Complex Power Series and Analytic Functions

If $f_n\to f$ uniformly, then for $\epsilon >0$ choose $N$ with $\mathrm{\mid }{f}_n(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ for every $n\ge N$ and $x\in X$; for $m,n\ge N$, [L2] gives $\mathrm{\mid }{f}_m(x)-f_n(x)\mathrm{\mid }<\epsilon$, so $(f_n)$ is uniformly Cauchy.

Conversely, suppose $(f_n)$ is uniformly Cauchy. For each $x\in X$, the sequence $(f_n(x))$ is Cauchy and hence has a limit $f(x)\in \mathbb{C}$ by [L1]; this defines $f:X\to \mathbb{C}$, including the unique empty function when $X=\mathrm{\varnothing }$.

Given $\epsilon >0$, choose $N$ such that $\mathrm{\mid }{f}_m(x)-f_n(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $m,n\ge N$ and $x\in X$. Fixing $n\ge N$ and passing $m\to \mathrm{\infty }$ in the continuous modulus gives $\mathrm{\mid }f(x)-f_n(x)\mathrm{\mid }\le \epsilon \mathrm{/}2<\epsilon$ for every $x$, so $f_n\to f$ uniformly.

Write $f_n=u_n+i{v}_n$ and $f=u+iv$. The componentwise dictionary in Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary shows that $u_n\to u$ and $v_n\to v$ uniformly; [L2] shows each $u_n,v_n$ is continuous.

By [L1], both $u$ and $v$ are continuous, including when $X$ is empty.

The componentwise continuity criterion [L2] now makes $f=u+iv$ continuous.

For partial sums $S_N(x)={\sum }_{n<N}{f}_n(x)$ and $q>p$, [L1] gives $\mathrm{\mid }{S}_q(x)-S_p(x)\mathrm{\mid }\le {\sum }_{p\le n<q}{M}_n$ for every $x$.

Since the real series $\sum M_n$ is Cauchy, its tails make the bound in step 1.1 uniformly small; thus $(S_N)$ is uniformly Cauchy and converges uniformly by [L2].

For each $x$, the nonnegative series $\sum \mathrm{\mid }{f}_n(x)\mathrm{\mid }$ is bounded termwise by $\sum M_n$, exactly the comparison used in [L3], and therefore converges. Zero majorants and the empty set require no separate choice.

By [L1], the real series $\sum \mathrm{\mid }{c}_n\mathrm{\mid }{r}^n$ converges, since it is the modulus series at any point whose distance from $a$ is $r$; for $r=0$ it has only the constant contribution.

If $\mathrm{\mid }z-a\mathrm{\mid }\le r$, then [L3] gives $\mathrm{\mid }{c}_n(z-a{)}^n\mathrm{\mid }\le \mathrm{\mid }{c}_n\mathrm{\mid }{r}^n$.

Apply [L2] to the majorants of step 1.1 and the bound of step 1.2. This also covers $R=+\mathrm{\infty }$ and makes no assertion when $r=R$.

The modulus coefficient of the formal derivative is $(n+1)\mathrm{\mid }{c}_{n+1}\mathrm{\mid }$, and that of the formal antiderivative is $\mathrm{\mid }{c}_n\mathrm{\mid }\mathrm{/}(n+1)$.

By [L1], the three complex radii are precisely the radii of the three real power series with the modulus coefficients described in step 1.1.

Apply [L2] to those real series. The conclusion includes radii $0$ and $+\mathrm{\infty }$ and uses ordinary embedded-number notation only.

Choose $r$ with $\mathrm{\mid }z-a\mathrm{\mid }<r<R$. For $w$ near $z$, the finite identity $w^n-z^n=(w-z){\sum }_{k<n}{w}^{n-1-k}{z}^k$ follows by expanding and telescoping, so the difference quotient of each monomial tends to $n{z}^{n-1}$.

On $\mathrm{\mid }w-a\mathrm{\mid },\mathrm{\mid }z-a\mathrm{\mid }\le r$, the quotient in step 1.1 is bounded in modulus by $n{r}^{n-1}$ after translating the centre to $a$. The series $\sum n\mathrm{\mid }{c}_n\mathrm{\mid }{r}^{n-1}$ converges by [L1], so its tails are uniformly small.

Split the difference quotient of $f$ into a finite head and a tail. The finite head tends termwise to its derivative by step 1.1, while step 2.1 bounds the tail uniformly; hence the quotient tends to ${\sum }_{n\ge 1}n{c}_n(z-a{)}^{n-1}$.

Since $z$ was arbitrary in the open disc, the derivative exists at every such point, which is holomorphy by Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions. If $R=0$ the disc is empty and the assertion is vacuous; the constant term differentiates to $0$.

For $k=0$, the displayed formula is the original series because $n!\mathrm{/}(n-0)!=1$.

Assume the formula holds for $k$, with its series having radius $R$. [L2] is stated for one series and its formal derivative, so it is applied once at this induction step, to the $k$th series: its formal derivative again has radius $R$. That is exactly what the induction needs, and no claim about "every successive derivative" is taken from [L2] at once. Then [L1] differentiates termwise and changes the coefficient $n!\mathrm{/}(n-k)!$ into $n!\mathrm{/}(n-k-1)!$ for $n\ge k+1$.

