Uniform Convergence. If a sequence of functions $f_j$ on a domain $S \subseteq \mathbb{R}$ has the property that $f_j \rightarrow f$ uniformly on $S$ then does it follow that $(f_j)^2 \rightarrow f^2$ uniformly on $S$?
I know this to be false. Suppose $f_j(x) = x + (1/j)$ for example.
But what would make this statement true? What if $f$ is bounded? Does anyone have a proof to show that if $f$ is bounded then the above is true?
 A: It is sufficient to assume that $f$ is bounded. Suppose $f_j\to f$ uniformly and that $|f|$ is bounded by $M\geq 1/2$. Then for any $\epsilon>0$, we have some $N\in \mathbb N$ such that 
$$n\geq N\implies |f_n(x)-f(x)|<\frac{\epsilon}{2M+\epsilon},\forall x\in S$$
thus for $x\in S,n\geq N$ we have
$$\begin{eqnarray}
|f_n^2(x)-f^2(x)|&=&|f_n(x)-f(x)||f_n(x)+f(x)|\\
&\leq& |f_n(x)-f(x)|(|f_n(x)-f(x)|+2|f(x)|)\\
&<&\frac{\epsilon}{2M+\epsilon}(2M+\epsilon)=\epsilon\\
\end{eqnarray}$$
so $f_j^2\to f^2$ uniformly. 
As a partial converse, if $|f_n(x)-f(x)||f(x)|$ is unbounded for infinitely many $n$ then
$$\begin{eqnarray}
|f_n^2(x)-f^2(x)|&=&|f_n(x)-f(x)||f_n(x)+f(x)|\\
&\geq& |f_n(x)-f(x)|(2|f(x)|-|f_n(x)-f(x)|)\\
&\geq& 2|f_n(x)-f(x)||f(x)|-|f_n(x)-f(x)|^2\\
\end{eqnarray}$$
can be made arbitrarily large for arbitrarily large $n$ by choosing $n$ such that $|f_n(x)-f(x)|<1$ for all $x$ and $|f_n(x)-f(x)||f(x)|$ is unbounded. Thus in this case $(f_j^2)$ does not converge uniformly to $f^2$.
Note that if $S$ is closed and bounded and $f$ is continuous then since $S$ is compact $f$ will be bounded, giving us another sufficient hypothesis.
