Uniform Convergence on an open interval

Question

My textbook says that for the sequence of functions $\{f_n\}$ given by $$f_n(x)=1+x+x^2+\dots+x^{n-1}=\frac{1-x^n}{1-x},$$ $f_n(x)$ converges pointwise to $f$ for each $x$ in $(-1,1)$.

However, when talking about uniform convergence, the book says that $f_n$ converges uniformly in $[-c, c]$ where $0<c<1$.

It seems unintuitive to state the range of uniform convergence like this. Is there a reason why the book doesn't say it converges uniformly for each $x$ in $(-1, 1)$?

Answer 1

Suppose that $f_n(x)=\frac{1-x^n}{1-x}$ converges uniformly on $(-1,1)$. Then, it must also be uniformly Cauchy, i.e., for all $\epsilon>0$, we can find some $N\in\mathbb{R}$ such that $\big|\frac{1-x^n}{1-x}-\frac{1-x^m}{1-x}\big|<\epsilon$ whenever $n,m\geq N$ and $x\in(-1,1)$. Simplifying this and using the fact that $x\in(-1,1)$, we find that $\big|x^m-x^n\big|<\big|\frac{x^m-x^n}{1-x}\big|<\epsilon$.

But by choosing $x=(\frac{1}{2})^\frac{1}{m}$, we have that $|\frac{1}{2}-(\frac{1}{2})^\frac{n}{m}|<\epsilon$. But then for large enough $n$ (which we supposed will satisfy the inequality), this second term will approach $0$, leaving us with $|\frac{1}{2}|<\epsilon$. However, this implies that no matter our choice of $N$, we can always find some $x\in(-1,1)$ such that if $|f_n(x)-f_m(x)|\geq\frac{1}{2}$, so $\{f_n(x)\}$ is not uniformly Cauchy on $(-1,1)$, so it cannot be uniformly convergent here either.

The reason it is uniformly convergent on $[-c,c]$ for $0<c<1$ is because we cannot choose $x$ to be arbitrarily close to 1, as we did at the beginning of the second paragraph.

Answer 2

Suppose that $f_n$ is convergent uniformly on $(-1,1)$. Then it is convergent uniformly to $f(x)=(1-x)^{-1}.$ Therefore for $\varepsilon =1$ there is $ n_0$ such that $$|f_{n_0}(x)-f(x)|={|x|^{n_0}\over 1-x}<1,\quad |x|<1$$ This leads to a contradiction as $$\lim_{x\to 1^-}{|x|^{n_0}\over 1-x}=\infty$$

Answer 3

Uniform convergence applies to sets (a function is uniformly convergent in an interval, not at a point).

The uniform convergence in $[-c,c]$, with $0<c<1$ is given by the fact that the set $[-c,c]$ is compact and the functions $f_n$ are continuous in $[-c,c]$. This result does not extend to intervals that are not closed (because they are not compact sets).

To show that $f_n$ does not converge uniformly in $(-1,1)$, reason by contradiction and assume it does. This means that for all $\epsilon>0$, there is $N_{\epsilon}$ such that, for all $x\in(-1,1)$

$$|f_n(x)-f(x)|<\epsilon\quad\forall n\geq N_{\epsilon}.$$

Now, $|f_n(x)-f(x)|=\frac{|x|^n}{1-x}$, hence, taking $\epsilon=1$, the uniform convergence implies that there exists $N_1$ such that for all $x\in(-1,1)$

$$\frac{|x|^{N_1}}{1-x}<1$$

which is a contradiction because $\frac{|x|^{N_1}}{1-x}\to\infty$ when $x\to ±1$.