Proof of Power Series Are Holomorphic and Are Differentiated Term by Term

Routine arithmetic and bookkeeping facts are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity), and a convergent power series defines a function on its set of convergence (Complex Power Series: the Sequence of Terms and the Sum as a Function on the Set of Convergence (iii)).

Both clauses are proved together. Let $\Omega$ be $D$ in (i) and $\mathbb{C}$ in (ii). For each $z\in\Omega$ we fix a real number $\rho$ with $|z|<\rho$, and with $\rho<R$ in (i); for instance $\rho=(|z|+R)/2$, respectively $\rho=|z|+1$. In particular $0<\rho$, so $\rho$ is admissible in the hypothesis of the clause being proved. In the case of (ii), every citation of The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (ii) below is to be read as one of The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (iii).

Step 1: $\Omega$ is open. The disk $D$ is the open ball $B_{d_{\mathbb{C}}}(0,R)$ of the metric space $(\mathbb{C},d_{\mathbb{C}})$ (Open Ball in a Metric Space (i), Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (v)), and open balls are open (Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (vi)). The whole plane $\mathbb{C}$ is open by Open Subset of a Metric Space (i), since about every $z\in\mathbb{C}$ there is an open ball, for instance $B_{d_{\mathbb{C}}}(z,1)$, and it is a subset of $\mathbb{C}$.

Step 2: absolute convergence. Let $w\in\mathbb{C}$ with $|w|\le\rho$. For every $k\in\mathbb{N}$,

$$|c_{k}w^{k-1}|=|c_{k}|\,|w|^{k-1}\le|c_{k}|\,\rho^{k-1},$$

by multiplicativity of the modulus (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv)), the identity $|w^{k-1}|=|w|^{k-1}$ (Continuity of Complex-Valued Functions, and the Modulus of a Power (i)) and monotonicity of powers (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (viii)), both sides of the last inequality being $|c_{1}|$ when $k=1$. The series $\sum_{k}|c_{k}|\rho^{k-1}$ converges by hypothesis, so $\sum_{k}c_{k}w^{k-1}$ converges absolutely by the comparison test (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (ii)). In the same way, using $|k|=k$ for the natural number $k$ read in $\mathbb{C}$ (The Modulus of a Complex Number (ii)),

$$|k\,c_{k+1}w^{k-1}|=k\,|c_{k+1}|\,|w|^{k-1}\le k\,|c_{k+1}|\,\rho^{k-1},$$

and the series of the right-hand sides converges by The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (ii); so $\sum_{k}k\,c_{k+1}w^{k-1}$ converges absolutely as well. Taking $w=z$ proves the convergence assertions of both clauses, and so $f$ is defined on $\Omega$. We write

$$g(z)=\sum_{k=1}^{\infty}k\,c_{k+1}z^{k-1}\qquad(z\in\Omega)$$

for the sum of the derived series.

Step 3: the second derived series of moduli. Apply The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (ii) to the sequence $(k\,c_{k+1})_{k}$ in place of $(c_{k})$. Its hypothesis, the convergence of $\sum_{k}|k\,c_{k+1}|\,\sigma^{k-1}$ for every admissible radius $\sigma$, is exactly the conclusion of The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (ii) (respectively The Derived Series of Moduli of a Power Series Converges at Every Smaller Radius, Below a Radius of Convergence, and at All Radii (iii)) for $(c_{k})$ at every such radius, since $|k\,c_{k+1}|=k\,|c_{k+1}|$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), The Modulus of a Complex Number (ii)). Its conclusion, at our radius $\rho$, is that

$$M=\sum_{k=1}^{\infty}k(k+1)\,|c_{k+2}|\,\rho^{k-1}$$

converges. Note that $M\ge0$, and that $M$ depends on $z$ only through $\rho$.

