TITLE: Proof that the space of vector valued function is complete
QUESTION [5 upvotes]: Let $\Omega \subseteq \mathbb{R}^{2}$ be open  let  $C^{\infty}(\Omega, \mathbb{R}^{2})$  denote the space of all smooth functions from $\Omega$ to $\mathbb{R}^{2}$.
We define  a norm in $R^2$ given by  $\left\| x\right\|=\max(|x_1|,|x_2|)$ where $x=(x_1,x_2)\in \mathbb{R}^{2}$.
This turn $\mathbb{R}^{2}$ into a Banach space.
Defining the seminorms $\mathfrak{p}_{K, k}$ (for $K \subseteq \Omega$ compact and $k \in \mathbb{N}_{0}$ ) on $C^{\infty}(\Omega, \mathbb{R}^{2})$ by
$$
\mathfrak{p}_{K, k}:=\sup _{|\alpha| \leq k, x \in K}\left\|\partial^{\alpha} f(x)\right\|
$$
My goal is to proof that $C^{\infty}(\Omega, \mathbb{R}^{2})$ is complete space witch means that for a cauchy sequence $f_n$ we have that  for all $K, k$ we have $\mathfrak{p}_{K, k}(f_n-f)\rightarrow 0$ for some fixed  $f \in C^{\infty}(\Omega, \mathbb{R}^{2})$.
Since $f_n$ is Cauchy we have   $\sup _{|\alpha| \leq k, x \in K}\left\|\partial^{\alpha} f_i(x)-\partial^{\alpha} f_j(x)\right\|\leq\varepsilon$ for $i,j \ge N$ and so for each $x$,
$\partial^{\alpha} f_i(x)$ is Cauchy in $\mathbb{R}^{2}$ and since $\mathbb{R}^{2}$ is complete we have that $\partial^{\alpha} f_i(x)$ has pointwise limit $v^{\alpha}(x)$.
So from this we have that $\sup _{|\alpha| \leq k, x \in K}\left\|\partial^{\alpha} f_i(x)-v^{\alpha} (x)\right\|\leq\varepsilon$ for $i \ge N_{sup}$ and so $f_i(x)$ converges to $v^{\alpha}$
Now to complete the proof I need  to show that $v^\alpha=\partial^{\alpha}v^0$ with $v^0 \in C^{\infty}(\Omega, \mathbb{R}^{2})$.
Let $e_1$ and $e_2$ be a basis of $\mathbb{R}^{2}$. We decompose .
$v^\alpha=v^0_1e_1+v^0_2e_2$ and $f_i=f_i^1e_1+f_i^2e_2$
Now suppose that $l^\alpha=\partial^{\alpha}v^0_1e_1+\partial^{\alpha}v^0_2e_2$ ,by the properties of norms we have
$$\sup _{|\alpha| \leq k, x \in K}\left\|\partial^{\alpha} f_i(x)-l^{\alpha} (x)\right\|\leq\ \sup _{|\alpha| \leq k, x \in K} \left(|f_i^1 -\partial^{\alpha}v^0_1|+ |f_i^2-\partial^{\alpha}v^0_2|\right)$$
Now it is clear that $f_i^1$ and$f_i^2$ is Cauchy so we know that $f_i^1$converges to  $v^0_1$ and   $f_i^2$converges to  $v^0_2$ and so
$\sup _{|\alpha| \leq k, x \in K}\left\|\partial^{\alpha} f_i(x)-l^{\alpha} (x)\right\|\leq\varepsilon$ for $i \ge N_{\sup}$
By uniqueness of limit we have then $l^{\alpha}=v^{\alpha}$.
Is my proof correct?

REPLY [2 votes]: Use Taylor's formula with integral remainder. If you take $x_{0}\in\Omega$ you
can find a ball $\overline{B(x_{0},r)}\subseteq\Omega$. The ball is actually a
cube. Then for every $x,y\in B(x_{0},r)$,
$$
f_{n}(x)=f_{n}(y)+\sum_{|\alpha|=1}^{k}\frac{1}{\alpha!}\partial^{\alpha}%
f_{n}(y)(x-y)^{\alpha}+\sum_{|\alpha|=k+1}c_{\alpha}(x-y)^{\alpha}\int_{0}%
^{1}(1-t)^{k}\partial^{\alpha}f_{n}(tx+(1-t)y)\,dt.
$$
By uniform convergence in the compact set $\overline{B(x_{0},r)}$,
$$
f(x)=f(y)+\sum_{|\alpha|=1}^{k}\frac{1}{\alpha!}v^{\alpha}(y)(x-y)^{\alpha
}+\sum_{|\alpha|=k+1}c_{\alpha}(x-y)^{\alpha}\int_{0}^{1}(1-t)^{k}v^{\alpha
}(tx+(1-t)y)\,dt.
$$
Taking $x=y+te_{i}$, dividing by $t$ and letting $t\rightarrow0$ gives
$\partial_{\iota}f(y)=v^{e_{i}}(y)$. You can now use an induction argument, or
apply the same reasoning with $f$ replaced by $\partial^{\beta}f$ to conclude
that $\partial^{\alpha}f(y)=v^{\alpha}(y)$ for every $y\in B(x_{0},r)$ and every multi-index $\alpha$.