Derivation on real analytic manifolds Let $M$ be a real analytic manifold. By $C^{\omega}(M)$ we mean the algebra of all analytic functions from $M$ to $\mathbb{R}$. Assume that $D$ is  a derivation on  $C^{\omega}(M)$ .

Is there  a  global real analytic vector field $X$ on $M$ such that $D(f)=X.f$ for all $f\in C^{\omega}(M)$?  

The motivation:
The smooth version of this statment is true but the proof is based on usage of non analytic functions
 A: $\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\NN{\mathbb{N}}$I'm going to try once more. As my previous flawed answers should have made obvious, the real analytic category is not my home, so read with caution.
Conveniently, the two big results I need are stated as Corollary 5.43 and 5.44 here:
Cor. 5.43 $M$ has a proper real analytic embedding into some $\RR^n$. I'll write $(z_1, \ldots, z_n)$ for the coordinates on $\RR^n$.
Cor. 5.44 (specialized and rephrased) Let $p$ be a point of $M$ and let $M$ be embedded in $\RR^n$ as above. Let $U$ be a  neighborhood of $p$ in $\RR^n$. Let $f: M \cup U \to \RR$ be a function which is real analytic on $U$ and on $M$. Let $d$ be a nonnegative integer. Then there is a real analytic function $g: \RR^n \to \RR$ such that $g|_M=f$ and $(g-f)|_U$ vanishes to order $d$ at $p$.
We also need the following:
Easy Lemma: Let $f: \RR^n \to \RR$ be a real analytic function for which $f(z_1, \ldots, z_{n-1},0)$ is identically zero. Then $f(z_1, z_2, \ldots, z_{n-1}, z_n)/z_n$ extends to a real analytic function $\RR^n \to \RR$.
Proof This is a local statement, and can be easily checked locally using convergent power series. $\square$
Key Lemma: Let $g: \RR^n \to \RR$ be a real analytic function vanishing to order $d$ at $0$. Then we can write $g(z)$ as a finite sum $\sum z^a p_a(z)$ where $z^a$ are monomials of degree $d$ and $p_a(z)$ are real analytic functions.
Proof
By induction on $d$ and $n$. The base case $d=0$ is trivial: Write $g(z) = 1 \cdot g(z)$; the base case $n=0$ is vacuous. We assume $d$ and $n>0$. Define $g(z_1, \ldots, z_n)$ by
$$f(z_1, \ldots, z_n) = f(z_1, \ldots, z_{n-1}, 0) + z_n g(z_1, \ldots, z_{n-1}).$$
By the Easy Lemma, $g$ is analytic and, by a basic computation, it vanishes to order $d-1$. By induction on $d$, we can write $g = \sum z^c r_c(z)$ where the monomials $z^c$ have degree $d-1$. By induction on the number of variables, we can write $f(z_1, \ldots, z_{n-1},0)$ as $\sum z^b p_b(z_1, \ldots, z_{n-1})$ where the monomial $z^b$ have degree $d$. So
$$f(z_1, \ldots, z_n) = \sum z^b p_b(z_1, \ldots, z_{n-1}) + \sum (z^c z_n) q_c(z_1,\ldots, z_n),$$
an expression of the desired form. $\square$
We now prove the result.
Lemma Let $f$ be a real analytic function on $M$ that vanishes to second order at $p$. Then $(D f)(p)=0$.
Proof Embed $M$ in $\RR^n$ by Cor 5.43. Extend $f$ to a function on $\RR^n$ also vanishing to order $2$ at $p$ and, without loss of generality, translate $p$ to $0$. By the Key Lemma, we can write $f(z) = \sum z_i z_j c_{ij}(z)$. Then
$$(D f)(0) = \sum (D z_i)(0) \cdot 0 \cdot c_{ij}(0) + \sum 0 \cdot (D z_j)(0) \cdot c_{ij}(0) + \sum 0 \cdot 0 \cdot (D c_{ij})(0) =0. \quad \square$$
Lemma $(D f)(p)$ depends only on $(df)(p)$.
Proof Suppose that $(d f_1)(p) = (d f_2)(p)$. Define $q(z)$ by the equation $f_2(z) = f_1(z) + (f_2(p)-f_1(p)) + q(z)$, then $q$ vanishes to order $2$ at $p$ so $(D q)(p) =0$. Also, $D$ of a constant function is $0$. So $(D f_1)(p) = (D f_2)(p)$. $\square$
The previous lemma shows that there is a well defined map $v(p) : T^{\ast}_p M \to \RR$
so that $(D f)(p) = v(p)(d f)$. (Well, we also need to see that $C^{\omega}(M) \to T^{\ast}_p M$ is surjective, but that is obvious because the functions $z_i$ span the cotangent space.) It is easy to see that $v(p)$ is linear. So $v(p)$ is an element of $(T_{\ast})_p M$. We have now created a section $v: M \to T_{\ast} M$ so that $(D f)(p) = \langle v(p), df \rangle$. It remains to see that $v$ is real analytic.
This question is local. Let $(z_1, \ldots, z_d)$ give local coordinates on $M$ near some point $p$. Then we can write
$$v(p) = \sum_{i=1}^d v_i(p) \frac{\partial}{\partial z_i}.$$
We need to show that the functions $v_i$ are real analytic. Since $v_i = D z_i$, this is clear. QED.
