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Intuitively how can the Laplace transform be injective? You are taking an integral with limits $0$ and $\infty$. So you don't care about the function before $0$. Define $g(x)=x^2$ for $x>0$ and $g(x)=0$ for $x \leq 0$ Define $f(x)=x^2$ for $x>0$ and $f(x)=-x$ for $x \leq 0$

$f$ and $g$ must have the same Laplace transform by definition right? Yet they are not equal. Can someone please explain?

Arcane1729
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    You are certainly correct - uniqueness refers to a class of functions with domain the range of integration defining the Laplace transform. For a quick argument that seems right, see http://www.ctr.maths.lu.se/media/MATC12/2013ht2013/uniqueness.pdf – peter a g Feb 13 '16 at 04:36

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  • $\displaystyle \int_0^\infty f(t) e^{-st} dt$ should be thought as the Laplace transform of a function defined only on $t \in [0;\infty[$ or as the bilateral Laplace transform of a function $f(t)$ being $0$ for $t<0$

  • the Laplace transform $$\int_a^\infty f(t) e^{-st} dt$$ is injective (in the almost everywhere sense) on functions defined on $[a;\infty[$ for which it converges when $Re(s)\to \infty$

    (often $a=0$ but depending on the function it can be another value, and it can even be $-\infty$ but in that case it becomes the bilateral Laplace transform)

  • the bilateral Laplace transform is injective on functions defined on $\mathbb{R}$ for which it converges when $Re(s) = c$ (for a fixed $c$) and there it coincides with the Fourier transform of $f(t) e^{-ct}$ .

  • note that $-\int_{-\infty}^0 e^{-st} dt = 1/s$ for $Re(s) < 0$ and $\int_{0}^\infty e^{-st} dt = 1/s$ for $Re(s) > 0$ :

    $\implies$ many functions have different inverse Laplace transforms, depending on the assumed domain of convergence.

so it is not injective anymore without the convergence assumption, this is why the domain of convergence is so important.

reuns
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