Does $\sum_{n \geq 0} a_n x^n=\sum_{n \geq 0} b_nx^n$ imply $a_n=b_n$ for vector-tuple power series?

Question

My reference is Infinite series in p-adic fields by Keith Conrad.

Corollary 5.6. If $f(x)=\sum_{n≥0} a_nx^n$ has a positive radius of convergence in the $p$-adic field $\mathbb Q_p$ then $f$ is infinitely differentiable on its disc of convergence in $K$ and for all $n ≥ 0$ and $a_n=\frac{f^{(n)}(0)}{n!}$. In particular, $f$ has only one choice of power series coefficients.

Proof. Let $D$ be the disc of convergence of $f(x)$. By Theorem 5.3 (of the same note), $f'(x)=\sum_{n \geq 0} n a_n x^{n-1}$ for all $x \in D$. This is again a power series, so by Theorem 5.3, it is differentiable on $D$ with derivative $f^{''}(x)=\sum_{n \geq 2}n(n-1)x^{n-2}$. By induction, the $k$th derivative is $$f^{(k)}(x)=n(n − 1) \cdots (n − (k − 1))a_nx^{n−k}$$ for all $x \in D$. In particular, the constant term is $f^{(k)}(0)= k(k−1) \cdots (k−(k−1))a_k= k!a_k$, so $a_k =\frac{f^{(k)}(0)}{k!}$. This shows the coefficients of the power series are determined by $f$ as a function around $0$.

Uniqueness follows from Corollary 3.15 of the same Note, and it is not required to use derivatives of $f$.

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My question: Is there a generalization of the above results for vector-tuple power series $f: \mathbb Q_p^n \to \mathbb Q_p^n$?

For example, take the simple case $n=2$ and consider the 2-tuple power series $f(x,y)=(f_1(x,y),f_2(x,y))$ defined in the following way $$f(x,y)=\begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix}=\sum_{m,n \geq 0} \begin{pmatrix} a_{mn}^{11} & a_{mn}^{12} \\ a_{mn}^{21} & a_{mn}^{22} \end{pmatrix} X^{m+n}, $$ where $X^{l+m}=\begin{pmatrix} x^i y^j \\ x^r y^s \end{pmatrix}$ is column vector and $i,j,r,s \geq 0$ such that $i+j=m$ and $r+s=n$. Also note that $a_{mn}^{uv}$ is the $(u,v)$-th entry of the 2-by-2 matrix corresponding to the coefficient of $X^{l+m}$.

Of course, there can be more different expositions of $f(x,y)=(f_1(x,y),f_2(x,y))$, for example: $$f(x,y)=\begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix}=\begin{pmatrix} \sum_{m,n \geq 0} a_{m,n} x^my^n \\ \sum_{m,n \geq 0} a'_{m,n}x^my^n \end{pmatrix};~a_{m,n}, a_{m,n}^{'} \in \mathbb Q_p.$$

Can we generalise Corollary 5.6?

I see the challenge in handling the derivatives of $f(x,y)$. The coefficients $a_{mn}^{11}, a_{mn}^{12}, a_{mn}^{21}, a_{mn}^{22}$ are determined by the mixed partial derivatives of $f(x,y)$ at $(0,0)$.

Can someone shed some light on it? Thanks

Answer 1

Since no answer has been provided so far, I am composing a self-answer. I appreciate suggestions and comments on how to improve it.

We have $$ f(x,y)=\begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix} = \begin{pmatrix} \sum_{m,n \geq 0} a_{m,n} x^my^n \\ \sum_{m,n \geq 0} b_{m,n}x^my^n \end{pmatrix};~a_{m,n}, b_{m,n} \in \mathbb Q_p. $$ In the region of convergence of $f(x,y)$, each component $f_1(x,y)$ and $f_2(x,y)$ is infinitely differentiable with respect to both variables $x$ and $y$. So partial derivatives of $f(x,y)$ exist of all orders.

