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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?

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Well it would certainly be true for $S$ compact, as on a compact set convergence $\implies$ uniform convergence. The same proof should work for bounded. – Alex Becker Feb 29 '12 at 3:38

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.

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The example $S = (0,1)$, $f_n(x) = \frac{1}{x} + \frac{1}{n}$ shows that some kind of hypothesis (e.g. "$|f|$ is bounded by $M$") is needed, even for continuous functions on bounded intervals. Clearly this sequence $f_n$ converges uniformly to $f(x) = \frac{1}{x}$ on $S$, but the fact that $|f_n(n^{-2})^2 - f(n^{-2})^2| = \frac{2n^3 + 1}{n^2} > 1$ holds for all $n$ shows that the sequence $f_n^2$ does not converge uniformly to $f^2$. – leslie townes Feb 29 '12 at 3:58
Yes, that was mentioned in the OP and is why I required that $f$ be bounded. Note that $S$ closed and bounded and $f$ continuous implies $f$ bounded, as $S$ is then compact. – Alex Becker Feb 29 '12 at 4:00
I don't quite follow why... $$\begin{eqnarray} |f_n(x)-f(x)|(|f_n(x)-f(x)|+2|f(x)|)&<&\frac{\epsilon}{2M+\epsilon}(2M+\epsilon)‌​=\epsilon\\ \end{eqnarray}$$ Shouldn't it be... $$\begin{eqnarray} |f_n(x)-f(x)|(|f_n(x)-f(x)|+2|f(x)|) &<&\frac{\epsilon}{2M+\epsilon}(\frac{\epsilon}{2M+\epsilon} + 2M) \end{eqnarray}$$ Am I missing something obvious? – Jason P. Feb 29 '12 at 4:21
@JasonP. I implicitly assumed that $M$ is chosen to be at least $1/2$. I've now made that explicit. – Alex Becker Feb 29 '12 at 4:24
The hypothesis that $f$ be bounded is not actually necessary for $f_n^2$ to converge uniformly to $f^2$. The silliest example is $f_n(x) = f(x) = x$ for all $x \in \mathbb{R}$, and there are many variations on this theme (e.g. nothing prevents all of the differences $|f(x) - f_n(x)|$ from being supported on the same compact set, even if $f$ is unbounded). The point of my example was mainly that boundedness of the function is definitely much more relevant to making the squares converge than boundedness of the domain (the OP's counterexample is on an unbounded domain). – leslie townes Feb 29 '12 at 4:27

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