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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:

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Also an equivalent definiton was given: enter image description here

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  • $\begingroup$ Comparison test. $\endgroup$ – copper.hat Nov 11 '18 at 6:49
  • $\begingroup$ @copper.hat Yes but how do you apply it what is your comparison sequence here. g_n(x) is a sequence of functions not a sequence of integers whose sum converges. How doe one get such a sequence? $\endgroup$ – Jhon Doe Nov 11 '18 at 6:54
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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,

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

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    $\begingroup$ $\log(1+g_n(x))$ need not be defined for $n \le n_0$ since you may have $1+g_n(x) = 0$. See the second part of the theorem. A better argument is this: $\prod_{n=1}^\infty (1+g_n(x))$ converges absolutely and uniformly if and only if $\prod_{n=n_0+1}^\infty (1+g_n(x))$ converges absolutely and uniformly. $\endgroup$ – Paul Frost Nov 11 '18 at 9:39
  • $\begingroup$ @RRL the proof for iii) makes use of cauchy criterion to proof i.e. we can get rid of the initial terms. But does how do i prove it with products. We can't minus of the initial terms like in the cauchy criterion $\endgroup$ – Jhon Doe Nov 11 '18 at 11:36
  • $\begingroup$ @PaulFrost pls refer to the above comment $\endgroup$ – Jhon Doe Nov 11 '18 at 12:07
  • $\begingroup$ I have to correct my comment: If $\prod_{n=n_0+1}^\infty(1+\lvert g_n(x) \rvert)$ converges absolutely and uniformly, then also $\prod_{n=1}^\infty(1+ g_n(x))$ converges absolutely and uniformly. The converse fails. Take e.g. $g_1(x) \equiv -1$. $\endgroup$ – Paul Frost Nov 11 '18 at 16:45
  • $\begingroup$ If for every $n$ there exist $k > n$ such that $g_k(x) = -1$ then trivially the product diverges to $0$. There is no point pursuing the question of convergence for such a product. If only a finite number of factors are $0$ then convergence is determined from the product with these factors removed. A product also can diverge to $0$ non-trivially. We say $\prod(1+a_n)$ diverges to $0$ if $\sum \log(1+a_n)$ diverges to $-\infty$. $\endgroup$ – RRL Nov 11 '18 at 16:59
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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.

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