Uniform Convergence Example

Question

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?

Answer 1

The problem is that $x_0$ may well be $\sqrt[n]{\frac12}$, in which case $f_n(x_0)=\frac12\geqslant\varepsilon.$ But we were supposed to have $\bigl\lvert f_n(x)-f(x)\bigr\rvert<\varepsilon=\frac14$ for each $x\in[0,1]$ and, in particular,$$\bigl\lvert f_n(x_0)-f(x_0)\bigr\rvert=\bigl\lvert f_n(x_0)\bigr\rvert<\varepsilon=\frac14.$$

Answer 2

Note that for uniform convergence convergence we need to find $N$ s.t. $n \ge N$ implies that $|f_n(x)-f(x)|< \epsilon$ for all $x \in [0,1]$. Note that we need to choose $N$ that will work for all $x \in [0,1]$. As the function is continuous it's not hard to see that for each $x$ we can find such $N$, however here $N$ is dependant on $x$. If we want to conclude uniform convergence $N$ shouldn't be dependant on $x$.

In the given example the author proves that for $\epsilon = \frac 14$ such $N$ doesn't exist. This $N$ should work for all $x \in [0,1]$. However the author explicitly states that $x_0 = \frac{1}{\sqrt[N]{2}}$ is a point at which the conclusion is wrong. Hence the function isn't uniformly continuous.

Answer 3

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.