Suppose I have a multivariate function $f$ from $\mathbb{C}^d$ to $\mathbb{C}$ that accepts a Taylor expension of the form


$$f(\mathbf x) = \sum\limits\_{\mathbf k \in \mathbb N^d} a\_{\mathbf k} \mathbf x^\mathbf k.$$


I do have a closed form expression for this function (which is a little complex to expose), allowing me to extract some information: e.g. I know that $f$ is $\mathcal{C}^{\infty}$, but I could extract more info if needed.


I want to bound the remainder of the approximation, by showing that $f$ belongs to some smooth functional ball like:


$$B\_1(s,L) = \left\{ f: \sum\limits\_{\mathbf k \in \mathbb N^d} a\_{\mathbf k}^2 \mathbf k^s \le L\right\} \text{for positive $s,L$}$$


or:


$$B\_2(r,L) = \left\{ f: \sum\limits\_{\mathbf k \in \mathbb N^d} a\_{\mathbf k}^2 e^{<\mathbf r,\mathbf k>} \le L\right\} \text{for positive $r,L$}$$


**Question 1:** Have these balls names? Are they known things, and is there some theory about them?


**Question 2:** What would it mean for $f$ to belong to one of these balls? What property of $f$ is necessary? Sufficient? Both?


**Question 3:** More directly, the quantity I really need to bound is, for a given $p \in \mathbb{N}^d$, the error $E\_{\mathbf p} = \sum\limits\_{\mathbf k \ge \mathbf p} a\_{\mathbf k}^2$. Is there a way to bound this quantity from information about $f$ ?


[Please retag if not tagged correctly]



Some multivariate Taylor series and corresponding smoothness balls
You can relate $B\_2$ balls to the domain of analyticity of your function $f$. For instance, if $d=1$, $f\in B\_2(r,L)$ implies that $f$ is analytic on $\{z : \vert z\vert < \sqrt{e^r}\}$ and reciprocally, if $f$ is analytic on $\{z : \vert z\vert < \sqrt{e^\rho}\}$ then $f\in B\_2(r,L)$ for all $r<\rho$.


To put it differently, if you can say something about the domain of analyticity of your function $f$ you can then use Cauchy's formula to control the decay of the coefficients $a\_k$, and therefore you could directly get an answer to your Question 3.


You should have similar relations in higher dimension. Indeed, the only thing you really needs is Cauchy integral formula, which can be generalized in higher dimension (see e.g. Hörmander's book "An Introduction to Complex Analysis in Several Variables"). From that formula, if you assume that $f$ is analytic on $\mathcal{B}\_\rho := \{ z\in \mathbb{C}^d : \vert z\_i \vert < \rho\_i,\ i=1,\ldots,d \}$, then for any $s=(s\_1,\ldots,s\_d)$ s.t. $0<s\_i<\rho\_i$ for all $i$, you get something like
$$
\vert a\_k \vert = \left\vert \frac{\partial\_k f (0)}{k!} \right\vert \leq \frac{\sup\_{\mathcal{B}}\vert f\vert}{(2\pi)^d} \frac{1}{s^k}.
$$
As soon as you can take each $s\_i>1$, then you get a control on the decay of your coefficients, and you can say that $f$ belongs to some of your $B\_2$ balls, or directly try and answer your question 3.


