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Show that convergence of a sequence of power-series is equivalent to convergence of their respective coefficients.

More precisely, let $\mathbf{a}:=\left[a_{i,j}\right]_{i,j\in\mathbb{N_0}}$ be a real matrix, such as a stochastic matrix, such that

i) $\forall i\in\mathbb{N}_0,\space\mathbf{L}\leq\mathbf{a}_i$ for some real row vector $\mathbf{L}$ that sums to a real number ($\mathbf{a}_i$ being $\mathbf{a}$'s $i$th row),

ii) $\sup_{i\in\mathbb{N}_0} \sum \mathbf{a}_i<\infty$ (Given a real vector $\mathbf{v}=\left[v_m\right]_{m\in\mathbb{N}_0}$, we define $\sum\mathbf{v}:=\sum_{m=0}^\infty v_m$ whenever the series on the right converges, possibly to $\pm\infty$.)

Additionally, let $\mathbf{b}:=\left[b_j\right]_{j\in\mathbb{N}_0}$ be a real row vector that sums to a finite number and define $\mathbf{x}:=\left[x_j\right]_{j\in\mathbb{N}_0}$ to be the column vector of monomial functions: $$x_j:\left(-1,1\right)\rightarrow\left(-1,1\right),\space\space x_j\left(y\right):=y^j$$

Lastly, set $\mathbf{f}:=\mathbf{a}\mathbf{x}$, $F:=\mathbf{b}\mathbf{x}$ and $$\mathbf{S}:=\left\{S\subseteq\left(-1,1\right):\space S\mathrm{\, has\, an\, accumulation\, point\, }\in \left(-1,1\right)\right\}$$ So $\mathbf{f}$ is a column vector whose components are power series and $F$ is a power series, and the radii of convergence of all these power series are $\geq1$.

Show that $$\begin{align}\mathbf{a}_i\underset{i\rightarrow\infty}{\rightarrow}\mathbf{b} &\iff f_i\underset{i\rightarrow\infty}{\rightarrow}F\mathrm{\, on\, }\left(-1,1\right)\\ &\iff\exists S\in\mathbf{S},\space f_i\underset{i\rightarrow\infty}{\rightarrow}F\mathrm{\, on\, }S\end{align}$$


Notes

  1. We take $0^0$ to be $1$, so $x_0\equiv 1$ on $\left(-1, 1\right)$.

  2. Conditions $\left(i\right)$ and $\left(ii\right)$ above will be required for the application of Fatou's lemma in the proof. In $\left(ii\right)$, $\sup$ is used rather than the laxer $\limsup$ required by Fatou's lemma, to ensure that the $f_i$'s radii of convergence are $\geq1$.

  3. The claim is adapted from Klenke, where an equivalent claim is stated without proof (Lemma 3.6, $\left(i\right)\iff\left(iii\right)\iff\left(iv\right)$).

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Convergence of a power series in what sense? –  Qiaochu Yuan Sep 5 '12 at 16:38
    
@QiaochuYuan: Thanks for your interest. Pointwise convergence of a sequence of power series (in plural). –  Evan Aad Sep 5 '12 at 16:44
    
Pointwise convergence where? –  Qiaochu Yuan Sep 5 '12 at 18:49
    
@QiaochuYuan: Consider the following sequence of functions $$\mathbf{f}:=\left(f_i\right)_{i\in\mathbb{N}_0}$$ where $$f_i:\left[0,1\right]\rightarrow\left[0,1\right],\space\space f_i\left(y\right):=\sum_{j=0}^\infty a_{i,j}y^j$$ Each $f_i$ is a power series that is convergent on $\left[0,1\right]$. –  Evan Aad Sep 5 '12 at 19:07
    
@QiaochuYuan: Assume that the sequence of functions $\mathbf{f}$ converges pointwise on $\left[0,1\right]$ to the function $$F:\left[0,1\right],\space\space F\left(y\right):=\sum_{k=0}^\infty b_k y^k$$ (which is also a power series convergent on $\left[0,1\right]$). Is it true that the coefficients of $f_0, f_1, \dots$ converge term-by-term to the corresponding coefficients of $F$: $$a_{i,0}\underset{i\rightarrow\infty}{\rightarrow}b_0, a_{i,1}\underset{i\rightarrow\infty}{\rightarrow}b_1, \dots$$ –  Evan Aad Sep 5 '12 at 19:08

1 Answer 1

up vote 0 down vote accepted

The following proof was suggested to me by Prof. Klenke himself in an email. His original message follows the proof.

$\boxed{\mathbf{a}_i\underset{i\rightarrow\infty}{\rightarrow}\mathbf{b} \implies f_i\underset{i\rightarrow\infty}{\rightarrow}F\mathrm{\, on\, }\left(-1,1\right)}$ Suppose $\mathbf{a}_i\underset{i\rightarrow\infty}{\rightarrow}\mathbf{b}$. Let $r\in\left(0,1\right)$. We will show that $$f_i\rightarrow F\mathrm{\, on\,}(-r,r) \space\space\left(*\right)$$ Let $z\in\left(-r,r\right)$ and fix $\epsilon>0$. We proceed to define a number $N_{r,\epsilon}\in\mathbb{N}_{0}$, such that for all $N_{r,\epsilon}\leq i\in\mathbb{N}_{0}$, $|f_i(z)-F(z)|<\epsilon$, thus proving $\left(*\right)$. Choose $n_{r,\epsilon}\in\mathbb{N}_{0}$ such that $\sum_{j>n_{r,\epsilon}}r^j<\epsilon/2$. Choose $N_{r,\epsilon}\in\mathbb{N}_{0}$ such that for all $N_{r,\epsilon}\leq i$, $\left|a_{i,j}-b_i\right|<\frac{\epsilon/2}{n_{r,\epsilon}+1}$, $j\in\left\{ 0,1,2,\dots,n_{r,\epsilon}\right\} $. Then for all $N_{r,\epsilon}\leq i$, $|f_i(z)-F(z)|<\epsilon$.

