It is a very nice result in complex analysis that uniform convergence on compact subsets preserves complex-analyticity. This is enabled by the fact that complex-analyticity is really a nice and very rigid property. It is natural, then, to ask: is there a mode of convergence for functions defined on some domain $D \subset \Bbb R^n$ that preserves real-analyticity?
I was thinking of
$$f_k \to f \iff \sup _{x \in K} |\partial _\alpha (f_k (x) - f (x))| \to 0 \quad \forall \text{ multiindex } \alpha \in \Bbb N^n \text{ and } \forall K \subset D \text{ compact} .$$
Is my attempt correct? Is there any conceptually simpler mode of convergence preserving real-analyticity?
You seem to be asking the following: If $f_1,f_2,\dots$ and $f$ belong to $C^\infty(D),$ if $D^\alpha f_k \to f$ uniformly on compact subsets of $D$ for each $\alpha,$ and if each $f_k$ is real analytic on $D,$ is $f$ is real analytic on $D?$
The answer is no, even if $n=1.$ To prove this we use the following: For every bounded interval $[a,b],$ the space of polynomials is dense in each $C^k([a,b]), k = 0,1,2,\dots$ This follows from the Weierstrass approximation theorem.
So now take $f\in C^\infty([a,b])$ that is not real analytic on $[a,b].$ For $k = 0,1,2,\dots $ we can choose polynomials $P_k$ such that $\|P_k-f\|_k < 1/k.$ Here $\|\,\|_k$ is the usual norm on $C^k([a,b]).$ We then have $D^\alpha (P_k-f) \to 0$ uniformly on $[a,b]$ for every $\alpha,$ giving us a counterexample.