# In Theorem 8.1 of Baby Rudin, why interval of uniform convergence must be closed subset of (-R,R)?

In Theorem of Baby Rudin 8.1, Rudin said that

If series $$\sum_{n=0}^\infty c_nx^n$$ converges for $$|x|

then $$\sum_{n=0}^\infty c_nx^n$$ converges uniformly on $$[-R+\epsilon, R-\epsilon]$$, no matter which $$\epsilon > 0$$ is chosen.

and I don't know why I can't use the open interval (-R, R) instead of $$[-R+\epsilon, R-\epsilon]$$

• Your question is not clear right now in my opinion. Are you asking how to conclude that the series converges uniformly on $(-R,R)$? Because that is not necessarily true. Apr 1 '20 at 12:37
• I believe OP is asking why the theorem cannot be strengthened to say that the series converges uniformly on $(-R,R)$. Apr 1 '20 at 12:43

$$\sum x^{n}$$ converges for $$|x| <1$$ but the convergence is not uniform on $$(-1,1)$$. Reason: $$x^{n}$$ does not tend to $$0$$ unifiormly on this interval.

Consider the harmonic series $$1+z+z^2+\cdots$$ You know that this series converge if $$|z|<1$$ and it fail to converge if $$z=1$$.

The point is that you don't have necessarily a good behavior in the extremes.

The difference is the word uniform and the placement of quantifiers. If $$(f_n)$$ is a sequence of functions on an interval $$I$$, we say $$(f_n)$$ converges to $$f$$ pointwise if: $$\forall x \in I,\forall \eta > 0, \exists N\in \mathbb N, \forall n \in \mathbb{N}, n > N \implies |f_n(x) - f(x) | < \eta$$ We say $$(f_n)$$ converges to $$f$$ uniformly on $$I$$ if: $$\forall \eta > 0, \exists N \in \mathbb{N},\forall x \in I,\forall n \in\mathbb{N}, n > N \implies |f_n(x) - f(x)| < \eta$$ In the first convergence, $$N$$ may depend on $$x$$ and $$\eta$$. But in the second, $$N$$ needs to work for all $$x$$ simultaneously (or “uniformly”).

As an example, consider the sequence of functions $$f_n(x) = x^n$$ on the interval $$(-1,1)$$. The pointwise limit of the sequence is $$0$$.

We claim that $$(f_n)$$ does not converge uniformly to $$0$$ on $$(-1,1)$$. To show this, let $$\eta = \frac{1}{2}$$. Given $$N \in \mathbb{N}$$, let $$x$$ be any number in the interval $$(\eta^{1/(N+1)},1)$$. Then $$|f_{N+1}(x)| = |x^{N+1}| > \eta$$.

However, for any $$\epsilon > 0$$ $$(f_n)$$ does converge to $$0$$ uniformly on $$(1-\epsilon,1-\epsilon)$$. Given $$\eta > 0$$, choose $$N$$ such that $$(1-\epsilon)^N < \eta$$. Then for any $$x$$ with $$|x| < 1-\epsilon$$, and any $$n > N$$, $$|x|^{n} < (1-\epsilon)^n < (1-\epsilon)^N < \eta$$