I am trying to determine whether a space is complete. I'll begin by setting up some notation. Let $\Omega \subseteq \mathbb{R}^N$ be an open set. For each $n\in \mathbb{N}$, define $C^n\left(\overline{\Omega}\right)$ be the space of $n$-times continuously differentiable functions endowed with the norm
$$
\lVert f \rVert_{C^n} = \sum_{\lvert \alpha \rvert \leq n} \sup_{x\in \overline{\Omega}}\lvert \partial^\alpha f(x)\rvert.
$$
Given a sequence $(c_n)$ of positive numbers, set
$$
\lVert f \rVert_c = \sum_{n = 1}^\infty c_n \lVert f\rVert_{C^n}.
$$
I wish to show that the normed space
$$
E = \left\{ f \in C^\infty\left(\overline{\Omega}\right) : \lVert f \rVert_c < \infty \right\}
$$
is a Banach space.
To show this, I can begin by considering a Cauchy sequence $(f_\alpha)$ in $E$. One can easily show that there exists a function $f\in C^\infty(\overline{\Omega})$ such that $f_\alpha \to f$ in $C^n$ for every $n\in \mathbb{N}$. If I can also show that $f_\alpha \to f$ with respect to $\lVert \cdot \rVert_c$ then the proof is complete. In order to prove this, I would like to interchange between the infinite sum and the limit. The dominated convergence theorem does not seem to be sufficient to establish this.
Am I missing something? Is there some kind of condition that I could impose on the sequence $(c_n)$ so that $E$ becomes a Banach space or is this a hopeless endeavour. If $E$ is not a Banach space, is there a way I could go about proving that?
Best Answer
It is a Banach space. Proof is by repeated application of the following one dimensional result:
Suppose $f_n$ is continuously differentiable for each $n$, $f_n \to f$ uniformly and $f_n' \to g$ uniformly. Then $f$ is differentiable and $g=f'$ ( so $f_n' \to f'$ uniformly).