Laguerre convolution truncation error

Question

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).

Answer 1

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.

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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 $min(r(f),r(g))$.