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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.2012-02-29

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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 $f^2$.2012-02-29
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    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.2012-02-29
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    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?2012-02-29
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    @JasonP. I implicitly assumed that $M$ is chosen to be at least $1/2$. I've now made that explicit.2012-02-29
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    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).2012-02-29
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    @leslietownes Note that I edited my question already to fix that.2012-02-29
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    And I should add that I mean "uniformly eventually constant".2012-02-29
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    The condition that $f$ is bounded on the set where $(f_n)$ is not eventually constant is **not necessary** for $(f_n^2)$ to converge uniformly to $f^2$. Consider $f_n(x)=x+(n(x^2+1))^{-1}$.2012-02-29
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    @DidierPiau Thanks for pointing that out, corrected.2012-02-29