Power Series and Real-Analytic Functions

Fix $x\in\mathbb R$ and put $d:=|x-c|$. The shifted roots of the terms $a_n(x-c)^n$, $n\ge1$, are $|a_{k+1}(x-c)^{k+1}|^{1/(k+1)}=q_kd$.

If $L=0$, then for $d=0$ every root in step 1.1 is $0$, while for $d>0$ and any $\eta>0$, [L2] applied with $\varepsilon=\eta/d$ makes $q_kd<\eta$ eventually. Thus $\limsup_k(q_kd)=0<1$ for every $x$.

Suppose $0<L<+\infty$. If $d<1/L$, choose a real $t$ with $L<t<1/d$ (with the second inequality omitted when $d=0$). By [L2], $q_k<t$ eventually, so $\limsup_k(q_kd)\le td<1$. If $d>1/L$, choose $t$ with $1/d<t<L$; [L2] gives $q_k>t$ frequently, so $\limsup_k(q_kd)\ge td>1$.

If $L=+\infty$ and $d>0$, then for every real $M>0$ and every index $N$ there is $k\ge N$ with $q_k>M$: otherwise $M$ would bound a tail and its supremum, forcing the infimum of the tail suprema to be finite. Taking $M>1/d$ shows $q_kd>1$ arbitrarily late, hence $\limsup_k(q_kd)>1$.

By [L3] and [L4], step 2.1 gives absolute convergence at every real $x$ when $L=0$; step 2.2 gives absolute convergence for $d<1/L$ and divergence for $d>1/L$ when $0<L<+\infty$; and step 2.3 gives divergence at every $x\ne c$ when $L=+\infty$, while the series converges at $c$ to $a_0$.

Reading these three alternatives through the definition of the radius yields $R=+\infty$, $R=1/L$, and $R=0$, respectively, which is the stated convention-complete formula.

The assertions for $|x-c|<R$ and $|x-c|>R$ are exactly the two strict alternatives supplied by [L1], including the cases $R=0$ and $R=+\infty$.

For endpoint behaviour at radius $1$, the series with coefficients $1/(n+1)^2$ converges absolutely at both $x=1$ and $x=-1$. The series with coefficients $1/(n+1)$ diverges at $x=1$, while the series with coefficients $(-1)^n/(n+1)$ diverges at $x=-1$. All three have radius $1$ by [L2].

Replacing $x$ by $(x-c)/R$ and multiplying coefficients by the corresponding powers of $R^{-1}$ transports the two radius-one examples to any finite $R>0$ and centre $c$. Thus either behaviour may occur at an endpoint, while no assertion has been made when the endpoints are not real.

Choose a real $\rho$ with $r<\rho<R$, or merely $\rho>r$ when $R=+\infty$. Then the scalar series $\sum |a_n|\rho^n$ converges by [L1], applied at $x=c+\rho$.

For every $x\in[u,v]$, order-convexity gives $|x-c|\le\max\{|u-c|,|v-c|\}=r<\rho$, and hence $|a_n(x-c)^n|\le |a_n|\rho^n$ for every $n$.

Apply [L2] to $f_n(x)=a_n(x-c)^n$ and $M_n=|a_n|\rho^n$. The series is absolutely convergent at each $x\in[u,v]$ and uniformly convergent on the whole interval.

Choose $\delta>0$ so small that $[x_0-\delta,x_0+\delta]$ lies strictly inside $|x-c|<R$.

The polynomial partial sums are continuous on this interval by [L2] and converge uniformly there to $f$ by [L1].

By [L3], $f$ is continuous on that interval, and in particular at $x_0$.

Fix distances $0\le r<s$ and put $q=r/s$ when $s>0$. By [L2], the series with nonnegative terms $\iota(n+1)q^n$ converges. Its terms tend to $0$ and hence form a bounded sequence by [L3], say with bound $M$.

Conversely, if the derivative series converges absolutely at a distance $s>0$, then $|a_{n+1}|s^{n+1}\le s\,\iota(n+1)|a_{n+1}|s^n$ because $\iota(n+1)\ge1$. Comparison gives absolute convergence of the original series there, after adjoining its first term.

If the original series converges absolutely at distance $s>0$, then the antiderivative terms satisfy $|a_n|s^{n+1}/\iota(n+1)\le s|a_n|s^n$, so the antiderivative converges absolutely at $s$.

Suppose the original series converges absolutely at distance $s>0$. Its shifted absolute terms $u_n:=|a_{n+1}|s^{n+1}$ form a convergent series. At distance $r<s$, the derivative's absolute terms satisfy $\iota(n+1)|a_{n+1}|r^n=s^{-1}\iota(n+1)q^n u_n\le (M/s)u_n$, so the derivative series converges absolutely there by [L3].

Conversely, if the antiderivative converges absolutely at distance $s>0$, put $v_n:=|a_n|s^{n+1}/\iota(n+1)$. At every $r<s$, $|a_n|r^n=s^{-1}\iota(n+1)q^n v_n\le(M/s)v_n$, so the original series converges absolutely at $r$ by [L3].

