Let $f_n(x)=x^n$ on $[0, 1]$. The pointwise limit of this sequence is $f(x)=\left\{\begin{matrix}
1,& \text{if $x=1$} \\
0,& \text{if $0\leq x<1$.}
\end{matrix}\right.$
Now, it is said that the convergence is not uniform here "examples: non-uniformity of convergence". The definition of uniform given for this example would be as follows $\forall \epsilon>0$, $\exists N$, $\forall n\geq N$, $\forall x\in [0, 1]$, $|f_n(x)-f(x)|<\epsilon$.
I have read the explanation again and again, and every time I read it I get more confused. In the article, it particularly considers $\epsilon:=\frac{1}{4}$ and $x\in [0, 1)$. So, I suppose the definition of uniform convergence holds true and aim to show a contradiction. Say $N$ is a natural number to hold true, so that $\forall n\geq N$, $|f_n(x)-f(x)|<\epsilon=\frac{1}{4}$. As $x\in [0, 1)$, $|f_n(x)-f(x)|=|x^n-0|=x^n<\frac{1}{4}$. I get to this point and think $x$ could just be $0.5$ so that $(0.5)^{100}<\frac{1}{4}$. I think at this point that the convergence is uniform. What am I doing wrong?
Best Answer
If you can always find an $x\in [0,1)$ such that $f(x)=1/2$, then you can not get closer to $0$ than $1/4$ uniformly on $[0,1)$
That is what the author is trying to say.