Hello, I was trying to prove that the Laplace transform is unique and was wondering if anyone could tell me if I've made any errors in my attempt. Here it is:(adsbygoogle = window.adsbygoogle || []).push({});

Suppose L(f) = L(g), where L() denotes the Laplace transform. We want to show that f = g. By linearity of the transform, L(f - g) = 0; so it suffices to show that the only continuous function whose transform is 0 is the 0 function.

O.K., so suppose f is continuous and h(s) = L(f) = 0. By definition,

[tex]L(f) = \int_{0}^{\infty}e^{-st}f(t)\mathrm dt.[/tex]

[tex] = \int_{0}^{\infty}(1 - st + (st)^{2}/2! - (st)^{3}/3! + ...)f(t)\mathrm dt[/tex]

[tex] = \int_{0}^{\infty}f(t)\mathrm dt - s\int_{0}^{\infty}tf(t)\mathrm dt + s^{2}/2! \int_{0}^{\infty}t^{2}f(t) \mathrm dt - s^{3}/3! \int_{0}^{\infty} t^{3}f(t)\mathrm dt + ...[/tex]

Since h(s) = 0, all of h's derivatives at 0 are equal to 0. Thus, the above identity tells us that

[tex](-1)^{n}\int_{0}^{\infty}t^{n}f(t)\mathrm dt = h^{(n)}(0) = 0.[/tex]

This means the moments of f are all 0. But the only continuous function with all 0 moments is the 0 function, so f(t) = 0 for all t >= 0.

Is this correct? Thanks in advance.

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# Uniqueness of Laplace Transform

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