Write $R_0,R_D,R_I$ for the three radii. If $0\le r<R_0$, the supremum definition supplies an admissible distance $s>r$ for the original series; choosing $u$ with $r<u<s$, the original series is absolutely convergent at $u$, and step 2.1 makes the derivative absolutely convergent at every distance below $r$. Thus $r$ is admissible for the derivative and $R_0\le R_D$. Conversely, if $0\le r<R_D$, choose an admissible derivative distance $s>r$ and then $u$ with $r<u<s$. The derivative converges absolutely at $u$, so step 1.2 and direct comparison make the original series absolutely convergent at every distance below $r$; hence $R_D\le R_0$. The same argument with steps 1.3 and 2.2 gives $R_0=R_I$. Therefore all three extended radii are equal, including $0$ and $+\infty$.

Fix $x_0$ with $|x_0-c|<R$ and choose a closed interval $J$ containing both $c$ and $x_0$ strictly inside the radius.

Each $p_N$ is continuously differentiable on $J$, and [L4] gives $p_N'(x)=\sum_{n<N-1}\iota(n+1)a_{n+1}(x-c)^n$. The derivative partial sums converge uniformly on $J$ by [L1] and [L2].

The sequence $p_N(c)$ converges to $a_0$, since it equals $a_0$ for every $N\ge1$. Thus [L3] applies and says that the uniform limit of $(p_N)$ on $J$ is differentiable with derivative equal to the uniform limit of $(p_N')$.

The uniform limit of $(p_N)$ is $f$, and the limit of $(p_N')$ is the displayed differentiated series. Hence the formula holds at $x_0$; since $x_0$ was arbitrary it holds throughout $|x-c|<R$, and [L1] supplies the equality of radii.

By [L1] and [L2], $G$ and $f$ converge uniformly on $[u,v]$; in particular $f$ is integrable there, and [L3] gives $G'=f$.

Define $H(x):=\int_u^x f(t)\,dt$. The sum $f$ is continuous as a differentiable function by [L3] and [L5], so [L4] gives $H'=f$ on $(u,v)$, while the integral construction makes $H$ continuous on $[u,v]$.

The functions $H$ and $G$ are continuous and have the same derivative on the interval. By [L5], $H-G$ is constant; evaluating at $u$ gives $H(v)=G(v)-G(u)$.

Subtracting the two convergent series for $G(v)$ and $G(u)$ term by term is licensed by their convergence, and gives the displayed series. This is also the limit of the integrals of the polynomial partial sums by [L2].

For $m=0$, the formula reads $f(x)=\sum_{j\ge0}\iota(j^{\underline0})a_j(x-c)^j=\sum_{j\ge0}a_j(x-c)^j$, so the claim holds.

Fix $m$ and assume the displayed formula for $f^{(m)}$, with radius $R$.

By [L1], differentiate the series in step 1.2 term by term. Reindexing $j+1$ as $j$ gives $f^{(m+1)}(x)=\sum_{j\ge0}\iota(j+1)\iota((m+j+1)^{\underline m})a_{m+j+1}(x-c)^j$.

The falling-factorial recursion with $n=m+j+1$ and $k=m$ gives $(m+j+1)^{\underline{m+1}}=(m+j+1)^{\underline m}(j+1)$. Since $\iota$ preserves products, step 2.1 is precisely the asserted formula with $m+1$ in place of $m$.

By [L3], the derivative formula holds for every $m$. At each induction step [L1] also preserves the radius $R$, so every derived series has radius $R$.

At $x=c$, every term with $j\ge1$ vanishes and the $j=0$ term is $\iota(m^{\underline m})a_m=\iota(m!)a_m$; since $m!\ne0$, division gives the coefficient formula.

Put $\rho:=|d-c|+|h|<R$. By [L2], the sum of the absolute values in row $n$ is $|a_n|\rho^n$.

The series $\sum_n|a_n|\rho^n$ converges by [L1], so the triangular double series is absolutely convergent.

Apply the binomial theorem before summing and [L3] to regroup the absolutely convergent double series by $k$. This yields the displayed identity.

Fix $x$ satisfying the stated inequality and set $h:=x-d$. By [L1], $f(x)=\sum_{k\ge0}b_kh^k$, where $b_k:=\sum_{n\ge k}\iota\!\binom nk a_n(d-c)^{n-k}$.

By [L2] and [L3], $\iota(k!)b_k=f^{(k)}(d)$ for every $k$.

Substituting the coefficient identity from step 2.1 into the series in step 1.1 proves the formula.

Fix $d$ in the open radius interval. Then $R-|d-c|>0$, and [L1] represents $f$ by a power series about $d$ whenever $|x-d|<R-|d-c|$.

The neighbourhood in step 1.1 lies inside the open radius interval, so [L2] applies at every $d$ and proves that $f$ is real analytic there.

Fix $c\in U$ and choose the local representing power series supplied by [L1].

By [L2], that series, hence $f$, has derivatives of every order on a neighbourhood of $c$. Since $c$ was arbitrary, $f$ is smooth on $U$.