|f|+\frac{1}{n}} -1 \right| = |f| \left( 1- \frac{|f|}{|f|+\frac{1}{n}} \right)=|f| \left( \frac{|f|+\frac{1}{n} - |f|} {|f|+\frac{1}{n}} \right)=|f| 
\left( \frac{\frac{1}{n}}{|f|+\frac{1}{n}} \right) =|f| \frac{1}{n|f|+1}
$$ 
Then, assuming $|f|>0$ (since the case for $f=0$ is trivial) 
$$
|f_n-f| = |f| \frac{1}{n|f|+1} < \frac{|f|}{n|f|} = \frac{1}{n}
$$
which of course goes to zero as $n \rightarrow \infty$ so the absolute convergence is demonstrated.
A: So far we have $$\left|f_n-|f|\right|=\left|\frac{|f|^2}{|f|+1/n}-|f|\right| \\ = \left|\frac{|f|^2}{|f|+1/n}-\frac{|f|(|f|+1/n)}{|f|+1/n}\right| \\ = \left|\frac{|f|^2-|f|^2-|f|\frac{1}{n}}{|f|+1/n}\right| \\= \left|\frac{-|f|\frac{1}{n}}{|f|+1/n}\right| \\ = \left|\frac{|f|}{n|f|+1}\right| \\ <  \left|\frac{|f|}{n|f|}\right| \\= \left|\frac{1}{n}\right| \\ = \frac{1}{n}$$ WHEW! Now we know $$\lim_{n \to \infty}\left|f_n-|f|\right| < \lim_{n \to \infty} \frac{1}{n} =0 $$ So we have uniform convergence.
