# Proof that the space of vector valued function is complete

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?

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 $$\alpha$$.

• @io67 why the second taylor expansion is true? Jan 30, 2022 at 20:00
• If you have uniform convergence, you can pass to the limit inside integrals. So you just let $n\to\infty$ in the Taylor formula for $f_n$. Is this your question? Jan 31, 2022 at 1:19
• @io67 Could you give a reference for the second taylor expansion? Is the first time that I see this formula Jan 31, 2022 at 2:44
• math.stackexchange.com/questions/2923981/… Jan 31, 2022 at 10:50
• The second taylor series how did you got it? Feb 5, 2022 at 10:38

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.

• I do not understand how did you get in the last inequality Jan 29, 2022 at 7:33

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.