As suggested by @Nandor in their comment, take the projection of $f(x,y)$ to the first coordinate so that $f(x,y)$ is of the form $f(x,y)=\sum_{m,n \geq 0} a_{m,n} x^my^n$. Keep unchanging $y$ and define the map $x \mapsto f(x,y)$ by $g(y)=f(\alpha,y):=\alpha^m \sum_{n \geq 0} a_{n}y^n$ for some constant $\alpha \in \mathbb Q_p$ and $a_n$ is adjusted with $a_{m,n}$. Then, using Corollary 5.6 (see in the question), we get $a_{n}=\frac{\partial^{(n)}}{n!}f(\alpha,0)$. Again, vary $x$ and apply the map $y \mapsto f(x,y)$ we get the coefficient $$ a_{m,n}=\frac{\partial^{(m) \partial^{(n)}} \partial}{m! n!}f(x,y)|_{x=y=0}. $$ Now we prove the uniqueness of the coefficients $a_{m,n}$. Let us assume $f(x,y)=h(x,y)$ i.e., $$ \begin{align} f(x,y) =\begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix} & =\begin{pmatrix} \sum_{m,n \geq 0} a_{m,n} x^my^n \\ \sum_{m,n \geq 0} b_{m,n}x^my^n \end{pmatrix}\\ &=\begin{pmatrix} \sum_{m,n \geq 0} a'_{m,n} x^my^n \\ \sum_{m,n \geq 0} b'_{m,n}x^my^n \end{pmatrix}=\begin{pmatrix} h_1(x,y) \\ h_2(x,y) \end{pmatrix}=h(x,y) \end{align} $$ where $a_{m,n}, b_{m,n}, a_{m,n}', b'_{m,n} \in \mathbb Q_p$, and this implies the vanishing of the difference between the coefficients, i.e. $$ \implies \begin{pmatrix} \sum_{m,n \geq 0} (a_{m,n}-a'_{m,n}) x^my^n \\ \sum_{m,n \geq 0} (b_{m,n}-b'_{m,n})x^my^n \end{pmatrix}=0 $$ We just need to show $c_{m,n}:=a_{m,n}-a'_{m,n}=0$ and $c._{m,n}=b_{m,n}-b'_{m,n}=0$ for all $m,n$. We have the power series $$ \begin{align} \sum_{m,n \geq 0} c_{m,n}x^my^n&=0, \\ \text{ and }\\ \\ \sum_{m,n \geq 0} c'_{m,n}x^my^n &=0. \end{align} $$ Put $x=0=y$, we get $c_{0,0}=0=c'_{m,n}$. Suppose for small non-zero $x$ and $y$, we have $c_{M,N}=0=c'_{M,N}$ for $m<M$ and $n<N$. Then, $$ \begin{align} \sum_{m \geq M, \\ n \geq N} c_{m,n}x^my^n& =0, \\ \text{and} \\ \\ \sum_{m \geq M, \\ n \geq N} c'_{m,n}x^my^n& =0. \end{align} $$ We want to show that $c_{M,N}=0=c'_{M,N}$. For, $$ \begin{align} \sum_{m \geq M, \\ n \geq N} &c_{m,n}x^my^n =0 \\ \implies &c_{M,N}x^My^N+c_{M+1, N+1}x^{M+1}y^{N+1}+c_{M+2,N+2}x^{M+2}y^{N+2}+\cdots=0 \\ \implies & c_{M,N}+c_{M+1, N+1}x^{M}y^{N}+c_{M+2,N+2}x^{M+1}y^{N+1}+\cdots=0\\ & \text{and since $x \neq 0 \neq y$} \\ \implies & c_{M,N}=0 \end{align} $$ Thus, $c(x,y):=\sum_{m,n \geq 0} c_{M+m,N+n}x^my^n=0$ in $\mathbb Q_p$ for all small non-zero $x,y$. Note that $c(x,y)$ is also convergent at $x=0=y$ with value $c_{M,N}=0$. By the convergence property, $$c_{M,N}=\lim_{x \to 0 \\ y \to 0} \sum_{m,n \geq 0} c_{M+m,N+n}x^my^n=\lim_{x \to 0 \\ y \to 0} 0=0,$$ which implies $c_{M,N}=0$. Thus, by induction, $c_{m,n}=0$. Similarly, we get $c'_{m,n}=0$. Therefore, $a_{m,n}=a'_{m,n}$ and $b_{m,n}=b'_{m,n}$. Hence unique coefficients of $f(x,y)$. This completes the rough demonstration.

Conclusion: If $f,g: \mathbb Q_p^r \to \mathbb Q_p^r$ be two vector $r$-tuple p-adic power series with positive radius of convergence in $\mathbb Q_p$ and they are equal, then their coefficients are equal, i.e., assumed $a_{m,n}, b_{m,n}, a_{m,n}', b'_{m,n} \in \mathbb Q_p$ $$ \begin{align} f(x,y) & = \begin{pmatrix} \sum_{m,n \geq 0} a_{m,n} x^my^n \\ \sum_{m,n \geq 0} b_{m,n}x^my^n \end{pmatrix} = \begin{pmatrix} \sum_{m,n \geq 0} a'_{m,n} x^my^n \\ \sum_{m,n \geq 0} b'_{m,n}x^my^n \end{pmatrix}=h(x,y)\\ \\ & \qquad\implies a_{m,n}=a'_{m,n},~ b_{m,n}=b'_{m,n}. \end{align} $$