Functions of the form $f(x)=\int_{0}^{\infty}F_t \exp(-tx) dt$

75 Views Asked by At

(1) Can each $f\in L^1([0,\infty))$ be written as

$f(x)=\int_{0}^{\infty}F_t \exp(-tx) dt$ for a function $F:[0,\infty)\to\mathbb{R}$?

I believe yes, because the Laplace transform restricted to real exponents is injective on $L^1$ (because the full Laplace transform on $L^1$ is injective into the holomorphic functions on the right half plane and because holomorphic functions are uniquely determined by their values on the reals) and hence the transpose, which formally is again the Laplace the transform, should be surjective.

(2) How can you find $F_t$ practically?

Surprisingly, none of the notes on Laplace transforms that I found online discussed these questions.

1

There are 1 best solutions below

1
On

If $\mu$ is a measure on $(0,\infty)$ such that $\int \frac 1 t d\mu(t) <\infty$ then the function $f(x)=\int_0^{\infty} e^{-tx}d\mu(t)$ is integrable but it cannot be written as $\int_0^{\infty} F_t e^{-tx}dt$ for any function $F_t$ unless $\mu$ is absolutely continuous. We can even have a singular measure $\mu$ such that $\int \frac 1 t d\mu(t) <\infty$.