hcju/proofwikiqa huggingface dataset:
From Quasiamicable Numbers: $48$ and $75$ we have that $48$ and $75$ are quasiamicable numbers. It remains to be demonstrated that these are indeed the smallest such. Let $n$ be the smaller number of the quasiamicable pair. Then we must have $\map \sigma n - n - 1 > n$. Since $\map \sigma n > 2 n$, $n$ is abundant. The abundant numbers for $n < 48$ are: :$12, 18, 20, 24, 30, 36, 40, 42$ And we have: {{begin-eqn}} {{eqn | l = \map \sigma {12} - 13       | r = \paren {1 + 2 + 4} \paren {1 + 3} - 13 }} {{eqn | r = 15 }} {{eqn | l = \map \sigma {15} - 16       | r = \paren {1 + 3} \paren {1 + 5} - 16 }} {{eqn | r = 8       | rr = \ne 12 }} {{eqn | l = \map \sigma {18} - 19       | r = \paren {1 + 2} \paren {1 + 3 + 9} - 19 }} {{eqn | r = 20 }} {{eqn | l = \map \sigma {20} - 21       | r = \paren {1 + 2 + 4} \paren {1 + 5} - 21 }} {{eqn | r = 21       | rr = \ne 18 }} {{eqn | l = \map \sigma {21} - 22       | r = \paren {1 + 3} \paren {1 + 7} - 22 }} {{eqn | r = 10       | rr = \ne 20 }} {{eqn | l = \map \sigma {24} - 25       | r = \paren {1 + 2 + 4 + 8} \paren {1 + 3} - 25 }} {{eqn | r = 35 }} {{eqn | l = \map \sigma {35} - 36       | r = \paren {1 + 5} \paren {1 + 7} - 36 }} {{eqn | r = 12       | rr = \ne 24 }} {{eqn | l = \map \sigma {30} - 31       | r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 5} - 31 }} {{eqn | r = 41 }} {{eqn | l = \map \sigma {41} - 42       | r = 0       | rr = \ne 30, 36 }} {{eqn | l = \map \sigma {36} - 37       | r = \paren {1 + 2 + 4} \paren {1 + 3 + 9} - 37 }} {{eqn | r = 41 }} {{eqn | l = \map \sigma {40} - 41       | r = \paren {1 + 2 + 4 + 8} \paren {1 + 5} - 41 }} {{eqn | r = 49 }} {{eqn | l = \map \sigma {49} - 50       | r = \paren {1 + 7 + 49} - 50 }} {{eqn | r = 7       | rr = \ne 40 }} {{eqn | l = \map \sigma {42} - 43       | r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 7} - 43 }} {{eqn | r = 53 }} {{eqn | l = \map \sigma {53} - 54       | r = 0       | rr = \ne 42 }} {{end-eqn}} hence $n$ is at least $48$, proving the result. {{qed}}
By definition of complex logarithm: :$-\map \ln {1 + z} = \displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {-z}^n} n$ Thus {{begin-eqn}} {{eqn  | l = \cmod {1 - \frac {\map \ln {1 + z} } z}        | r = \cmod {\sum_{n \mathop = 2}^\infty \frac {\paren {-1}^n z^{n - 1} }n}        | c = Linear Combination of Convergent Series }} {{eqn  | o = \le        | r = \sum_{n \mathop = 2}^\infty \cmod {\frac{\paren {-z}^{n - 1} } n}        | c = Triangle Inequality for Series }} {{eqn  | o = \le        | r = \sum_{n \mathop = 2}^\infty \frac {\cmod z^{n - 1} } 2        | c = $n \ge 2$ }} {{eqn  | r = \frac {\cmod z / 2} {1 - \cmod z}        | c = Sum of Geometric Sequence }} {{eqn  | o = \le        | r = \frac 1 2        | c = $\cmod z \le \dfrac 1 2$ }} {{end-eqn}} By the Triangle Inequality: :$\displaystyle \frac 1 2 \le \cmod {\frac {\map \ln {1 + z} } z} \le \dfrac 3 2$ {{qed}}
By continuity of $x \mapsto 1 / x$: :$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = 1}^N \frac 1 {1 + a_n} = \frac 1 a$ It remains to prove the absolute convergence. Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$  converges absolutely, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely. By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} \le \dfrac 1 2 $ for $n$ sufficiently large. Thus $\norm {\dfrac 1 {a_n + 1} - 1} = \norm {\dfrac {a_n} {a_n + 1} } \le 2 \norm {a_n}$ for $n$ sufficiently large. By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 {a_n + 1} - 1}$ converges absolutely. Thus $\ds \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$  converges absolutely. {{qed}} Category:Infinite Products bvj860sg8ra24s02t8zbbtpfd9gxcwc
