i) I have been stuck on this for quite some time. Can anyone explain how $h(x)$ converges uniformly(and absolutely) given the inequality. I don't think I can use Weierstrass M-Test.
ii) Secondly, when they say that a function converges absolutely and uniformly are they saying that it $\sum |f_n(x)|$ converges uniformly or $\sum f_n(x) $ converges uniformly and absolute convergence point wise?
iii) Thirdly, how do we know that f(x) converges uniformly given that we have only shown h(x) converges uniformly which is defined for only $n>n_0$
This was the definiton given in conway regarding the absolute convegrence:
Definition of absolute convergence of product:
Also an equivalent definiton was given:
Best Answer
For (i), given $\epsilon > 0$ there exist $N \in \mathbb{N}$ such that for all $m > n > N$ we have for all $x \in X$
$$\left|\sum_{k=n+1}^m \log (1 + g_k(x)) \right| \leqslant\sum_{k=n+1}^m |\log (1 + g_k(x))| \leqslant \frac{3}{2}\sum_{k=n+1}^m|g_k(x)| < \epsilon,$$
since the RHS series is uniformly convergent.
For (iii), the series $\sum_{n \geqslant 1} \log(1+g_n(x))$ converges uniformly if and only if $\sum_{n \geqslant n_0+1} \log(1+g_n(x))$ converges uniformly. We can add or subtract a finite number of terms without consequence.
Also if $S_n(x) \to S(x)$ uniformly then $\exp(S_n(x)) \to \exp(S(x))$ since the exponential function is continuous everywhere. Thus, uniform convergence of $h(x)$ imples uniform convergence of $f(x)$.
Addendum: Absolute convergence of an infinite product implies convergence
Let $P_n = \prod_{k=1}^n (1+a_k)$ and $Q_n = \prod_{k=1}^n (1+|a_k|)$. We have
$$P_n - P_{n-1} = (1+a_1) \ldots (1+a_{n-1}) a_n, \\ Q_n - Q_{n-1} = (1+|a_1|) \ldots (1+|a_{n-1}|) |a_n|,$$
and it follows that $|P_n - P_{n-1}| \leqslant Q_n - Q_{n-1}$.
If $\prod(1+|a_n|)$ is convergent then the series $\sum(Q_n- Q_{n-1})$ converges since
$$\lim_{N \to \infty}\sum_{n=2}^N (Q_n - Q_{n-1})= Q_1 + \lim_{N \to \infty}Q_N = \prod_{n=1}^\infty (1 + |a_n|)$$
By the comparison test, the series $\sum(P_n- P_{n-1})$ is convergent and, therefore, the product $\prod(1+a_n)$ is convergent since
$$\prod_{n=1}^\infty(1+a_n) = \lim_{N \to \infty} P_N = P_1 + \sum_{n=2}^\infty (P_n - P_{n-1})$$
A final but important detail is to show that $\lim_{N \to \infty}P_N \neq 0$. This follows from the convergence of $\sum|a_n|$ which implies $1 + a_n \to 1$. It follows that the series $\sum |a_n(1+a_n)^{-1}|$ and, hence, the product $\prod(1 - a_n(1+a_n)^{-1})$ are convergent. Thus,
$$\lim_{N\to \infty} \frac{1}{P_N} = \prod_{n=1}^\infty \frac{1}{1+a_n} = \prod_{n=1}^\infty \left(1 - \frac{a_n}{1+a_n}\right) \neq \infty$$