Uniform Convergence and LimSup's I am new to MSE, so apologies for any inconveniences regarding my question. I am studying for the GRE's Math Subject and decided to go back and study certain important concepts from the Analysis series. I am reading Elementary Analysis by Ross, particularly section 24, Uniform Convergence.
We are shown the sequence of real valued functions 
$$f_n(x) = nx^n, \quad x \in [0,1),$$ and told that it does not converge uniformly on said interval. However, $\ f_n \rightarrow f$ point-wise on $[0,1)$ where $\ f(x) = 0, \ \forall x \in [0,1).$  
This all makes sense and I am fine with it. However, it is later remarked that a sequence $\ (f_n)$ of functions on a set $\ S \in \mathbb{R}$ converges uniformly if and only if 
$$lim[sup\{|f(x)-f_n(x)|:x \in S\}] = 0.$$
This is what confuses me, because I am pretty sure (not $100 \%$ sure), that since $limsup = L = liminf$ for any convergent sequence (where $L$ is the limit), then for the example $f_n(x) = nx^n$, shouldn't this sequence of functions technically be uniformly convergent? 
I'm most definitely missing something, though hopefully not anything glaringly obvious, and would appreciate some feedback. Thanks!
 A: For bounded functions $f:S\to \Bbb R$ and $f_n:S\to \Bbb R$ let $$\|f-f_n\|=\sup_{x\in S}|f(x)-f_n(x)|.$$ Consider $\|f-f_n\|$ to be a measurement of how closely $f_n$ approximates $f,$ not at any one $x\in S,$ but as an over-all measurement over all of $S.$  Consider $f$ and $f_n$  as single objects, and $\|f-f_n\|$ as the distance from $f$ to $f_n.$
The sequence $(f_n)_n$ converges uniformly iff $\|f-f_n\|\to 0$ as $n\to \infty.$
If $(f_n)_n$ converges uniformly to  $f$ then for any $y\in S$ we have $$|f(y)-f_n(x)|\leq \|f-f_n\|,$$ so $f_n(y)\to f(y)$  as $n\to \infty.$
If $f_n(y)\to f(y)$ for each $y\in S$ we say that $f_n$ converges point-wise to $f.$ So uniform convergence implies point-wise convergence (but not vice-versa).  If we are testing whether $(f_n)_n$ converges uniformly to some (any) $f$, we can, as a first step, identify the "candidate" $f$ by evaluating $\lim_{n\to \infty}f_n(y)$ for each $y\in S.$
With $S=[0,1)$ and $f_n(x)=nx^n,$ then $f_n(y)\to 0$ for each $y\in S.$  By the previous paragraph, if $(f_n)_n$ did converge uniformly to a function $f,$ then we would have $f(y)=\lim_{n\to \infty}f_n(x)=0$ for each $y\in S$.....   But for any $n\in \Bbb N$ we can find some $y_n \in S$ with $y_n$ close enough to $1$ that $(y_n)^n>1/2.$ Then $f(y_n)>n/2$ and $f(y)=0$ so $$\|f-f_n\|\geq |f(y_n)-f_n(y_n)|=|0-f_n(y_n)|>n/2.$$ So the sequence $(f_n)_n$ in the Q is not a  uniformly convergent sequence..
BTW . An important result that holds for all metric spaces $S, T$ (not just sub-spaces of $\Bbb R$ ), is that if $(f_n:S\to T)_{n\in \Bbb N}$ is a sequence of continuous functions converging uniformly to $f:S\to T$, then $f$ is continuous. 
A: The thing is, for uniform convergence, the function sequence $(f_n)$ must converges for all $x$ in the interval. Pointwise convergence gives convergence of the sequence of numbers $(f_n(x))$, where $x$ is a pre-fixed point in the interval. When we compute $$\lim_{n\to\infty}\left[\sup\{|f(x)-f_n(x)|:x\in S\}\right]$$the point $x$ is not fixed, as the supremum of the limiting set can change (and in your example does change) as $n\to\infty$. 
Pointwise convergence says that, for a fixed $x\in S$ we have $$\lim_{n\to\infty}|f(x)-f_n(x)|=0$$and in this case, we can use the result that says $\liminf=\lim=\limsup$. However, for uniform convergence we consider the limit of the supremum of the set $\{|f(x)-f_n(x)|:x\in S\}$, which is not the same as $\limsup |f(x)-f_n(x)|$ for a fixed $x$. 
Essentially, this test for uniform convergence is seeing whether or not the maximum distance between the sequence functions and the pointwise limit function goes to zero. If this distance goes to zero, then the sequence functions approach the limit function uniformly, otherwise they do not. Pointwise convergence just shows that, at a specific point $x$, the distance between the value of the sequence functions and the value of the limit function at this point $x$ goes to zero. This does not guarantee that the maximum distance between the limit function and the sequence functions goes to zero, as the point $x$ that produces the maximum distance need not be the same point as $n\to\infty$.