Step 4: the key estimate for a single power. Let $w\in\mathbb{C}$ with $w\neq z$ and $|w|<\rho$. For $n=1$ it vanishes:

$$\frac{w^{n}-z^{n}}{w-z}-n\,z^{n-1}=\frac{w-z}{w-z}-1=0$$

(Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii), Powers with Exponents in the Natural Numbers with Zero (ii)). Let now $n\in\mathbb{N}$ with $n\ge2$. The factorisation $w^{n}-z^{n}=(w-z)\sum_{j=1}^{n}w^{j-1}z^{n-j}$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (iii)), the identity $n\,z^{n-1}=\sum_{j=1}^{n}z^{j-1}z^{n-j}$ (Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iv), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i)), and the rule for sums of differences (Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (ii)) give the first equality below; the factorisation of $w^{j-1}-z^{j-1}$ for $j\ge2$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (iii) again), the term $j=1$ being $0$, gives the second:

$$\frac{w^{n}-z^{n}}{w-z}-n\,z^{n-1}=\sum_{j=1}^{n}\bigl(w^{j-1}-z^{j-1}\bigr)z^{n-j}=(w-z)\sum_{j=2}^{n}\sum_{l=1}^{j-1}w^{l-1}z^{j-1-l}\,z^{n-j}.$$

The double sum has at most $n(n-1)$ terms (at most $n-1$ indices $j$, each with at most $n-1$ indices $l$), and each has modulus at most $\rho^{(l-1)+(j-1-l)+(n-j)}=\rho^{n-2}$, because $|w|,|z|<\rho$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), Continuity of Complex-Valued Functions, and the Modulus of a Power (i), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (viii), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i)). By the triangle inequality (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (viii)) and the comparison of finite sums with a constant (Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iv), Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (v)), therefore,

$$\Bigl|\frac{w^{n}-z^{n}}{w-z}-n\,z^{n-1}\Bigr|\le|w-z|\,n(n-1)\,\rho^{n-2}\qquad(n\ge2).$$

Step 5: the estimate for the sum. Let $w$ be as in Step 4. Then $w\in\Omega$, since $|w|<\rho<R$ in the case of (i), so $f(w)$ is defined. All series involved converge by Step 2, so they may be combined termwise and shifted (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (iii), Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (iv)): the terms of $f(w)$ and $f(z)$ with $k=1$ are both $c_{1}$ and cancel, and shifting the index by one,

$$\frac{f(w)-f(z)}{w-z}-g(z)=\sum_{k=1}^{\infty}c_{k+1}\Bigl(\frac{w^{k}-z^{k}}{w-z}-k\,z^{k-1}\Bigr).$$

The term with $k=1$ is $0$ by Step 4, so, shifting once more (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (iv)), the right-hand side is the sum over the exponents $n=k+1\ge2$,

$$\sum_{k=1}^{\infty}c_{k+2}\Bigl(\frac{w^{k+1}-z^{k+1}}{w-z}-(k+1)\,z^{k}\Bigr).$$

By Step 4 with $n=k+1$ and multiplicativity of the modulus (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv)), the $k$-th term of this series has modulus at most $|w-z|\,k(k+1)\,|c_{k+2}|\,\rho^{k-1}$, and these bounds have the convergent sum $|w-z|\,M$ by Step 3. By the comparison test (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (ii)) we arrive at the key estimate

$$\Bigl|\frac{f(w)-f(z)}{w-z}-g(z)\Bigr|\le M\,|w-z|\qquad(w\neq z,\ |w|<\rho).$$

Step 6: conclusion. The order of choice is: first $z\in\Omega$, then $\rho$ and with it $M$, which depend only on $z$; then an arbitrary $\varepsilon>0$; and then

$$\delta=\min\Bigl(\rho-|z|,\ \frac{\varepsilon}{M+1}\Bigr)>0.$$

Let $w\in\Omega$ with $0<|w-z|<\delta$. Then $|w|\le|z|+|w-z|<|z|+(\rho-|z|)=\rho$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vii)), so Step 5 applies and gives

$$\Bigl|\frac{f(w)-f(z)}{w-z}-g(z)\Bigr|\le M\,|w-z|\le\frac{M\,\varepsilon}{M+1}<\varepsilon .$$

Hence $g(z)$ is a complex derivative of $f$ at $z$ (The Complex Derivative at a Point (i)), and $f'(z)=g(z)$ (The Complex Derivative at a Point (ii)). As $z\in\Omega$ was arbitrary and $\Omega$ is open by Step 1, $f$ is holomorphic on $\Omega$ with derivative $f'=g$ (Holomorphic Functions and Their Derivatives (i), Holomorphic Functions and Their Derivatives (ii)).