Let $X \subset \mathbb C$ be perfect and compact. Consider the algebra $D^1(X)$ of the functions with cotninuous derivative in $X$ with the norm $\|f\|_1 = |f|_X + |f'|_X$. For a sequence of positive real numbers $M= (M_n)$ such that
$$ M_0 = 1 \quad, \frac{M_n}{M_{k} M_{n-k}} \geq \binom{n}{k},\, \forall \, 0 \leq k \leq n $$
define the algebra $D(X,M)$ of the functions on $X$ with derivatives of all orders satisfying $\sum_{k = 0}^\infty \frac{|f^{(k)}|_X}{M_k} < \infty$. So we can define the norm $\|f\| = \sum_{k = 0}^\infty \frac{|f^{(k)}|_X}{M_k}$ in $D(X,M)$.
I'm trying to proof the following result:
If $D^1(X)$ is complete, then $D(X,M)$ is complete.
My attempt:
Take $(f_n)$ a Cauchy sequence in $D(X,M)$, in particular, for each $k$, $(f_n^{(k)})$ is a Cauchy sequence in $D^1(X)$, so exists $g_k \in D^{1}(X)$ such that $f_n^{k}$ converges to $g_k$ in $D^1(X)$. Since
$$ \|f_n ^{(k)} - g_k \|_1 = |f_n ^{(k)} - g_k|_X + |(f_n ^{(k)})' - g_k'|_X $$
follows that $(f_n ^{(k)})'$ converges uniformly to $g_0'$, however $(f_n ^{(k)})' = f_n^{(k+1)}$ that converges to $g_{k+1}$ in $D_1(X)$. Hence, $g_{k+1} = g_k '$ for all $k$.
We conclude that $f_n^{ (k)}$ converges to $g_0^{(k)}$ in $D^1(X)$, for all $k$.
I still need to proof that $g_0 \in D(X,M)$ and $f_n$ converges to $g_0$ in the norm $\| \cdot \|$.
If I assume $g_0 \in D(X,M)$, the convergence holds because
$$ \|f_n - g_0 \| = \sum_{k = 0}^\infty \frac{|f_n^{(k)} - g_0^{(k)}|_X}{M_k} \text{ and } |f_n^{(k)} - g_0^{(k)}|_X \to 0, \, \forall k $$
My problem is to show that $\sum_{k = 0}^\infty \frac{|g_0^{(k)}|_X}{M_k} < \infty$.
Help?