# If $f_n{\rightarrow} f$ uniformly and $f$ is bounded then prove that $f_n^{2}{\rightarrow}f^{2}$ uniformly

If $f_n{\rightarrow} f$ uniformly and $f$ is bounded, how can we prove that $f_n^{2}{\rightarrow}f^{2}$ uniformly? I am revising for my analysis final and got stuck on this problem, any help is really appreciated.

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yes uniformly, i forgot to put that in> – UH1 Dec 7 '12 at 23:21

1. You can use the fact that $f$ is bounded and $f_n\to f$ uniformly to show that there is a uniform bound $M>0$ such that $|f_n(x)|<M$ and $|f(x)|<M$ for all $n$ and all $x$.

2. You can use the identity $a^2-b^2 = (a+b)(a-b)$ to bound the difference of squares in terms of $M$ and the difference of the original functions.

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Thanks, got it! if u have time, can you also look at one other problem i posted,thanks!math.stackexchange.com/questions/253434/… – UH1 Dec 7 '12 at 23:36

I think more assumptions are needed if you mean $f_n\rightarrow f$ in the pointwise sense. A counter-example is taking $f_n$ to be the positive spike of height 1 and width $2/n$ centred at $x=1/n$ and zero outside $(0,2/n)$. This function tends to zero pointwise whilst the sup of its square is always 1.

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It's true if $f_n\to f$ uniformly, which is probably intended (unless it was a "prove or disprove" type problem). Edit: The question has been updated to state the hypothesis that $f_n\to f$ uniformly. – Jonas Meyer Dec 7 '12 at 23:18

Hint: $f_n^2-f^2=(f_n-f)(f_n+f)$.

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