Thus the formula holds for $k+1$, and induction gives it for every $k$. The cases $k>n$ contribute no term and $k=0$ was the base case.

Evaluate [L1] at $z=a$, so every remaining power is $0^m$ with $m=k-n\ge 0$. From the recursion of [L2], $0^{m+1}=0^m\cdot 0=0$ for every $m\in \mathbb{N}$, so $0^m=0$ for every positive $m$, while $0^0=1$ by the base clause of [L2]. Hence every term with a positive remaining power vanishes and the term indexed by $n$ is $n!c_n$.

Thus $f^{(n)}(a)=n!c_n$; divide by the nonzero factorial from [L3]. For $n=0$, this reads $c_0=f(a)$.

Both series represent the same function on a neighbourhood, so their derivatives of every order at $a$ are the same.

Applying [L1] to both representations gives $c_n=f^{(n)}(a)\mathrm{/}n!=d_n$ for every $n$, including $n=0$.

Let $a\in U$. By [L1], $f$ agrees near $a$ with a convergent power series.

By [L2], that power-series sum is complex differentiable at $a$, hence so is $f$.

Since $a$ was arbitrary, $f$ is holomorphic on $U$; if $U$ is empty, this conclusion is vacuous.

Apply [L1] to the real coefficient sequence $\mathrm{\mid }{c}_n\mathrm{\mid }$ and the nonnegative numbers $\mathrm{\mid }b-a\mathrm{\mid },\mathrm{\mid }h\mathrm{\mid }$; this gives the displayed finite total majorant.

By [L2], $c_n((b-a)+h{)}^n$ is the finite sum over $k\le n$ of the corresponding complex terms, each bounded by the majorant term in step 1.1.

Absolute convergence now permits regrouping by $k$ under [L3]. The cases $h=0$ and $b=a$ merely make some terms vanish and are included.

Put $z=b+h$. The binomial expansion and [L1] give $f(b+h)={\sum }_{k\ge 0}{d}_k{h}^k$, where $d_k={\sum }_{n\ge k}\left(\genfrac{}{}{0px}{}{n}{k}\right)c_n(b-a{)}^{n-k}$.

By [L3], evaluating the $k$th derivative at $b$ gives $f^{(k)}(b)=k!d_k$.

By [L2], $d_k=f^{(k)}(b)\mathrm{/}k!$, proving the stated expansion for $\mathrm{\mid }h\mathrm{\mid }<R-\mathrm{\mid }b-a\mathrm{\mid }$. The bound remains valid when $R=+\mathrm{\infty }$.

Let $b\in D$. Its distance to the original boundary is positive, and [L1] supplies a power-series representation of $f$ on a disc about $b$.

This is exactly analyticity at $b$ by [L2]. Since $b$ was arbitrary, $f$ is analytic on $D$, including the entire-radius case.

For every finite $N$, exact distributivity gives $\alpha {\sum }_{n<N}{a}_n(z-c{)}^n+\beta {\sum }_{n<N}{b}_n(z-c{)}^n={\sum }_{n<N}(\alpha a_n+\beta b_n)(z-c{)}^n$.

On a closed subdisc inside both radii, [L1] makes both sequences of partial sums uniformly convergent. If their limits are $A,B$, then [L2] bounds the error after taking the linear combination by $\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }{A}_N-A\mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }{B}_N-B\mathrm{\mid }$, which tends uniformly to $0$.

Passing to the limit in step 1.1 proves the formula and its local uniform convergence. Zero scalars and unequal radii are included by taking the common disc.

At a fixed point in the common disc, [L2] gives absolute convergence of both numerical series.

Apply [L1]; multiplying $(z-c{)}^k(z-c{)}^{n-k}$ gives $(z-c{)}^n$, so the Cauchy coefficient is the displayed finite convolution. For $n=0$ this is the one-term sum with $k=0$, not an empty sum.

Fix a radius $r$ inside both original radii. The absolute Cauchy convolution has total sum $(\sum \mathrm{\mid }{a}_n\mathrm{\mid }{r}^n)(\sum \mathrm{\mid }{b}_n\mathrm{\mid }{r}^n)<\mathrm{\infty }$ by [L1] and [L2], so [L3] gives uniform convergence of the product power series on $\mathrm{\mid }z-c\mathrm{\mid }\le r$. This includes $r=0$ and either input series being identically zero.

Choose $\rho >0$ inside the radius of $H$. By [L2], $A={\sum }_{n\ge 1}\mathrm{\mid }{h}_n\mathrm{\mid }{\rho }^n$ is finite. Since $R>0$, choose $0<r\le \rho$ and $0\le q<R$ with $(r\mathrm{/}\rho )A\le q$; then ${\sum }_{n\ge 1}\mathrm{\mid }{h}_n\mathrm{\mid }{r}^n\le (r\mathrm{/}\rho )A\le q$. If $A=0$, then $H\equiv 0$ and this estimate holds with $q=0$.

