Suppose $f_n$ converges uniformly to $f$. Do the positive (i.e. $f_n^+=\max\{f(x),0\}$) and negative parts (i.e. $f_n^-=\max\{-f(x),0\}$) converge uniformly to the positive and negative parts of $f$ respectively. I don't think so but can’t come up with any counter examples.
Tried to proof but was stuck: For any $\epsilon>0$ $\exists N$ s.t for all $n\geq N$ for $x$ $|f_n(x)-f(x)|<\epsilon$. Suppose we consider just the convergence of the positive parts. It can be shown to converge pointwise but I think we need a uniform $N'$ s.t for all $n\geq N'$ $f_n(x)$ is positive for all $x$ where $f(x)$ is positive. But that need not be true.
Does anyone know how to proceed?
As per the comments above we note that for any real number $2a^{\pm}=|a| \pm a$ so for example $2f_n^+(x)=|f_n(x)|+f_n(x)$ and same for $f^+$.
Using the inequality $||a|-|b|| \le |a-b|$ we get that uniform convergence of $f_n$ to $f$ implies uniform convergence of $|f_n|$ to $|f|$ hence the uniform convergence of $|f_n|+f_n$ to $|f|+f$ so the required uniform convergence of $f_n^+$ to $f^+$. Similarly by taking differences we get the uniform convergence of $f_n^-$ to $f_n^-$.