Theorem 9.5 Cauchy Condition for uniform convergence of series - Math Analysis 2nd ed - Apostol

Question

Theorem 9.5 (Cauchy condition for uniform convergence of series) The infinite series $\sum f_n(x)$ converges uniformly on $S$ if, and only if, for every $\epsilon>0$ there is an $N$ such that $n>N$ implies $$\left |\sum_{k=n+1}^{n+p}f_k(x)\right|<\epsilon \qquad \text{for each $p=1,2,...,$ and every $x$ in $S$.}$$

Proof. Let's define $s_n(x)=\sum_{k=1}^nf_k(x)$ for $n=1,2,...$ . If $\sum f_n(x)\to f(x)$ uniformly on $S$, then there is an $\epsilon>0$ and an $N$ such that $n>N$ implies
$$f(x)-\epsilon<\sum_{k=1}^{n+p}f_k(x)<f(x)+\epsilon\qquad (1)\\ \text{so we have also}\qquad f(x)-\epsilon<f_1+f_2+...+f_n+\sum_{k=n+1}^pf_k<f(x)+\epsilon$$

We know that the $\lim_{p\to\infty}f_1+f_2+...+f_n+\sum_{k=n+1}^pf_k=f_1+f_2+...+f_n+f(x)$.

Now we can write $$f(x)-\epsilon<f_1+f_2+...+f_n+f(x)<f(x)+\epsilon\\-\epsilon<f_1+f_2+...+f_n<+\epsilon\\\text{now I can write}\qquad \left |\sum_{k=n+1}^{n+p}f_k(x)\right|<\epsilon.$$

This is my attempt to prove the necessity part. The sufficiency part should follow step by step the inverse path. Please could you check my proof and correct me if I'm wrong somewhere? I'm not convinced at all. Thank you in advance.

Edit I substituted $\sum_{k=1}^{n+p}f_k(x)$ to $s_n(x)$ in (1).

Answer 1

For your proof:

We know that the $\lim_{p\to\infty}f_1+f_2+...+f_n+\sum_{k=n+1}^pf_k=f_1+f_2+...+f_n+f(x)$.

That's not correct.

Everything below this line is just for the proof of sufficency ---------------

For the sufficiency: there's probably more than one way to do it, but here's a neat way to show sufficiency (stolen from Ahlfors' Complex Analysis p.36).

Pick $\epsilon$, and let $N$ be large enough so that $| \sum_{k=n+1}^{n+p}f_k(x) | < \epsilon$ whenever $n > N$.

Key point: The series is uniformly Cauchy, so is Cauchy for a fixed $x$, and so converges pointwise for each $x$

Keeping $n$ and $x$ fixed, but letting $p \to \infty$ we see that $| \sum_{k=n+1}^{\infty}f_k(x) |\le \epsilon$.

Note that $$\sum_{k=n+1}^{\infty}f_k(x) = \sum_{k=0}^{\infty}f_k(x) - \sum_{k=0}^{n}f_k(x) = f(x) - \sum_{k=0}^{n}f_k(x) $$

$x$ was arbitrary, so we just found $N$ such that $n > N$ implies $| f(x)- \sum_{k=0}^{n}f_k(x) | \le \epsilon$

Now replace $\epsilon$ with $\frac{\epsilon}{2}$.