Let $s \left({n}\right)$ denote the sum of the proper divisors of (strictly) positive integer $n$. The sum of all the divisors of a (strictly) positive integer $n$ is $\sigma \left({n}\right)$, where $\sigma$ is the $\sigma$ function. The proper divisors of $n$ are the divisors $n$ with $1$ and $n$ excluded. Thus: :$s \left({n}\right) = \sigma \left({n}\right) - n - 1$ Suppose: :$s \left({n}\right) = m$ and: :$s \left({m}\right) = n$ Then: {{begin-eqn}} {{eqn | l = \sigma \left({n}\right) - n - 1       | r = m       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({n}\right)       | r = m + n + 1       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \sigma \left({m}\right) - m - 1       | r = n       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({m}\right)       | r = m + n + 1       | c =  }} {{end-eqn}} Thus: :$s \left({n}\right) = s \left({m}\right) = m + n + 1$ The argument reverses. {{qed}} Category:Quasiamicable Numbers o4dzfu4ftg5w34ol2ejglp25odrs98d
=== 1 implies 2 === By the Monotone Convergence Theorem, it suffices to show that the partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ are bounded. Because $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \norm{a_n}}\right)$ converges, its partial products are bounded. By Bounds for Finite Product of Real Numbers, $\displaystyle \sum_{n \mathop = 1}^N \norm{a_n} \leq \prod_{n \mathop = 1}^N\left({1 + \norm{a_n}}\right)$. {{qed}} === 2 implies 1 === ==== Proof 1 ==== By the Monotone Convergence Theorem, it suffices to show that the partial products of $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \norm{a_n}}\right)$ are bounded. By Bounds for Finite Product of Real Numbers: :$\displaystyle \prod_{n \mathop = 1}^N\left({1 + \norm{a_n}}\right) \leq \exp\left( \sum_{n \mathop = 1}^N \norm{a_n}\right)$ Because $\displaystyle \sum_{n \mathop = 1}^\infty \norm{a_n}$ converges, its partial sums are bounded. {{qed}} ==== Proof 2 ==== By the Monotone Convergence Theorem, it suffices to show that the partial products of $\displaystyle \prod_{n \mathop = 1}^\infty\left({1 + \norm{a_n}}\right)$ are bounded. By the AM-GM Inequality: :$\displaystyle \prod_{n \mathop = 1}^N\left({1 + \norm{a_n} }\right) \leq \left(\frac{N + \sum_{n \mathop = 1}^N \norm{a_n}}N \right)^N\leq \left(1 + \frac MN\right)^N$ where $M>0$ is such that $\displaystyle \sum_{n \mathop = 1}^N \norm{a_n}\leq M$ for all $N$. By definition of the real exponential, $\displaystyle\left(1 + \frac M N\right)^N \to \exp(M)$ as $N\to\infty$. By Convergent Sequence in Metric Space is Bounded, $\left(1 + \frac MN\right)^N$ is bounded. {{qed}}
By definition of convergent product, there exists $n_0 \in \N$ such that: :$a_n \ne 0$ for $n \ge n_0$ :the sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty a_n$ has a nonzero limit. Let $p_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = n_0}^\infty a_n$. For $n > n_0$: :$a_n = \dfrac {p_n} {p_{n - 1} }$ By the Combination Theorem for Sequences: :$a_n \to 1$ {{qed}}
