Proof of non-uniform convergence of $x^n$

Question

The wikipedia page on uniform convergence indicates that $f_{n}:[0,1]\to[0,1]$ with $f_{n}(x):=x^{n}$ converges pointwise but not uniformly to $$ f(x)= \begin{cases} 0,\quad x \in [0,1)\\ 1,\quad x=0. \end{cases} $$ Unfortunately, it does not give a proof or any hint.

One could prove this either using the definition (there exists no $N$ such that $\forall \epsilon$, $|f_n(x)-f(x)|<\epsilon$. Or the theorem that $\sup_{x}|f_{n}(x)-f(x)|\to 0$, showing it does not hold.

So my question is:

1. I could not find a way to prove from absence of uniform convergence based on the definition. Any hint?

2. Is my proof based on the theorem correct?

Proof: let us look at $x=1-1/n$, clearly, $x\in [0,1)$ so $x^n$ should converge to zero. However, by Euler: $\lim_{n\to\infty} (1-1/n)^n=1/e\neq 0.$

Thanks!!

Answer 1

A Sameer Kulkarni noted, I don't think your proof works because $x$ is also changing. Fortunately, there's a nice conceptual way to prove that the convergence isn't uniform:

An important property of uniform convergence is that if each function $f_n(x)$ is continuous, and if $f_n\to f$ uniformly, then $f(x)$ is continuous as well.

In your case, $f_n(x)=x^n$ is continuous but $f(x)$ isn't, so the convergence must not be uniform.

Answer 2

Let $\epsilon$ be a positive number less than $1$, let $N$ be a positive integer, and let $x$ be the number $\epsilon^{1/N}$. Then $\lvert f_N(x)\rvert=\epsilon$, which proves nonuniformity.