Smooth functions on a fixed compact is a Banach space Let $\Omega \subset \mathbb{R}^n$ be an open set and $K \subset \Omega$ be a compact set. Define
$$
C^{\infty}_K(\Omega) = \{ u \in C^{\infty}(\Omega) : \text{supp}(u) \subset K \text{ is compact}\}.
$$
For a fixed $N \in \mathbb{N}$, consider the norm
$$
||u||_{N} = \max\{|D^{\alpha} u(x)| :  x \in K, |\alpha| \leq N\}, \forall u \in C^{\infty}_K(\Omega) .
$$
Is it true that $(C^{\infty}_K(\Omega), ||\cdot||_N)$ is a Banach space ?
 A: In general it is not true. In particular, if $K$ has empty interior, then the space $C^\infty_K(\Omega)$ contains only a zero function, so it is a Banach space in this case. On the other hand, if the interior of $K$ is not empty, then this space is not complete. The construction of a Cauchy sequence in this space that diverges is somewhat tedious, so I shall just sketch it. Also, note that if you consider only $N$ times continuously differentiable functions, then the analogous space (i.e. $C^N_K(\Omega)$) is complete.
At first, solve this problem in one-dimensional space, i.e. construct a sequence of smooth functions $f_m$ on an interval that converges to a function $f$ uniformly with $N$ its first derivatives, and $f$ is not smooth. Then, assuming that $K$ has nonempty interior, construct a smooth non-trivial function $\phi \in C^\infty_K(\Omega)$. Then consider $g_m(x_1,\dots,x_n) = \phi(x_1, \dots,x_n) f_m(x_1)$. This function converges uniformly with all derivatives of order $\le N$ to $\phi(x_1, \dots, x_n)f(x_1)$ and if $f$ is not smooth on the interior of the projection of $\mathrm{supp} \phi$ on the first coordinate, then $\phi(x_1, \dots, x_n)f(x_1)$ is not smooth. Thus, $g_m$ is a divergent Cauchy sequence in $C^\infty_K(\Omega)$.
A: It is not trivial to prove/see, but constraints on finitely-many derivatives do not constrain higher derivatives at all. Thus, to make a space of smooth functions (on a reasonable space, let's also say compact) complete (in a metric sense, for example), we "probably" need all those (semi-) norms, for all $N$. In fact, this can be proven. Still, we have countably-many semi-norms, and the metric completeness can be proven, so that space of smooth functions is a Frechet space.
If the physical space is non-compact, then the family of seminorms needs to be larger, namely, sups of derivatives on compact subsets...
But, no, no finite collection of derivatives constrains the higher ones. In fact, in all reasonable situations, the completion of smooth functions in the normed space indicated in the question is just the collection of $N$-fold differentiable functions...
A: Here is another way to see why your space is not complete. Assume first $\Omega = B(0,2)$, $K = \overline B(0,1)$ and pick $u \in C^N(\Omega)$ such that $\operatorname{supp} u \subset B(0,1/2)$ and $u \notin C^\infty$. For instance we may think of $u(x) = \phi(||x||^2)$ with $\phi \in C^N([0,\infty)) \setminus C^{N+1}([0,\infty))$ vanishing outside of $[0,1/\sqrt 2]$. Then introduce a mollifier $(\rho_\epsilon)$ on $\mathbb R^d$, you can show that $u * \rho_\epsilon \rightarrow u$ in $C^N_K$ but the limit is not in $C^\infty$. Then the sequence $(\rho_\epsilon)$ is Cauchy in $C_K^\infty$ but has no limit in this space because its limit is necessarily $u$ that is not $C^\infty$.
In the general case this contstruction can be adapted by translations if $K$ has no empty interior. In case $K$ has empty interior someone already said that $C_K^N$ was trivial.