Let $p_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$. Let $s_n$ denote the $n$th partial sum of $\displaystyle \sum_{n \mathop = 1}^\infty \ln a_n$. By Sum of Logarithms, $s_n = \map \ln {p_n}$. === 1 implies 2 === Let $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ converge to $a>0$. Then $p_n \to a$. By Natural Logarithm Function is Continuous, $s_n \to \ln a$. {{qed|lemma}} === 2 implies 1 === Let $\displaystyle \sum_{n \mathop = 1}^\infty \ln a_n$ converge to $\ln a$. Then $s_n \to \ln a$. By Exponential Function is Continuous, $p_n \to a$. Because $a \ne 0$, $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ converges to $a$. {{qed}}
Because $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ is absolutely convergent, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent. By Terms in Convergent Series Converge to Zero, $a_n\to0$. {{qed}}
Because $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ is absolutely convergent, the real product $\displaystyle \prod_{n \mathop = 1}^\infty (1+ \norm{a_n})$ is convergent. By Factors in Convergent Product Converge to One, $\norm{a_n} \to 0$. Thus $a_n\to0$. {{qed}}
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$. Let $n_1 \ge n_0$. {{AimForCont}} the product diverges to $0$. Then: :$\displaystyle \prod_{n \mathop = n_1}^\infty \paren {1 + a_n} = 0$ By Norm of Limit: :$\displaystyle \prod_{n \mathop = n_1}^\infty \norm {1 + a_n} = 0$ By the Triangle Inequality and Squeeze Theorem: :$\displaystyle \prod_{n \mathop = n_1}^\infty \paren {1 - \norm {a_n} } = 0$ By the Weierstrass Product Inequality, we have for $N \ge n_1$: :$\displaystyle \prod_{n \mathop = n_1}^N \paren {1 - \norm {a_n} } \ge 1 - \sum_{n \mathop = n_1}^N \norm{a_n}$ Taking limits: :$0 \ge 1 - \displaystyle \sum_{n \mathop = n_1}^\infty \norm {a_n}$ Because $\displaystyle \prod_{n \mathop = 1}^\infty \paren  {1 + a_n}$ is absolutely convergent: :$\displaystyle \sum_{n \mathop = n_1}^\infty \norm {a_n} < 1$ for $n_1$ sufficiently large. This is a contradiction. {{qed}}
We have that $\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \norm {a_n} }$ is absolutely convergent. By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$. Thus $\displaystyle \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$ is absolutely convergent. {{AimForCont}} the product diverges to $0$. Then: :$\displaystyle \prod_{n \mathop = n_0}^\infty \paren {1 + a_n} = 0$ By Norm of Limit: :$\displaystyle \prod_{n \mathop = n_0}^\infty \norm {1 + a_n} = 0$ By the Triangle Inequality and Squeeze Theorem: :$\displaystyle \prod_{n \mathop = n_0}^\infty \paren {1 - \norm {a_0} } = 0$ By Logarithm of Infinite Product of Real Numbers: :$\displaystyle \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$ diverges to $-\infty$. This is a contradiction. {{qed}}
By absolute convergence of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$, $\displaystyle \sum_{n \mathop = 1}^\infty \left({a_n - 1}\right)$ converges absolutely. By the Triangle Inequality: : $\size{\norm{a_n} - 1} \le \norm{a_n - 1}$ By the Comparison Test, $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$  converges absolutely. By Absolute Value is Continuous, its limit is $\norm{a}$. {{qed}}
