Is My Proof of Laplace Transform Uniqueness Correct?

[mag487]

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:

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,

$$L(f) = \int_{0}^{\infty}e^{-st}f(t)\mathrm dt.$$
$$= \int_{0}^{\infty}(1 - st + (st)^{2}/2! - (st)^{3}/3! + ...)f(t)\mathrm dt$$
$$= \int_{0}^{\infty}f(t)\mathrm dt - s\int_{0}^{\infty}tf(t)\mathrm dt + s^{2}/2! \int_{0}^{\infty}t^{2}f(t) \mathrm dt - s^{3}/3! \int_{0}^{\infty} t^{3}f(t)\mathrm dt + ...$$

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

$$(-1)^{n}\int_{0}^{\infty}t^{n}f(t)\mathrm dt = h^{(n)}(0) = 0.$$

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.

 

[Mute]

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.

Not true. The function $f(t) = 0, t=0; \exp[-1/t^2]~\mbox{otherwise}$ is smooth (and hence continuous), but has moments about t = 0 all equal to zero.

Of course, I'm not sure if that function qualifies as being of "exponential type", so this particular example may not matter, but it serves to show your claim that "the only function with zero moments is 0 is false", so you need to somehow prove that is true (if it is) for functions of exponential type in order for your proof to hold.

 

[mag487]

Not true. The function $f(t) = 0, t=0; \exp[-1/t^2]~\mbox{otherwise}$ is smooth (and hence continuous), but has moments about t = 0 all equal to zero.

Of course, I don't think that function qualifies as being of "exponential type", so this particular example may not matter, but it serves to show your claim that "the only function with zero moments is 0 is false", so you need to somehow prove that is true (if it is) for functions of exponential type in order for your proof to hold.

Thanks, but I'm not sure I understand. I'm looking at the integral from 0 to infinity of x^n*f(x). In this case, it would be $$\int_{0}^{\infty}x^{n}e^{\frac{-1}{x^{2}}} \mathrm dx$$, which is infinite for each n rather than 0.

I realize now that I may have been misusing the term "moment;" I was only intending to refer to this integral.

 

[Mute]

Thanks, but I'm not sure I understand. I'm looking at the integral from 0 to infinity of x^n*f(x). In this case, it would be $$\int_{0}^{\infty}x^{n}e^{\frac{-1}{x^{2}}} \mathrm dx$$, which is infinite for each n rather than 0.

I realize now that I may have been misusing the term "moment;" I was only intending to refer to this integral.

I think I muddled my point by discussing whether or not the function I quoted had a laplace transform.

The point I was trying to convey was that there exists a non-zero function for which every term in its taylor series is zero. Hence, zero is not the only function whose Taylor series is all zeros, so your proof does not hold. You would need to first prove that any function with a laplace transform does not yield one of these "non-analytic smooth functions" whose taylor series converges to zero at every point.

This may not be the only issue with the proof. I haven't given much thought to whether or not there are problems with your interchange of summation and integration (requires uniform convergence, etc).

 

[mag487]

The point I was trying to convey was that there exists a non-zero function for which every term in its taylor series is zero. Hence, zero is not the only function whose Taylor series is all zeros,

Right, but I didn't entirely understand how this pertained to the question of whether $$\int_{0}^{\infty}x^{n}f(x)\mathrm dx = 0$$ for each integer n >= 0 implies that f(x) = 0 for x nonnegative (which is what my attempted proof used, perhaps incorrectly referring to those integrals as "moments").

This may not be the only issue with the proof. I haven't given much thought to whether or not there are problems with your interchange of summation and integration (requires uniform convergence, etc).

Yeah, I'll have to think about that. Thank you.

 

[Mute]

Right, but I didn't entirely understand how this pertained to the question of whether $$\int_{0}^{\infty}x^{n}f(x)\mathrm dx = 0$$ for each integer n >= 0 implies that f(x) = 0 for x nonnegative (which is what my attempted proof used, perhaps incorrectly referring to those integrals as "moments").

Ah, I get what you're saying now. I was thinking in terms of you constructing the Taylor series expansion of h(s), finding all terms were zero and concluding that h(s) was zero because its taylor series expansion was zero, which is why I brought up the smooth nonanalytic functions. You're thinking of the reserve way, though - you know h(s) is zero and want to argue that this implies all the moments are zero.

In terms of saying the integrals

$$\int_0^{\infty}dt~t^n (f(t) - g(t)) = 0$$

(I put in both functions here: you switching notations to just f(t) may have been what confused me) for all n implies f(t) - g(t) = 0, I'm still not sure you can argue that.

For instance, what if

$$\int_0^{\infty}dt~t^n (f(t) - g(t)) = 0$$

but only for n >= 2, and it happens to be the case that

$$\int_0^{\infty}dt~(f(t) - g(t)) = -\int_0^{\infty}dt~t(f(t) - g(t))?$$

In this case h(s) is still zero, but not all of the 'moments' are zero.