Showing that a thin-plate spline RBF approximation is real analytic I am finishing my Ph.D. dissertation in engineering and I would like to show a simple proof. I  am having troubles formalizing my ideas into a proof though. I think in a mathematics paper this concept is obvious, but I think a simple proof will go a long way in an engineering paper. 
I need to show that a thin-plate spline RBF approximation:
\begin{equation}
 \mathcal{P}_f(\boldsymbol{x}) = \sum_{k=1}^N c_k r_k^2 \log (r_k) + P\left(\boldsymbol{x}\right) 
\end{equation}
is real analytic, where $r_k=||\boldsymbol{x}-\boldsymbol{x_k}||$, $c_k$'s are constant,  $\boldsymbol{x}_k$ are constant training nodes and $P\left(\boldsymbol{x}\right)$ is a polynomial in the coordinates of $\boldsymbol{x}$, e.g. for $\boldsymbol{x}\in\mathbb{R}^2$ , $P\left(\boldsymbol{x}\right)=d_0 + d_1x_1 +d_2x_2$, where the $d$'s are the coefficients and $x_1$ and $x_2$ are the coordinates of $\boldsymbol{x}$.
I know that the sums, products, and compositions of real analytic functions are real analytic , that the Logarithm is real analytic on $\left(0,\infty\right)$, and that any real polynomial is real analytic.
Would it be appropriate to do something along the following:
Theorem 1: $\mathcal{P}_f$ is a real analytic function on $\left(0,\infty\right)$
Axiom 1: the sums, products, and compositions of real analytic functions are real analytic
Axiom 2: the Logarithm is real analytic on $\left(0,\infty\right)$
Axiom 3: any real polynomial is real analytic.
...
 A: Well, the proof would look something like this:
To show $P_f$ analytic, we observe that $P$ is analytic (it's just an affine function!), so all that remains is to show that the summation is analytic. We do so by showing that each term is. To do this, we need only show that the function $g(x) = x^2 \log |x|$ is analytic. (Indeed, this is both sufficient and necessary, as the case where $N = 1$ and $x_1 = 0$ shows.)
The only problem remaining is that I'm not at all certain that the function $g$ really is analytic. I look at $g(1+x) = (1+x)^2 \log(|1 + x|)$, and for positive $x$, the Taylor series for that (around x = 0) is the product of the series for the two factors, i.e., it's
$$
(1 + 2x + x^2) * (x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots) = \\
(x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots) + 2x((x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots))+ x^2((x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots)).
$$
The $n+1$th term of that series (after the first few) has coefficient $c_{n+1} = \pm (\frac{1}{n+1} - \frac{2}{n} + \frac{1}{n-1}) = \pm \frac{2}{n(n - 1)(n+1)}$. If you apply the ratio test to try to find the radius of convergence, you end up with 
$$
\rho = x \frac{n (n-1) (n+1)}{(n+1) (n) (n+2)}= x \frac{ (n-1) }{(n+2)}
$$
which in the limit looks like just $x$. So you get a radius of convergence of just 1. That doesn't sound like a real analytic function to me. 
I may have done the computation wrong, of course. But in general, I wouldn't expect this result to be real-analytic, and my computation seems to suggest it's not.  