By Absolute Value of Limit of Sequence: :$\displaystyle \prod_{n \mathop = 1}^\infty \norm {a_n} = \norm a$ It remains to show that the product converges. By the convergence, there exists $n_0\in\N$ such that $a_n \ne 0$ for $n \ge n_0$. Then $\norm {a_n} \ne 0$ for $n \ge n_0$. Let $P_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = n_0}^\infty a_n$. Then $\norm {P_n}$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = n_0}^\infty \norm {a_n}$. Let $P_n$ converge to $b \in \mathbb K \setminus \set 0$. By Convergence of Absolute Value of Sequence: :$\norm {P_n} \to \norm b \in \mathbb K \setminus \set 0$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges. {{qed}}
=== 1 implies 2 === ==== Case 1 ==== Let there be infinitely many $n\in\N$ such that $a_n=0$. Then there are infinitely many $n\in\N$ such that $|a_n|=0$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ diverges to $0$. ==== Case 2 ==== Let $n_0\in\N$ such that $a_n\neq0$ for $n\geq n_0$. Then $\norm{a_n}\neq0$ for $n\geq n_0$. By the divergence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N a_n = 0$. By Absolute Value of Limit of Sequence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N \norm{a_n} = 0$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ diverges to $0$. {{qed|lemma}} === 2 implies 1 === ==== Case 1 ==== Let there be infinitely many $n\in\N$ such that $\norm{a_n} = 0$. Then there are infinitely many $n\in\N$ such that $a_n = 0$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$. ==== Case 2 ==== Let $n_0\in\N$ such that $\norm{a_n}\neq0$ for $n\geq n_0$. Then $a_n\neq0$ for $n\geq n_0$. By the divergence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N \norm{a_n} = 0$. By Absolute Value of Limit of Sequence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N a_n = 0$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$. {{qed}}
Let $p_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$. Let $s_n$ denote the $n$th partial sum of $\displaystyle \sum_{n \mathop = 1}^\infty\log a_n$. By Sum of Logarithms: :$s_n = \map \log {p_n}$ {{ProofWanted}}
Let $p_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$. Let $s_n$ denote the $n$th partial sum of $\displaystyle \sum_{n \mathop = 1}^\infty \log a_n$. By Sum of Logarithms: :$s_n = \log \left({p_n}\right)$ {{ProofWanted}}
Let $n_0 \in \N$ such that $a_n \ne 0$ for $n> n_0$. Let $n_1 \in \N$ such that $b_n \ne 0$ for $n> n_1$. Then $a_n b_n \ne 0$ for $n > n_2 = \max \set {n_0, n_1}$. Let $p_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2 + 1}^\infty a_n$. Let $q_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2 + 1}^\infty b_n$. Then $p_n q_n$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = n_2 + 1}^\infty a_n b_n$. Because $p_n$ and $q_n$ converge to a nonzero limit, so does $p_n q_n$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n b_n$ converges. Let $P_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n$. Let $Q_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty b_n$. Then $P_n Q_n$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty a_n b_n$. By Limit of Product of Sequences, $P_n Q_n\to ab$. Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n b_n$ converges to $a b$. {{qed}}
Take as an example $n = 49$. We have that: :$49^3 = 117 \, 6 \mathbf{49}$ demonstrating that $49$ is trimorphic. However, we also have that: :$49^2 = 2401$ demonstrating that $49$ is not automorphic. {{qed}}
