I'm considering the question: "If $f_n:[0,1] \rightarrow \mathbb{R}$ is a sequence of functions that converges uniformly to $f:[0,1] \rightarrow \mathbb{R}$, does $f_n^2$ converge uniformly to $f^2$ on $[0,1]$?"

From the examples I've come up with, I can certainly conclude that the answer is no, but I've noticed that the above does seem to hold if in addition we require that $f_n$ is a sequence of continuous functions.

For example, if $f_n(x) = \frac{x}{nx+1}$, then $f_n(x)$ converges uniformly to $f(x) = 0$, and $[f_n(x)]^2$ converges to $[f(x)]^2$ (also 0).

Am I correct in attributing this to continuity? If so, how would I go about proving this formally without specifically defining $f_n(x)$?

Thanks in advance.

  • $\begingroup$ Maybe of relevance is that $x \mapsto x^2$ is continuous, and since the interval is compact, uniformly continuous. So you have a uniformly convergent sequence of continuous functions, and you composed them with a uniformly continuous function. So morally what's (probably) happening is that uniform control lets you replace $\epsilon$ with $\epsilon / n$ for some appropriate $n$ corresponding to uniform continuity, and the proof goes through the same. $\endgroup$ – Alfred Yerger May 28 '18 at 2:48
  • $\begingroup$ Boundedness is enough to guarantee uniform convergence. $\endgroup$ – clark May 28 '18 at 3:03

If all $f_n$ are continuous, $f$ is continuous, hence there is $M \ge 0$ such that $|f(x)| \le M$ for all $x \in I$, where $I=[0,1]$.

There is $N \in \mathbb N$ such that $|f_n(x)-f(x)| \le 1$ for all $n>N$ and all $x \in I$, hence $|f_n(x)| =|f_n(x)-f(x)+f(x)| \le 1+M$ for all $n>N$ and all $x \in I$.

We get:

$|f_n(x)^2-f(x)^2|=|f_n(x)-f(x)| \cdot |f_n(x)+f(x)| \le |f_n(x)-f(x)| \cdot(|f_n(x)|+|f(x)| \le |f_n(x)-f(x)| \cdot(1+2M)$

for all $n>N$ and all $x \in I$.

Can you proceed ?

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  • $\begingroup$ I believe so. It looks like we would like to fix some $\epsilon >0$, and then choose $N$ such that $|f_n(x) - f(x)| < \frac{\epsilon}{1+2M}$, $\forall n \geq N$, which we are guaranteed exists via uniform continuity. Is this reasonable? $\endgroup$ – OGBerglemir May 29 '18 at 18:37
  • $\begingroup$ Yes, your argument is fine. $\endgroup$ – Fred May 30 '18 at 4:23

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