Working through Spivak's Calculus - Ch 24 - 2) (i) and (ii), I've been asked to determine whether the following functions converge pointwise and to determine if they are uniformly convergent. The functions are:
i) $f_{n}(x) = x^{n} - x^{2n}$ on $[0,1]$
ii) $f_{n}(x) = \frac{nx}{1 + n + x}$ on $[0, \infty)$
Hints were given to solve the expressions and they are the following:
So solving (i). The first thing to observe after a little playing around is that:
$$f(x) = \lim_{n \to \infty} f_{n}(x) = 0$$
Now using the hint one works out that the maximum of $|f-f_{n}|$ on $[0,1]$ occurs when:
$$x = \frac{1}{\sqrt[n]{2}}$$
If you plug this value into the expression of $|f(x) - f_{n}(x)|$ it will always work out to be:
$$|\frac{1}{2} - \frac{1}{4}| = \frac{1}{4}$$ for all values of $n$. From this I concluded that the sequence of functions does not converge uniformly to $f(x)$ because there will always be a distance of $\frac{1}{4}$ between the sequence and $f(x) = 0$.
So now working on (ii) with some fiddling around again one can conclude that:
$$f(x) = \lim_{n \to \infty}f_{n}(x) = x$$
Now using the hint given one works out:
$$\bigg|\frac{nx}{1+n+x} - x \bigg| \Rightarrow \bigg|\frac{-1(x^{2} + x)}{1 + n + x}\bigg| = \frac{(x^{2} + x)}{1 + n + x}$$
Now here is where my question lies. I was going to conclude that this illustrates that the sequence of functions does converge uniformly. I was basing this off of the fact that the distance between $f_{n}$ and $f(x)$ will always be dependent on $n$ and as $n \to \infty$ my expression will go to $0$ (i.e. the distance between $f(x)$ and $f_{n}(x)$). This was the reasoning I used behind why (i) does not uniformly converge because the distance is not dependent on $n$ and is instead always constant.
Looking at the solutions my conclusion for (ii) is wrong. I wanted to know why that is the case and I also wanted to know if my reasoning for why (i) is not uniformly convergent is the right approach as well.
The reason is that your distance depends on $x$. It is not true that you can make it uniformly small (by choosing $n$ large) for all $x\ge 0$. If you assume that you can make the distance say less than $1$ for $n\ge N>2$, you get a contradiction for $x=N$, where the distance is greater than $N/2>1$.