We have: : $a_n b_n - 1 = \left({a_n - 1}\right) \left({b_n - 1}\right) + \left({a_n - 1}\right) + \left({b_n - 1}\right)$ By the Triangle Inequality: :$\norm{a_n b_n - 1} \le \norm{a_n - 1} \norm {b_n - 1} + \norm{a_n - 1} + \norm{b_n -1}$ By the absolute convergence, $\displaystyle \sum_{n \mathop = 1}^\infty \norm{a_n - 1}$ and $\displaystyle \sum_{n \mathop = 1}^\infty \norm{b_n -1}$ converge. By Inner Product of Absolutely Convergent Series, $\displaystyle \sum_{n \mathop = 1}^\infty \norm{a_n - 1} \norm{b_n -1}$ converges. By the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty \norm{a_n b_n - 1}$ converges. {{qed}}
Let $\epsilon > 0$. Let $n_0 \in \N$ be such that $\displaystyle \prod_{n \mathop = n_0}^\infty a_n$ converges to some $a \in \mathbb K \setminus \set 0$. By Convergent Sequence is Cauchy Sequence, there exists $N_0 \ge n_0$ such that: :$\displaystyle \norm {\prod_{n \mathop = n_0}^k a_n - \prod_{n \mathop = n_0}^l a_n} \le \epsilon$ for $k, l \ge N_0$. By Sequence Converges to Within Half Limit, there exists $N_1 \ge n_0$ such that: :$\displaystyle \norm {\prod_{n \mathop = n_0}^M a_n} \ge \frac {\norm a}2$ for $M \ge N_1$. Let $N = \max \set {N_0, N_1}$. For $N + 1 \le k \le l$: {{begin-eqn}} {{eqn | l = \norm {\prod_{n \mathop = k}^l a_n - 1}       | r = \norm {\frac {\prod_{n \mathop = n_0}^l a_n - \prod_{n \mathop = n_0}^{k - 1} a_n} {\prod_{n \mathop = n_0}^{k - 1} } }       | c =  }} {{eqn | o = \le       | r = \frac {2 \epsilon} {\norm{a} }       | c = $l, k - 1 \ge N_0$ and $k - 1 \ge N_1$ }} {{end-eqn}} Hence the result. {{qed}}
Let $n_0 \in \N$ be such that the sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty f_n$ converges uniformly. By the Uniform Limit Theorem, $\displaystyle \prod_{n \mathop = n_0}^\infty f_n$ is continuous. Because $f_1, \ldots, f_{n_0 - 1}$ are continuous, so is $\displaystyle \prod_{n \mathop = 1}^\infty f_n$. {{qed}}
=== 1 implies 2 === It suffices to show that: :$(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty \map \Re {\ln f_n} = \map \Re {\ln f}$ locally uniformly :$(2): \quad \displaystyle \sum_{n \mathop = 1}^\infty \map \Im {\ln f_n} = \map \Im {\ln f} + 2 k \pi$ locally uniformly for some $k: K \to \Z$ ==== Real part ==== By Absolute Value of Uniformly Convergent Product, $\displaystyle \prod_{n \mathop = 1}^\infty \cmod {f_n} = \cmod f$ locally uniformly. By Logarithm of Infinite Product of Real Functions $\displaystyle \sum_{n \mathop = 1}^\infty \ln \cmod {f_n} = \ln \cmod f$ locally uniformly. Hence the result. ==== Imaginary Part ==== Let $K\subset X$ be compact. Let $P_n$ denote the $n$th partial product. Then $P_n \to f$ uniformly on $K$. Let $\theta$ be the argument of $f$. Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\hointl {\theta - \pi} {\theta + \pi}$. By the corollary to Uniform Convergence of Complex Functions in Polar Form: :$\theta_n \to \theta$ uniformly on $K$. Let $k_n: K \to \Z$ be such that: :$\displaystyle \sum_{j \mathop = 1}^n \map \Im {\ln \map {f_n} x} = \map {\theta_n} x + 2 \map {k_n} x \pi$ We show that $k_n$ is eventually equal to a fixed function $k: K \to \Z$. By the Triangle Inequality for Complex Numbers: :$2 \pi \sup_{x \mathop \in K} \size {\map {k_{n + 1} } x - \map {k_n} x} \le \sup_{x \mathop \in K} \cmod {\map {\theta_{n + 1} } x - \map {\theta_n} x} + \sup_{x \mathop \in K} \size {\map \Im {\ln \map {f_{n + 1} } x } }$ By Factors in Uniformly Convergent Product Converge Uniformly to One, $f_n \to 1$ uniformly on $K$. Let $n_0 \in \N$ be such that $\cmod {f_n - 1} \le \dfrac 1 2$ for $n \ge n_0$. By Complex Logarithm is Continuous Outside Branch and Heine-Cantor Theorem, $\ln$ is uniformly continuous on $\map {\overline B} {1, \dfrac 1 2}$. By Uniformly Continuous Function Preserves Uniform Convergence, $\ln f_n \to 0$ uniformly on $K$. Thus $\displaystyle 2 \pi \sup_{x \mathop \in K} \cmod {\map {k_{n + 1} } x - \map {k_n} x} \to 0$. So the sequence $k_n$ is eventually constant, say equal to $k: K \to \Z$. Then: :$\displaystyle \sum_{j \mathop = 1}^n \map \Im {\ln f_n} \to \theta + 2 k \pi$ uniformly on $K$. {{qed|lemma}} === 2 implies 1 === Follows from Complex Exponential is Uniformly Continuous on Half-Planes. {{ProofWanted|Explain how.}}
{{questionable|this needs a stronger version of Corollary to Logarithm of Infinite Product of Complex Functions, where the logarithms are analytic}} Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic. Let $z_0\in D$ with $f(z_0)\neq0$. By the Corollary to Logarithm of Infinite Product of Complex Functions, there exist $n_0\in\N$, $k\in\Z$ and a open neighborhood $U$ of $z_0$ such that: * $f_n(z)\neq0$ for $n\geq n_0$ and $z\in U$ * The series $\displaystyle \sum_{n \mathop = n_0}^\infty \log f_n$ converges uniformly on $U$ to $\log g + 2k\pi i$, where $g = \displaystyle \prod_{n\mathop =n_0}^\infty f_n$. We have, for $z\in U$: {{begin-eqn}} {{eqn | l = \frac {f'}{f}       | r = \frac {(f_1\cdots f_{n_0-1}\cdot g)'}{f_1\cdots f_{n_0-1}\cdot g}       | c =  }} {{eqn | l =        | r = \frac {f_1'}{f_1} + \cdots + \frac {f_{n_0-1}'}{f_{n_0-1} } + \frac{g'}g       | c = Logarithmic Derivative of Product of Analytic Functions }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \frac {g'}{g}       | r = (\log g)'       | c = Logarithmic Derivative is Derivative of Logarithm }} {{eqn | l =        | r = \sum_{n \mathop = n_0}^\infty (\log f_n)' - (2k\pi i)'       | c = Derivative of Uniform Limit of Analytic Functions }} {{eqn | l =        | r = \sum_{n \mathop = n_0}^\infty \frac {f_n'}{f_n}       | c = Logarithmic Derivative is Derivative of Logarithm }} {{end-eqn}} and the series converges locally uniformly on $U$. Thus $\frac{f'}f = \displaystyle \sum_{n \mathop = 1}^\infty \frac {f_n'}{f_n}$ converges locally uniformly on $U\setminus\{z\in U : f(z) = 0\}$. {{qed}}
=== Sufficient Condition === Assume that :$d$ is a surjection. By Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element :$\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$ By Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity: :$g \circ d = I_T$ Thus by Injection iff Left Inverse: :$d$ is an injection. {{qed|lemma}} === Necessary Condition === Assume that :$d$ is an injection. By definition of Galois connection: :$g$ and $f$ are increasing mappings. By Galois Connection Implies Order on Mappings: :$d \circ g \preceq I_S$ and $I_T \precsim g \circ d$ By Increasing and Ordering on Mappings implies Mapping is Composition: :$d = \paren {d \circ g} \circ d$ By Composition of Mappings is Associative: ::$= d \circ \paren {g \circ d}$ By Injection iff Left Cancellable: :$g \circ d = I_T$ By Identity Mapping is Right Identity: :$d = d \circ I_T$ Thus by Surjection iff Right Inverse: :$g$ is a surjection. {{qed}}
