---
tag: 3N8DRR7X92
kind: definition
name: Ill-posedness of real Laplace inversion
msc: [44A10, 47A52, 65R30]
defines: TRTN840ZB0
uses: [FZ64956KX2, 05XT5AVJBK, DWJF7ABPR1, V3YTF5WJHY]
sources:
  - url: https://en.wikipedia.org/wiki/Well-posed_problem
---

## Definition

Let $L^2(0, \infty)$ be the space of square-[[integrable|DWJF7ABPR1]] functions on $(0, \infty)$ with the norm $\norm{f}_2 = \left( \int_0^\infty \abs{f(t)}^2 \, dt \right)^{1/2}$.
For $f \in L^2(0, \infty)$, its [[Laplace transform|05XT5AVJBK]] $\mathcal{L} f(s) = \int_0^\infty f(t) e^{-s t} \, dt$ is defined for real $s > 0$, and $\mathcal{L}$ is a bounded linear operator from $L^2(0, \infty)$ to $L^2(0, \infty)$.
The *real Laplace inversion problem* is: given $F \in L^2(0, \infty)$, find $f \in L^2(0, \infty)$ with $\mathcal{L} f = F$ on $(0, \infty)$.
*Ill-posedness* of this problem means that it is not [[well posed|FZ64956KX2]] in the sense of Hadamard.
Here the solution is unique when it exists, but the inverse map $\mathcal{L} f \mapsto f$ is not continuous.
That is, there are $f_n$ with $\norm{f_n}_2$ bounded away from $0$ and $\norm{\mathcal{L} f_n}_2 \to 0$.

## Remarks

Uniqueness holds by [[Lerch's uniqueness theorem|V3YTF5WJHY]].
Instability, with the check: let $f_n(t) = e^{-t} \sin(n t)$.
Then $\norm{f_n}_2^2 = \int_0^\infty e^{-2t} \sin^2(n t) \, dt = \frac{n^2}{4(1 + n^2)} \ge \frac{1}{8}$ for $n \ge 1$.
Its transform is $\mathcal{L} f_n(s) = \frac{n}{(s + 1)^2 + n^2}$, and $\norm{\mathcal{L} f_n}_2^2 \le \int_{-\infty}^{\infty} \frac{n^2}{(u^2 + n^2)^2} \, du = \frac{\pi}{2 n} \to 0$.
So no estimate $\norm{f}_2 \le C \norm{\mathcal{L} f}_2$ can hold.
Existence also fails for most data: the range of $\mathcal{L}$ consists of restrictions of holomorphic functions, so a generic $F \in L^2(0, \infty)$ is not a transform at all.
Contrast: inversion from values of $F$ on a vertical line in the complex plane, by the Bromwich integral, is a Fourier inversion and is stable in $L^2$.