For each $n ∈ N$ and $x ∈ R$ let $f_n(x) = x + \frac{1}{n}$, and for each $x ∈ R$ let $f(x) = x$. Working directly from the definition of uniform convergence (i.e. without using the Uniform Norm Theo-rem), show that
(i) $f_n$ converges uniformly to f, but
(ii) $f^2_n$ does not converge uniformly to $f^2$.
Obviously the hypothesis is true but I'm having trouble showing this rigorously as required by the problem. This is what I have so far.
$f_n→f$ uniformly in $A$ if for any $ϵ>0$ there exists $N(ϵ)$ such that $n≥N(ϵ)⟹|f_n(x)−f(x)|<ϵ,∀x∈A$
So we can fix $\epsilon>0$ and then there exists $N(ϵ)$ such that $n≥N(ϵ)⟹|x + \frac{1}{n} - x|<ϵ,∀x∈A$. Then we have that $n≥N(ϵ)⟹|\frac{1}{n}|<ϵ,∀x∈A$. As $n \to \infty \frac {1}{n} \to 0$. Then we have that $\epsilon >0$ which we fixed earlier.
Now I need to show that there exists some $\epsilon$ for which $n≥N(ϵ)⟹|f_n^2(x) -f^2(x)|\geϵ,∀x∈A$. From this we get that $n≥N(ϵ)⟹|x^2 + \frac{2x}{n} + \frac{1}{n^2} -x^2|\geϵ,∀x∈A$ is what I have to show. Simplified I have to show that $n≥N(ϵ)⟹|\frac {1}{n}(x + \frac{1}{n})|\geϵ,∀x∈A$
This is where I have trouble. I thought that If I choose $\epsilon = \frac{\epsilon}{n}$ this would work but then I would have to show that $n≥N(ϵ)⟹|x+\frac{1}{n}| \ge \epsilon$.
I'm very confused as how to show that this is true rigorously. Thank You in advance.