Note that by Uniform Limit of Analytic Functions is Analytic, $f$ is analytic. Let $a\in U$. Let $D$ be a disk of radius $r$ about $a$, contained in $U$. We have Cauchy's Integral Formula for Derivatives :$\displaystyle f'_n \left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f_n \left({z}\right)} {\left({z - a}\right)^2} \, \mathrm d z$ and: :$\displaystyle f'\left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)}{\left({z - a}\right)^2} \, \mathrm d z$ Therefore: {{begin-eqn}} {{eqn | l = \left\vert{f'_n \left({a}\right) - f' \left({a}\right) }\right\vert       | r = \frac 1 {2 \pi} \left\vert{ \int_{\partial D} \frac {f_n \left({z}\right) - f \left({z}\right)} {\left({z - a}\right)^2} \, \mathrm d z }\right\vert       | c =  }} {{eqn | o = \le       | r = \frac r {r^2} \sup_{z \mathop \in \partial D} \left\vert{ f_n \left({z}\right) - f \left({z}\right) }\right\vert        | c = Estimation Lemma for complex integrals }} {{eqn | r = \frac 1 r \sup_{z \mathop \in \partial D} \left\vert{ f_n \left({z}\right) - f \left({z}\right) }\right\vert        | c =  }} {{end-eqn}} Now the $f_n$ tend uniformly to $f$, and we can bound $r$ away from zero. It follows that $f_n' \to f'$ uniformly in each compact disk contained in $U$. {{qed}}
Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic. We may suppose none of the $f_n$ is identically zero on any open subset of $D$. Let $E = D \setminus \left\{ {z \in D: f \left({z}\right) = 0}\right\}$. By Logarithmic Derivative of Infinite Product of Analytic Functions, $\displaystyle \frac{f'} f = \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$ converges locally uniformly in $E$. By Linear Combination of Convergent Series, $\displaystyle f' = \sum_{n \mathop = 1}^\infty f_n' \cdot \prod_{\substack {k \mathop = 1 \\ k \mathop \ne n}}^\infty f_k$ on $E$. By Uniformly Convergent Sequence Multiplied with Function, the series converges locally uniformly in $E$. By Uniformly Convergent Sequence on Dense Subset, the series converges locally uniformly in $D$. Let $g$ denote its limit. By Uniform Limit of Analytic Functions is Analytic, $g$ is analytic in $D$. By Uniqueness of Analytic Continuation, $f' = g$. {{qed}}
Denote $\norm {\, \cdot \,}$ as the norm on $V$. Let $\epsilon > 0$. By boundedness of $g$: :$\exists M \in \R: \forall x \in X: \norm {\map g x} < M$ By uniformly convergence of $\sequence {f_n}$: :$\exists f:X \to \mathbb K: \exists N \in \R: \forall x \in X: \norm {\map {f_n} x - \map f x} < \dfrac \epsilon M$ Pick any $x \in X$. Then: {{begin-eqn}} {{eqn | l = \norm {\map {f_n} x \map g x - \map f x \map g x}       | r = \norm {\map g x \paren {\map {f_n} x - \map f x} } }} {{eqn | r = \norm {\map g x} \norm {\map {f_n} x - \map f x}        | c = Norm Axiom $(\text N 2)$ (Multiplicativity) }} {{eqn | o = <       | r = M \cdot \dfrac \epsilon M }} {{eqn | r = \epsilon }} {{end-eqn}} As the choice of $x$ is arbitrary, $\sequence {f_n g}$ uniformly converges to $\sequence {f g}$. {{qed}}
Let $\epsilon>0$. Let $N \in\N$ be such that $\norm {f_n - f_m} < \epsilon$ for $n, m > N$ on $Y$. Let $x \in X$. Then there exists a sequence $\sequence {y_n} \in Y$ with $y_n \to x$. By continuity: :$\norm {\map {f_n} x - \map {f_m} x} \le \epsilon$ Because $V$ is complete, $\sequence {f_n}$ is uniformly convergent on $X$. {{qed}} Category:Uniform Convergence 4admzzw3uwxw37uthv76co97u9dce7t
Follows directly from Uniformly Convergent Product Satisfies Uniform Cauchy Criterion. {{qed}}