Question on uniform convergence of sum of continuous functions.

Question

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:

Answer 1

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,

$$\lim_{N\to \infty} \frac{1}{P_N} = \prod_{n=1}^\infty \frac{1}{1+a_n} = \prod_{n=1}^\infty \left(1 - \frac{a_n}{1+a_n}\right) \neq \infty$$

Answer 2

An infinite product $\prod_{n=1}^\infty x_n$ is said to converge to $x$ if $\xi_k = \prod_{n=1}^k x_n \to x$ as $k \to \infty$. Some authors require that all $x_n \ne 0$ and some authors regard $\prod_{n=1}^\infty x_n$ as divergent if $\xi_k \to 0$. The above question shows that the referenced textbook uses the most general interpretation.

I am not really sure what it means that an infinite product $\prod_{n=1}^\infty(1+a_n)$ converges absolutely. In

Trench, William F. "Conditional convergence of infinite products." The American mathematical monthly 106.7 (1999): 646-651

http://ramanujan.math.trinity.edu/wtrench/research/papers/TRENCH_RP_93.PDF

one can find the following definition: $\prod_{n=1}^\infty(1+a_n)$ converges absolutely if $\prod_{n=1}^\infty (1 + \lvert a_n \rvert)$ converges. That seems reasonable. Absolute convergence implies convergence: Let $u_k = \prod_{n=1}^k (1+a_n)$ and $v_k = \prod_{n=1}^k (1 + \lvert a_n \rvert)$. We have $\lvert 1 + a_n \rvert \le 1 + \lvert a_n \rvert $. Hence for $k < l$ we get $$\lvert u_l - u_k \rvert = \prod_{n=1}^k \lvert 1+a_n \rvert \cdot \lvert \prod_{n=k+1}^l (1+a_n) - 1 \rvert = \prod_{n=1}^k \lvert 1+a_n \rvert \cdot \lvert s(a_{k+1},\dots,a_l) \rvert ,$$ where $s(a_{k+1},\dots,a_l)$ is the sum of all finite products $a_{r_1} \dots a_{r_j}$ with $1 \le j \le l-k$ and $k+1 \le r_1 < \dots < r_j \le l$. We conclude $$\lvert u_l - u_k \rvert \le \prod_{n=1}^k (1 +\lvert a_n \rvert) \cdot s(\lvert a_{k+1} \rvert,\dots,\lvert a_l\rvert) = \prod_{n=1}^k (1 +\lvert a_n \rvert) \cdot (\prod_{n=k+1}^l (1+\lvert a_n \rvert ) - 1)$$ $$= v_l- v_k = \lvert v_l - v_k \rvert .$$ Therefore $(u_k)$ is a Cauchy sequence and converges.

RRL's proof shows that $\prod_{n=n_0+1}^\infty(1+\lvert g_n(x) \rvert)$ converges uniformly to some $G(x)$. Let $w_k(x) = \prod_{n=n_0+1}^k(1+\lvert g_n(x) \rvert)$ for $k > n_0$, $w(x) = \prod_{n=1}^{n_0}(1+\lvert g_n(x) \rvert)$ and $W = sup_{x \in X} w(x)$. For $k > n_0$ we get $$\lvert \prod_{n=1}^k (1+\lvert g_n(x) \rvert) - w(x)G(x) \rvert = w(x) \lvert w_k(x) - G(x) \rvert \le W \lvert w_k(x) - G(x) \rvert .$$ This shows that $\prod_{n=1}^\infty(1+\lvert g_n(x) \rvert)$ converges uniformly to $w(x)G(x)$. Therefore $\prod_{n=1}^\infty(1+ g_n(x))$ converges absolutely and uniformly.