Let $\alpha$ be a positive constant, $\mu$ be a Borel nonnegative measure in $\mathbb{R}^n_+$. We can define a transform $$ \tilde{L}\[\mu\](p) = \int\limits_{\mathbb{R}^n_+} e^{-(p_1 x_1 + \ldots + p_n x_n)^{\alpha}} \mu(dx). $$ This is a simple generalization of the Laplace transform for which $\alpha = 1$. Is there something in literature on this transform? I'm interested in questions of inversion, characterisation e.t.c.


2 Answers 2


In one dimension, if $\mu$ has a density $f$ with respect to the Lebesgue measure, $$ (\tilde L f)(p)=\int_0^{+\infty} e^{-p^\alpha y}f(y^{\frac{1}{\alpha}})\frac{1}{\alpha} y^{\frac{1}{\alpha}-1} dy=(Lg_\alpha)(p^\alpha), $$ that is the ordinary Laplace transform of $g_\alpha$, $g_\alpha(y)=f(y^{\frac{1}{\alpha}})\frac{1}{\alpha} y^{\frac{1}{\alpha}-1} $ evaluated at $p^\alpha$. The extension to a class of Borel measure is certainly possible.

The multidimensional case is not so clear.


Here is some hope. (For simplicity, when referring to the unit sphere, I really mean only the part of the unit sphere contained in the first quadrant $\mathbb{R}^n_+$.)

If $\mu$ is the Dirac delta concentrated at a point $\vec{x}$ then $L[\mu](\vec{p})= e^{-(\vec{p}\cdot\vec{x})^\alpha}$, which shows that $\mu$ is determined by this transform. More generally suppose $\mu$ is a linear combination of $\delta$-s

$$\mu=\sum_{k=1}^N c_k \delta(\vec{x}-\vec{x}_k). $$

In this case we have

$$L[\mu](\vec{p})=\sum_{k=1}^N c_k e^{-(\vec{p}\cdot\vec{x}_k)^\alpha}=:f(\vec{p}). $$

Consider the set $P$ of $\vec{p}$-s of length $1$ so that the numbers $\vec{p}\cdot\vec{x}_i$ are pairwise disjoint. $P$ is open and dense in the unit sphere because it's the complement in the unit sphere is the union of the great spheres $\vec{p}\cdot \vec{x}_i=\vec{p}\cdot\vec{x}_j $, $i\neq j$.

The set $P$ is made up of several connected components, chambers, $P_1,\dotsc, P_\nu$. For any chamber $P_s$, there exists $i(s)=1,\dotsc, N$ such that

$$\vec{p}\cdot \vec{x}_{i(s)} < \vec{p}\cdot \vec{x}_j,\;\;\forall j\neq i(s),\;\;\forall \vec{p}\in P_s. $$

Then as $t\to \infty$ we have

$$\frac{\log f(t\vec{p})}{t^\alpha} \sim -(\vec{p}\cdot\vec{x}_{i(s)})^\alpha. $$

This determines $\vec{p}\cdot\vec{x}_{i(s)}$ for any $p\in P_s$. This determines the point $\vec{x}_{i(s)}$.

The constant $c_{i(s)}$ is then determined from the equality

$$c_{i(s)}= \lim_{t\to\infty} e^{t^\alpha(\vec{p}\cdot\vec{x}_{i(s)})^\alpha} f(t\vec{p}),\;\; \vec{p}\in P_s.$$

The function

$$ g(\vec{p})= f(\vec{p})-c_s e^{-(\vec{p}\cdot \vec{x}_{i(s)})^\alpha} $$

is the generalized Laplace transform of a linear combination of $\delta$-s concentrated at $(N-1)$ points.

Thus, the generalized Laplace transform is injective when restricted to the vector space spanned by $\delta$-s. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\vp}{\vec{p}}$ $\newcommand{\vx}{\vec{x}}$ $\newcommand{\vy}{\vec{y}}$

Remark 1 Let me denote by $Q_x$ the quadrant $\mathbb{R}^n_+$ in the $x$-space and by $Q_p$ the same quadrant but viewed in the $p$-space. Consider the kernel $K_\alpha:Q_p\times Q_x\to\bR$

$$ K_\alpha(\vec{p},\vec{x}) = e^{-(\vp\cdot {\vx})^\alpha}. $$

and its dual $K_\alpha^*:Q_y\times Q_p\to\bR$

$$ K_\alpha^*(\vec{y},\vp) = e^{-(\vec{y}\cdot\vp)^\alpha}. $$

The kernel $K_\alpha$ defines the generalized Laplace transform $L_\alpha$ when applied to functions on $Q_x$ that decay sufficiently fast. The convolution of the $K_\alpha^*$ with $K_\alpha$ is the kernel

$$ N_\alpha(\vy,\vx)=K_\alpha^* \ast K_\alpha(\vy,\vx)=\int_{Q_p} e^{-(\vy\cdot\vp)^\alpha-(\vp\cdot\vx)^\alpha} d\vp. $$

Note that

$$ N_\alpha(\vx,\vy)=N_\alpha(\vy,\vx)>0, \;\; \forall \vx,\vy\in Q$$

and if $f, g:Q\to\bR$ decay sufficiently fast at $\infty$ we have

$$\int_Q\int_Q N_\alpha(\vx,\vy) f(\vx)g(\vy) d\vx d\vy = (L_\alpha[f],L_\alpha[g])_{L^2(Q_p)}. $$

The integral operator $T_\alpha$ defined by $N_\alpha$ is symmetric and nonnegative, i.e.,

$$ (T_\alpha f,f)\geq 0,\;\; \forall f\in L^2(Q). $$

To check the injectivity of $L_\alpha$ one needs to show that

$$ (T_\alpha f,f)=0,\;\; f\in L^2(Q) \Rightarrow f=0. $$

Comment There seems to be problems with MathJax because I get distorted display and I cannot find my TeX errors. I'll post as is and Edit later.


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