Repeated use of [L1] expands each $H(z{)}^m$ as a power series about $a$. Its absolute coefficient sum at radius $r$ is bounded by $q^m$ from step 1.1, and [L2] applied to $G$ gives convergence of the scalar majorant $\sum \mathrm{\mid }{g}_m\mathrm{\mid }{q}^m$.

The resulting double series is absolutely convergent, so [L3] regroups it by powers of $z-a$ and produces the desired local series. If $H\equiv 0$, it reduces to the constant $G(0)$.

Put $H(z)=1-f(z)\mathrm{/}{c}_0$, so $H(a)=0$. By [L2] and [L3], $f$ is continuous at $a$; since $\mathrm{\mid }H(z)-H(a)\mathrm{\mid }=\mathrm{\mid }f(z)-f(a)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{c}_0\mathrm{\mid }$ and $c_0\mathrm{\ne }0$, $H$ is continuous there. Hence on a sufficiently small disc one has $\mathrm{\mid }H(z)\mathrm{\mid }<1$.

The finite identity $(1-H){\sum }_{m=0}^N{H}^m=1-H^{N+1}$ and $\mathrm{\mid }H\mathrm{\mid }<1$ give $(1-H{)}^{-1}={\sum }_{m\ge 0}{H}^m$. By [L1], this composition has a local power series.

Hence $1\mathrm{/}f=c_0^{-1}(1-H{)}^{-1}$ has a local power series. Multiply it by $f$ using [L4] and compare with the constant series $1$ by [L5]; the constant coefficient gives $d_0=c_0^{-1}$, while for $n\ge 1$ the nonempty sum over $1\le k\le n$ gives the displayed recursion. The nonzero constant term licenses division. If $f$ is constant, the same recursion gives $d_n=0$ for every $n\ge 1$.

Fix a point in the relevant natural domain and choose local series for all participating functions by [L1], shrinking to a common disc when necessary.

Apply [L2] to finite linear combinations and [L3] to products.

If $g$ is nonzero at the point, its local series has nonzero constant term, so [L4] gives a reciprocal series and [L3] gives $f\mathrm{/}g$; for composition, recenter the outer series at the inner value and apply [L4].

Each construction supplies a convergent power-series representation near every point of its stated domain, so each result is analytic by [L1].

Choose a local representation from [L1] and define $F(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^{n+1}\mathrm{/}(n+1)$ on the same disc.

By [L3], $F^{\mathrm{\prime }}(z)=\sum c_n(z-a{)}^n=f(z)$ throughout the disc, so $F$ is a local primitive.

Substitute the series [L1] at $z,-z,iz,-iz$ into [L2]. Absolute convergence, which [L5] supplies for every complex argument, allows [L3] to separate the even and odd indices.

The identities $i^{2n}=(-1{)}^n$ and $i^{2n+1}=i(-1{)}^n$ simplify those even and odd parts to the four displayed series.

Their factorial coefficients have root limsup $0$: for $n\ge 2$ the factorial satisfies $n!\ge (n\mathrm{/}2{)}^{\lfloor n\mathrm{/}2\rfloor }$, since at least $\lfloor n\mathrm{/}2\rfloor$ of the factors $1,\dots ,n$ are at least $n\mathrm{/}2$, so $(1\mathrm{/}n!{)}^{1\mathrm{/}n}\le (2\mathrm{/}n{)}^{\lfloor n\mathrm{/}2\rfloor \mathrm{/}n}\to 0$. Hence [L4] gives infinite radius. The constant terms are retained in the even series and absent from the odd series.

Apply [L1] to each of the four infinite-radius series and cancel the positive integer factor against the factorial.

Shifting the resulting indices gives respectively the series for $\cos ,-\sin ,\cosh ,\sinh$. Infinite radius makes each function entire.

Substitute $z+w$ into the four formulas in [L1] and factor every exponential using [L2].

Expanding the right-hand sides in the Statement with [L1] gives the same symmetric or antisymmetric combinations, so the four identities follow.

Substituting the real $y$ into [L1] gives $\sin (iy)=i\sinh y$ and $\cos (iy)=\cosh y$.

By [L2] and [L3], both positive real quantities $\sinh y$ and $\cosh y$ tend to $+\mathrm{\infty }$ as $y\to +\mathrm{\infty }$.

Hence $\mathrm{\mid }\sin (iy)\mathrm{\mid }=\sinh y$ and $\mathrm{\mid }\cos (iy)\mathrm{\mid }=\cosh y$ are unbounded along the imaginary axis, so both complex functions are unbounded.

By [L1] and multiplication by the nonzero $\exp (iz)$, $\sin z=0$ is equivalent to $\exp (2iz)=1$. By [L2], this is equivalent to $2iz=2\pi ik$ for some integer $k$, hence to $z=k\pi$.

Similarly, $\cos z=0$ is equivalent to $\exp (2iz)=-1=\exp (i\pi )$ by [L3]. By [L2], $2iz-i\pi =2\pi ik$, so $z=(k+1\mathrm{/}2)\pi$.

Reversing each algebraic equivalence proves both converses, so the displayed descriptions are exact.