Norm turning $C^\infty(\overline{B(0,1)})$ into a Banach space? I've been trying to find a proof that for $K=\overline{B(0,1)} \subseteq \mathbb R^n$, we can find some norm or notion of convergence (I am indifferent as to how we specify the topology) on $\mathcal C^\infty (K, \mathbb R) = \mathcal C^\infty (K)$ such that it is a Banach space. To my surprise, I couldn't find anything of the sort. This fact is stated in the answer here, so I have reason to believe it's true. Also, I saw here how this works but only for $n=1$, and there the proof of the limit function also having a continuous derivative depends on being able to express the limit function with respect to the integral of the derivative. Can this approach be generalised to $\mathcal C^\infty(K)$ (there would be an obvious problem  of convergence with taking the sum of the supremum norms of all partial derivatives of all orders), or how would one endow $\mathcal C^\infty (K)$ with a topology such that it is complete?
 A: Part 1
If you can show that the algebraic dimension of the space $C^\infty(K)$ is equal to $\frak{c}$, then since $\frak{c}^{\aleph_0}=\frak{c}$, this post shows that there does exist a norm that makes it a Banach space:
Can every vector space (over $\mathbb{R}$ or $\mathbb{C}$) can be a Banach space (or Hilbert space)?
Part 2
Of course, the above is not practical at all and not constructive at all. Let's see a construction for $n=1$. The answer I am going to give yields a metric and not a norm, but it makes your space complete.
Let $f,g\in C^\infty([-1,1])$.
We define
$$d(f,g)=\sum_{n=0}^\infty\frac{\min\{1,\|f^{(n)}-g^{(n)}\|_\infty\}}{2^n}$$
where $f^{(n)}$ denotes the $n$-th derivative of $f$. It is immediately verified that $d$ defines a metric on our space. We show that the space is also complete.
Let $(f_n)\subset C^\infty([-1,1])$ be a Cauchy sequence with respect to $d$. by the inequality $\|f_n^{(k)}-f_m^{(k)}\|_\infty\leq2^k\cdot d(f_n,f_m)$ which holds for all $n,m,k\geq0$ we can easily deduce that for any $k\geq0$ the sequence $(f_n^{(k)})$ is Cauchy with respect to the supremum norm. Since uniformly Cauchy sequences of continuous functions converge to some continuous function, we have that each sequence $(f_n^{(k)})$ converges uniformly to some continuous function.
So, for $k=0$, let's say that $f_n\to f$ and for $k=1$, we have that $f_n'\to g$ uniformly.
Note that
$$\sup_{x\in[-1,1]}\bigg|\int_{-1}^xf_n'(t)dt-\int_{-1}^xg(t)dt\bigg|\leq\int_{-1}^1|f_n'(t)-g(t)|dt\leq2\|f_n'-g\|_\infty\to0 $$
but $\int_{-1}^xf_n'(t)dt=f_n(x)-f_n(-1)\longrightarrow f(x)-f(-1)$. We conclude that $f(x)=\int_{-1}^xg(t)dt+f(-1)$, so $f$ is differentiable and moreover $f'(x)=g(x)$.
Following a similar argument for higher derivatives, one can easily deduce that $f\in C^\infty([-1,1])$ and that $f_n^{(k)}\to f^{(k)}$ for all orders $k$. We will now show that $d(f_n,f)\to0$: Let $\varepsilon>0$. We find $k_0\in\mathbb{N}$ large enough so that $\sum_{k=k_0+1}^\infty\frac{1}{2^k}<\varepsilon/2$. Then
$$d(f_n,f)=\sum_{k=0}^\infty\frac{\min\{1,\|f_n^{(k)}-f^{(k)}\|_\infty\}}{2^n}\leq\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}+\sum_{k=k_0+1}^\infty\frac{1}{2^k}\leq$$
$$\leq\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}+\varepsilon/2 $$
But now we can take $n_0$ large enough so that $\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}<\varepsilon/2$ for all $n\geq n_0$, which will in turn yield $d(f_n,f)<\varepsilon$ for all $n\geq n_0$. As $\varepsilon>0$ was arbitrary, we conclude that $f_n\to f$ with respect to $d$. Thus $C^\infty([-1,1])$ is complete when equipped with the metric $d$.
Comment: The reason that $d$ fails to give a norm is because we apply these minima $\min\{1,-\}$. Those are necessary: if for example we tried to define a norm as $\|f\|=\sum_{n}\frac{\|f^{(n)}\|_\infty}{2^n}$, then the function $f(x)=e^{2x}$ will give $f^{(n)}(x)=2^ne^{2x}$, so the series will not converge. In order to avoid such pathologies we have to be more flexible and settle for a metric, using this $\min\{1,-\}$ trick that resolves such problems.
Comment: I am not completely sure but I guess that the analogue for greater dimensions $d\geq1$ would have to be something with multi-indexes, like
$$d(f,g)=\sum_{\alpha}\frac{\min\{1,\|\partial^\alpha f-\partial^\alpha g\|_\infty}{2^{|\alpha|}}$$
where $\alpha\in\mathbb{N}_0^d$, $\alpha=(\alpha_1,\dots,\alpha_d)$, $|\alpha|=\alpha_1+\dots\alpha_d$ and
$$\partial^\alpha f=\prod_{j=1}^d\frac{\partial^{\alpha_j}f}{\partial x_j^{\alpha_j}}.$$
I believe that the same arguments can be modified to get a complete metric on this space, but this is only my opinion, I am not sure.
