Proof that the space of vector valued function is complete

Question

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?

Answer 1

The idea is right, but I think there're many typos in the last few steps, and I cannot understand them.

Remember that in $\mathbb{R}^n,$ $l^p$ norm are equivalent with each other, for $p\in [1,\infty],$ so it doesn't matter that which norm you used. Especially, $\|x\|=\max(|x_1|,\cdots,|x_n|)$ is $l^\infty$ norm, and $\|x\|=|x_1|+\cdots+|x_n|$ is $l^1$ norm. As we can decompose vector valued function to each dimension, I believe there's no different with the case about $n>1$ with $n=1.$

To show that $v^\alpha=\partial^\alpha v^0,$ a classical method is to use the idea of test function/weak derivative. For a fixed continuous function $u,$ if for any $\varphi\in C_c^\infty(\Omega,\mathbb R^n),$ there exists a continuous function $v,$ such that $$\int_{\Omega}(v\varphi+u\partial_i\varphi)dx=0, \quad\forall \varphi\in C_c^\infty(\Omega,\mathbb R^n), $$ then call $v$ the weak derivative (on the direction of $x_i$) of $u.$ (And $\varphi$ is called test function.) Because (try to prove that)$$\int_{\Omega} u\varphi=0,\quad \forall \varphi\in C_c^\infty(\Omega,\mathbb R^n)\qquad\Rightarrow \qquad u\equiv 0\quad\text{in} \:\Omega,$$ the weak derivative is unique. Especially, when $u$ is derivable itself, it's not hard to see that $\partial_i u$ is its weak derivative by Green formula. So if you want to prove that $\partial_i u=v,$ just show that $v$ is its weak derivative. Then you get that by uniquness.

Now we show that $v^i=\partial_i v^0,$ for $|\alpha|=1.$ For arbitrary $\alpha,$ use induction. We need to prove that $$ \int_{\Omega} (v^i\varphi+v^0\partial_i \varphi)dx=0,\quad \forall \varphi\in C_c^\infty(\Omega,\mathbb R^n). $$ Let $K:=\operatorname{supp} \varphi\subset \Omega,$ then: $$ \begin{aligned} \left\|\int_{\Omega} (v^i\varphi+v^0\partial_i \varphi)dx\right\|&=\left\|\int_{\Omega} (v^i\varphi-(\partial_i f_k\varphi+f_k\partial_i \varphi)+v^0\partial_i \varphi)dx\right\|\\ &\le\int_{\Omega} \left\|v^i\varphi-\partial_i f_k\varphi\right\|+\left\|v^0\partial_i \varphi-f_k\partial_i \varphi\right\|dx\\ &\le 2|K|\mathfrak{p}_{1,K}(v^0-f_k)\mathfrak{p}_{1,K}(\varphi)\rightarrow 0 \end{aligned} $$ So $v^i$ is the weak derivative of $v^0,$ then $\partial^i v^0=v^i$ as $v^0\in C^\infty.$ By induction, $\partial^\alpha v^0=v^\alpha.$

Remark: $\partial^i$ means $\partial^\alpha,$ $\alpha=(0,\cdots,0,\stackrel{i^{th}}1,0,\cdots,0)$ here.

Answer 2

Just to clarify DreamAR's answer .

Let $\left\| x\right\|=\max(|x_1|,|x_2|)$ ,$K:=\operatorname{supp} \varphi\subset \Omega,$ and $K':=\operatorname{supp} \partial_i \varphi \subset \Omega,$ then we have \begin{aligned} \left\|\int_{\Omega} (v^i\varphi+v^0\partial_i \varphi)dx\right\|&=\left\|\int_{\Omega} (v^i\varphi-(\partial_i f_k\varphi+f_k\partial_i \varphi)+v^0\partial_i \varphi)dx\right\|\\ &\le\int_{\Omega} \left\|v^i\varphi-\partial_i f_k\varphi\right\|+\left\|v^0\partial_i \varphi-f_k\partial_i \varphi\right\|dx\\ &\le |K|\int_{\Omega}\left\|v^i-\partial_i f_k\right\|+|K'|\int_{\Omega}\left\|v^0-\ f_k\right\| \\ &\le |K|\operatorname{supp}\left\|v^i-\partial_i f_k\right\|\int_{\Omega}dV+|K'|\operatorname{supp}\left\|v^0-\ f_k\right\|\int_{\Omega}dV \left\|v^0-\ f_k\right\| \\ &=\Omega|K|\operatorname{supp}\left\|v^i-\partial_i f_k\right\|+\Omega|K'|\operatorname{supp}\left\|v^0-\ f_k\right\| \\ \end{aligned} and since we have that $ \operatorname{supp}\left\|v^i-\partial_i f_k\right\|\rightarrow 0$ and $ \operatorname{supp}\left\|v^0- f_k\right\|\rightarrow 0$ we obtain the desired result.

Answer 3

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 $\alpha$.