$\boxed{\exists S\in\mathbf{S},\space f_i\underset{i\rightarrow\infty}{\rightarrow}F\mathrm{\, on\, }S\implies\mathbf{a}_i\underset{i\rightarrow\infty}{\rightarrow}\mathbf{b}}$ Let $S\in\mathbf{S}$ and suppose $f_i\underset{i\rightarrow\infty}{\rightarrow}F$ on $S$. It is left for the reader to verify that it suffices to demonstrate that every strictly ascending sequence of indices $\left(i_k\right)_{k\in\mathbb{N}_{0}},\, i_k\in\mathbb{N}_{0}$ has a strictly ascending subsequence $\left(i_{k_n}\right)_{n\in\mathbb{N}_{0}}$ such that $$\mathbf{a}_{i_{k_n}}\underset{n\rightarrow\infty}{\rightarrow}\mathbf{b}\space\space\left(**\right)$$

Therefore let $\left(i_k\right)_{k\in\mathbb{N}_{0}}$ be strictly ascending. By the Bolzano-Weierstrass theorem there is a srtictly ascending subsequence $\left(i_{k_n}\right)_{n\in\mathbb{N}_{0}}$ such that $a_{i_{k_n}, 0}$ converges to some limit $c_0\in\mathbb{R}$. Appealing to the same theorem again we can further assume w.l.g. that also $\left(a_{i_{k_n}, 1}\right)_{n\in\mathbb{N}_{0}}$ converges to some $c_1\in\mathbb{R}$. Iterating the argument recursively, we may assume w.l.g. that $\left(a_{i_{k_n}, j}\right)_{n\in\mathbb{N}_{0}}$ converges to some $c_j\in\mathbb{R}$ for every $j\in\mathbb{N}_{0}$. Defining the row vector $\mathbf{c}:=\left[c_j\right]_{j\in\mathbb{N}_0}$, we have $\mathbf{a}_{i_{k_n}}\underset{n\rightarrow\infty}{\rightarrow}\mathbf{c}$. Since $\forall i\in\mathbb{N}_0,\space\mathbf{L}\leq\mathbf{a}_i$, $-\infty<\sum\mathbf{L}\leq\sum\mathbf{c}$ and by Fatou's lemma $\sum\mathbf{c}\leq\limsup_{i\in\mathbb{N}_0} \sum\mathbf{a}_i<\infty$. So $\sum\mathbf{c}\in\mathbb{R}$.

By the first part we see that $f_{i_{k_n}}\underset{n\rightarrow\infty}{\rightarrow}F_\mathbf{c}$ on $\left(-1,1\right)$, where $F_\mathbf{c}$ is the power series with coefficient vector $\mathbf{c}$. Now we use the underlying assumption that $f_i\underset{i\rightarrow\infty}{\rightarrow}F$ on $S$ to conclude that $F_\mathbf{c}=F$ on $S$. By the uniqueness theorem of power series (cf. Baby Rudin, Theorem 8.5) we get $\mathbf{c}=\mathbf{b}$, and so $\left(**\right)$ holds. $\square$


Note Here's Prof. Klenke's original message to me.

I do think that the equivalence of $(i)$ and $(iii)$ in Lemma 3.6 is standard, but I do not have a reference at hand at the moment. Let’s try the following argument.

$(i)\implies(iii)$. Fix $r\in(0,1)$ and $\epsilon>0$. Choose $N$ such that $\sum_{k>N} r^k < \epsilon/2$. Choose $n$ such that $\mu_n(k)<\epsilon/(N+1)$ for $k\leq N$. Then $|\psi_n(z)-\psi(z)|<\epsilon$ for all $0<z<r$. Hence $\psi_n$ converges to $\psi$ pointwise on $(0,r)$ for all $0<r<1$ and hence pointwise on $(0,1)$. Finally $\psi_n(1)=1=\psi(1)$ hence this is trivial.

$(iii)\implies(i)$ By the Bolzano-Weierstrass theorem, there is a subsequence such that $\mu_{n_k}(0)$ converges to some limit $\nu(0)$. Taking a further subsequence, also $\mu_{n_k}(1)$ converges to some $\nu(1)$. Iterating the argument, we get a subsequence such that $\mu_{n_k}(i)$ converges to some $\nu(i)$ for every $(i)$. By Fatou’s lemma, we have $\sum_i \nu(i)\leq 1$. By the argument of part $(i)\implies(iii)$, we see that $\psi_n(z)$ converges to $\phi(z)$ [the generating function of $\nu$] for all $0<z<1$ and hence $\phi_z=\psi(z)$ for all $0<z<1$. By the uniqueness theorem of power series, we get $\mu=\nu$. This construction shows that every subsequence of $\mu_n$ has a subsequence that converges to $\mu$. Hence $\mu_n$ converges to $\mu$.

I hope the argument is clear.

With the best regards, Achim.

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