When I can safely assume that a function is a Laplace transform of other function?

Question

If I have a function and I want to represent it as being the Laplace transform of another, that is, I want to be sure that there is $\hat{f}(s)$ such that my function $f(x)$ can be written as:

$$f(x) = \int ds \hat{f}(s) \exp(-sx)$$

what conditions should I impose over $f(x)$?

In other words, what are the conditions for the Fourier–Mellin–Bromwich integral

$$\hat{f}(s) = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} f(x) \exp(sx) dx$$

to exist?

Answer 1

The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\tag{1}\label{1} $$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\tag{2}\label{2}$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\tag{3}\label{3}$$ Now rewriting \eqref{2} as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended by $0$ to $t\leq 0$) belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$ e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\\ \\\ 0, & t\leq 0. \end{cases} $$

One can show also that the Parseval identity

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt$$ holds, so there is a complete analogy with the standard Fourier transform.

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Executive summary. A function $f$ is the Laplace transform of some function $\phi$ satisfying condition \eqref{1}, if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition \eqref{3} holds. This class of functions is known as the Hardy space on a (right) half-plane.

Answer 2

This kind of question is very interesting, and I too would like to know answers.

Sorry to self-publicise; I hope it's not regarded as impolite, but since I have also considered this exact kind of question, it's quickest just to refer to my own paper (and the references I give in there): Laplace Transform Representations and Paley–Wiener Theorems for Functions on Vertical Strips by Zen Harper, Documenta Math. 15 (2010) 235-254.

Given an analytic function on a vertical strip, I try to find conditions which guarantee it can be represented by a bilateral Laplace transform (in various senses). I definitely don't claim to have any kind of complete answer, but it's the best I know of (by definition! If I knew any better answers, I would have written them in my paper!).

It seems like there are still many open questions about this.

Answer 3

In Audrey Terras' "Harmonic Analysis on Symmetric Spaces and Applications, I" she has on p. 21 the following: Suppose $\exp(-cx)f(x)$ lies in $L^1(\mathbb R)$ and $f(x)$ vanishes for negative $x$. Assume also that $f(x)$ is piecewise differentiable. Then

$$(f(x+)+f(x-))/2=\lim_{T\to\infty}\frac{1}{2\pi i}\int_{c-i T}^{c+i T}\exp(sx)\mathcal Lf(s)ds.$$

(NB traditional Laplace transform notation seems to be the reverse of yours; $x>0$ and $s\in\mathbb C$.)

Answer 4

I would suggest the classic book "The Laplace transform" by David Vernon Widder. On its page 310, Theorem 14a states the following: A necessary and sufficient condition that $f(x)$ can be expressed in the form $f(x) = \int_{0}^{\infty} e^{-xt} d \alpha(t)$, where $\alpha(t)$ is non-decreasing and bounded for $t$ in $[0,\infty)$, is that $f(x)$ should be completely monotonic for $x$ in $[0,\infty)$.

For a given $\alpha(t)$ that is non-decreasing and bounded for $t$ in $[0,\infty)$, i.e., $\alpha$ is of bounded variation on $[0,\infty)$, if $f(z) = \int_{0}^{\infty} e^{-zt} d \alpha(t)$ converges for some $z \in \mathbb{C}$, then

1. there is a necessary condition on the order of $\alpha(t)$, e.g., given by Theorem 2.2a on page 39;
2. Theorem 5a on page 57 gives the analyticity of $f(z)$ in a suitable region;
3. if such a convergence holds absolutely on a vertical line in the complex plane $\mathbb{C}$, then Theorem 7.3 on page 66 provides the inversion formula.

Now for the two-sided integral $f(z) = \int_{-\infty}^{\infty} e^{-zt} d \alpha(t)$, just break it into the sum of $\int_{-\infty}^{0} e^{-zt} d \alpha(t)$ and $\int_{0}^{\infty} e^{-zt} d \alpha(t)$, and study both, we can obtain similar conclusions, some of which are directly stated in this book. The key to get these results is to use Dirichlet integrals, Riemann-Lebesgue lemma, and Weierstrass M-test, analytic continuation.

Finally, by a change of variable, we can study Fourier transform, Stieltjes transform, and Mellin transform of functions of bounded variation, using the same techniques. These transforms are also discussed in the book. Functions of bounded variation form a very important class of functions in probability theory, real analysis, functional analysis, and measure theory.

Answer 5

The question has already an accepted answer, but nevertheless I think that the recent result of Buriánková, Edmunds and Pick [1] can give more insight. Precisely, they prove the following result

Theorem 3.8 ([1], §3, p. 463) Assume that $p \in (1, \infty)$ and $q \in [1, \infty]$. Then $$ \mathcal L:L^{p,q} \to L^{p^\prime,q}. $$ Moreover, both the domain space and the target space are the optimal such rearrangement-invariant spaces (hence they form an optimal pair).

Here,

• $L^{p,q}$ is the classical Lorenz space whose quasi norm is defined as $$\DeclareMathOperator{\Dm}{d\!} \begin{split} \|f\|_{L^{p,q}(\mathbf R)} &=p^{\frac{1}{q}}\left(\int_0^\infty t^q \mu\left\{x : |f(x)| \ge t\right\}^{\frac{q}{p}}\,\frac{\Dm t}{t}\right)^{\frac{1}{q}}\\ &= \left(\int_0^\infty \bigl(r \mu_{\mathscr L}\left\{x : |f(x)|^p \ge r \right\}\bigr)^{\frac{q}{p}}\,\frac{\Dm r }{r}\right)^{\frac{1}{q}} \end{split} $$ where $\mu_{\mathscr L}\{A\}$ is the Lebesgue measure of the set $A\subset \mathbf R$.
• $p^\prime$ is defined as $$ p^\prime = \begin{cases} \dfrac{p}{p-1} & p \in (1, \infty)\\ \infty & p =1 \end{cases} $$
• $\mathcal L:L^{p,q} \to L^{p^\prime,q}$ means that the Laplace transform $\mathcal L$ is a bounded operator from $L^{p,q}$ to $L^{p^\prime,q}$

Moreover, in the same page they point out that

Remark 3.9 It follows from Theorem 3.8 that $\mathcal L:L^{p,q} \to L^{p,q}$ if and only if $p = 2$. In particular, the only $L^p$ space which $\mathcal L$ maps boundedly into itself is $L^2$.

The Authors prove their results by using rearrangement of functions and their associated inequalities.

Reference

[1] Eva Buriánková, David E. Edmunds, Luboš Pick, "Optimal function spaces for the Laplace transform" (English), Revista Matemática Complutense 30, No. 3, 451-465 (2017), MR3690749, Zbl 1400.46022.