Suppose i have extended two d-variate functions $f$ and $g$ (two densities: positives and integrate to one) supported on $\mathbb{R}\_{+}^d$ into the following (tensorised) Laguerre($\alpha = 0$) orthonormal basis of $L^2(\mathbb R\_{+}^d)$: $$\left(\varphi\_{\mathbf k}(\mathbf x) = \sqrt{2}^d e^{-\lvert \mathbf x \rvert} \sum\limits\_{\mathbf j \le \mathbf k} \binom{\mathbf k}{\mathbf j} \frac{(-2\mathbf x)^{\mathbf j}}{\mathbf j !}\right)\_{\mathbf k \in \mathbb N^d}.$$


I obtain coefficients $a\_{\mathbf p}$ for $f$ and $b\_{\mathbf p}$ for g, such that :


$$f(x) = \sum\limits\_{\mathbf p \in \mathbb{N}^d} a\_{\mathbf p} \varphi\_{\mathbf p}(\mathbf x) \text{ and } g(x) = \sum\limits\_{\mathbf p \in \mathbb{N}^d} b\_{\mathbf p} \varphi\_{\mathbf p}(\mathbf x)$$


I already know that the follwoing convolution formula holds :


$$(f \star g)(x) = \sum\limits\_{\mathbf p \in \mathbb{N}^d} c\_{\mathbf p} \varphi\_{\mathbf p}(\mathbf x)$$


where the coefficients are given by:


$$c\_{\mathbf p} = \sqrt{2}^{-d}\sum\_{\mathbf \epsilon \in \{0,1\}^d} (-1)^{\lvert \mathbf \epsilon \rvert} \sum\limits\_{\mathbf k \le \mathbf p - \mathbf \epsilon} a\_{\mathbf p} b\_{\mathbf p - \mathbf \epsilon - \mathbf k}$$


Suppose now that $f$ and $g$ belong to smooth laguerre balls, i.e there exists $\mathbf r(f),\mathbf r(g) \in \mathbb R\_{+}^d$ and $L(f),L(g) >0$ such that :


$$\sum\_\limits{\mathbf p \in \mathbb N^d} a\_{\mathbf p}^2 e^{\langle \mathbf r(f),\mathbf p\rangle} \le L(f)$$


and same thing for $g$. This ensure some bound on the truncation error, which is what I am after.


**Question:** Can I find constants $\mathbf r(f\star g)$ and $L(f\star g)$ such that the same bound applies to the convolution? Moreover for the n-convolutions of $f\_1,...,f\_n$ ( by recursion, or better if posible).



Laguerre convolution truncation error
A possible way to get such a bound is to use the following estimate (I write it for $d=1$, but you should be able to generalize to higher dimensions).


**Lemma. Let $\nu>1$ and $a$ and $b$ be sequences such that
$$
\Vert a\Vert\_{\ell^2\_\nu}^2 := \sum\_{p\geq 0} a\_p^2 \nu^p<\infty \quad \text{and}\quad \Vert b\Vert\_{\ell^2\_\nu}^2 <\infty.
$$
Define the sequence $c$ by $c\_p = \sum\_{k=0}^p a\_k b\_{p-k}$ (this is nothing but the usual convolution/cauchy product). Then, for any $\mu\in[1,\nu)$
$$
\Vert c\Vert\_{\ell^2\_\mu} \leq \kappa\_{\mu/\nu} \Vert a\Vert\_{\ell^2\_\nu}
 \Vert b\Vert\_{\ell^2\_\nu},
$$
where the constant $\kappa\_{\mu/\nu}$ is explicit.**


**Proof.**


\begin{align\*}
\sum\_{p\geq 0} \vert c\_p\vert^2 \mu^p &\leq \sum\_{p\geq 0} \left(\sum\_{k\leq p} \vert a\_k \vert \vert b\_{p-k}\vert \right)^2 \nu^p \\
&\leq \sum\_{p\geq 0} \left(\sum\_{k\leq p} \frac{\Vert a\Vert\_{\ell^2\_\nu}}{\nu^{\frac{k}{2}}} \frac{\Vert b\Vert\_{\ell^2\_\nu}}{\nu^{\frac{p-k}{2}}} \right)^2 \mu^p \\
&\leq \Vert a\Vert\_{\ell^2\_\nu}^2 \Vert b\Vert\_{\ell^2\_\nu}^2 \sum\_{p\geq 0} \left(\sum\_{k\leq p} \left(\frac{\mu}{\nu}\right)^{\frac{p}{2}} \right)^2 \\
&\leq \Vert a\Vert\_{\ell^2\_\nu}^2 \Vert b\Vert\_{\ell^2\_\nu}^2 \sum\_{p\geq 0} (p+1)^2 \left(\frac{\mu}{\nu}\right)^p.
\end{align\*}
This sum converges since $\mu<\nu$, and I let you compute the explicit expression if you need to. $\square$


This bound is probably not sharp, in particular we may lose a lot by bounding each $\vert a\_k\vert$ independently by $\frac{\Vert a\Vert\_{\ell^2\_\nu}}{\nu^{k/2}}$, but at least it is rather elementary.


Of course this does not exactly answer your question, but by applying this lemma to the various terms in your $\sum\_{\epsilon}$ you should get what you want.


I leave below the old par of the answer about the bound in $\ell^1$, but maybe you don't need it anymore.




---


Since what you seem to really be after is a bound on the truncation error, it may be enough for you to use the fact that $\ell^1$ is a Banach algebra for the usual discrete convolution product, i.e. $\sum \vert c\_p\vert \leq \sum \vert a\_p\vert \sum \vert b\_p\vert $. Since you have a more complicated convolution product you just get an additional constant in front of r.h.s..


I'll do the computation here for $d=1$ to simplify a bit the notations, but the generalization to $d>1$ should be straigthforward.


Let $\nu = \exp(\min (r(f),r(g)))$, so that


$$
\sum\_{p\geq 0} \vert a\_p\vert \nu^p \leq L(f) \quad\text{and}\quad \sum\_{p\geq 0} \vert b\_p\vert \nu^p \leq L(g).
$$


Then


\begin{align\*}
\sum\_{p\geq 0} \vert c\_p\vert \nu^p &\leq \frac{1}{\sqrt 2}\sum\_{p\geq 0} \left(\sum\_{k\leq p} \vert a\_k \vert \vert b\_{p-k}\vert + \sum\_{k\leq p-1} \vert a\_k\vert \vert b\_{p-1-k}\vert \right) \nu^p \\
&= \frac{1}{\sqrt 2}\sum\_{p\geq 0} \left(\sum\_{k\leq p} \vert a\_k \vert \nu^k \vert b\_{p-k}\vert \nu^{p-k} + \nu \sum\_{k\leq p-1} \vert a\_k\vert\nu^k \vert b\_{p-1-k}\vert\nu^{p-1-k} \right) \\
&= \frac{1}{\sqrt 2}\left(\sum\_{k\geq 0} \vert a\_k \vert \nu^k \sum\_{l\geq 0} \vert b\_{l}\vert \nu^{l} + \nu \sum\_{k\geq 0} \vert a\_k \vert \nu^k \sum\_{l\geq 0} \vert b\_{l}\vert \nu^{l}\right) \\
&\leq \frac{1+\nu}{\sqrt 2}L(f)L(g).
\end{align\*}


In general, the constant in front should depend on $d$ and on $min(r(f),r(g))$.


