Uniform convergence of $f^2_n$ when $f_n$ converges uniformly

Question

Let $(f_n)$ be a sequence of functions that converge uniformly to $f$ on the interval $I$.

Prove or disprove: $f^2_n \to f^2$ uniformly on I.

I was almost certain this claim is false but was unable to construct any example. After several hours I started to try and prove it.

What i have so far is that if we can prove that $\sup_{x \in I}|f_n(f_n-f)| \ \to 0$ we're done. But after playing for another couple of hours with different functions i'm starting to believe it could be wrong.

I'm really at a loss here and any intuition i had about sequences seems to have been lost with this problem.

Answer 1

If $f_n+f $ is bounded, then $f_n^2-f^2=(f_n-f)(f_n+f) \rightarrow 0$ .

Answer 2

Let $f_n(x) = x+{1 \over n} $, $f(x) = x$. Then $f_n \to f$ uniformly, but $f_n(n)^2-f(n)^2 = { 2 n^2 +1 \over n^2}$, hence the square does not converge uniformly.

By considering $f_n \circ \tan$, $f \circ \tan$ (and appropriate adjustments for the argument above), we can apply similar considerations to the open interval $(0,{\pi \over 2})$.