It's true that $\sum_{k\in \mathbb N}|f ^{(k)}|_X < \infty$?

Question

Let $X$ be a compact subset of $\mathbb C$ and $(f_n)$ be a sequence in $D^{\infty}(X) = \left \{ f:X \to \mathbb C: f^{(k)} \text{ is continuous for all k} \right \}$ such that $f_n \to f$ uniformly and $f_n^{ (k)} \to f^{(k)}$ uniformly.

Assuming that $ \sum_{k\in \mathbb N}|f_n ^{(k)}|_X < \infty $, for all $n$. It's true that $\sum_{k\in \mathbb N}|f ^{(k)}|_X < \infty$?

I'm trying the following argument: for each $k \in \mathbb N$,

$$ |f^{(k)}|_X \leq |f^{(k)} - f_n^{(k)}|_X + |f_n^{(k)}|_X $$

taking $n \to \infty$,

$$ |f^{(k)}|_X \leq \lim_{n\to \infty}|f^{(k)} - f_n^{(k)}|_X + \lim_{n\to \infty}|f_n^{(k)}|_X = \lim_{n\to \infty}|f_n^{(k)}|_X$$

then,

$$\sum_{r=0}^k |f^{(r)}|_X \leq \lim_{n\to \infty} \sum_{r=0}^k |f_n^{(r)}|_X \leq \lim_{n\to\infty} \sum_{r=0}^\infty |f_n^{(r)}|_X$$

My problem here is that the limit in the right side converges.

Help?

Answer 1

If there is an $M \in (0,+\infty)$ such that

$$\sum_{k = 0}^{\infty} \lvert f_n^{(k)}\rvert_X \leqslant M$$

for all $n$, then your argument shows that also

$$\sum_{k = 0}^{\infty} \lvert f^{(k)}\rvert_X \leqslant M.$$

By extracting a subsequence, we see that

$$\liminf_{n\to \infty} \sum_{k = 0}^{\infty} \lvert f_n^{(k)}\rvert_X < +\infty$$

suffices.

But if

$$\lim_{n\to \infty} \sum_{k = 0}^{\infty} \lvert f_n^{(k)}\rvert_X = +\infty,$$

then it does not follow that

$$\sum_{k = 0}^{\infty} \lvert f^{(k)}\rvert_X < +\infty.$$

For an example, let $0 \in X$, and

$$f_n(z) = \sum_{m = 0}^n \frac{z^m}{m!}.$$

Since each $f_n$ is a polynomial, clearly

$$\sum_{k = 0}^{\infty} \lvert f_n^{(k)}\rvert_X = \sum_{k = 0}^{n} \lvert f_n^{(k)}\rvert_X < +\infty,$$

but since $f = \exp$, we have $f^{(k)}(0) = 1$ for all $k$ and hence

$$\sum_{k = 0}^{\infty} \lvert f^{(k)}\rvert_X = +\infty.$$