"Proving" the definition of the Laplace Transform from two properties

Question

Suppose $\mathbf{G}$ is the class of piecewise continuous functions of exponential order from $[0,\infty)$ to $\mathbb{R}$ and that $\mathcal{L}$ is a transformation defined on elements of $\mathbf{G}$ satisfying the two properties

For $f,\,g\in\mathbf{G}$ and $c_1,\,c_2\in\mathbb{R}, \mathcal{L}\{c_1f+c_2g\}=c_1\mathcal{L}\{f\}+c_2\mathcal{L}\{g\}$

$\mathcal{L}\{f^\prime(t)\}=s\mathcal{L}\{f(t)\}-f(0)$

Certain elementary properties of Laplace transforms follow easily from these two.

For example:

$\mathcal{L}\{0\}$ follows from property ($1$).

Let $f(t)=1$ for $t\ge0$. Then $\mathcal{L}\{1^\prime\}=s\mathcal{L}\{1\}-1$ therefore $$\mathcal{L}\{1\}=\dfrac{1}{s}\tag{1}$$

Using this as a basis step and using property ($2$) to prove that for $n\ge0$ $$ \mathcal{L}\{t^{n+1}\}=\dfrac{s}{n+1}\mathcal{L}\{t^n\}$$ one may establish the inductive step showing that $$\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}\tag{2}$$

Also, since property ($2$) gives $\mathcal{L}\{\left(e^{at}\right)^\prime\}=s\mathcal{L}\{e^{at}\}-e^0$ we immediately get

$$ \mathcal{L}\{e^{at}\}=\dfrac{1}{s-a}\tag{3}$$

From $\mathcal{L}\{(\cos t)^\prime\}=s\mathcal{L}\{\cos t\}-1$ we obtain

$$ \mathcal{L}\{\sin t\}+s\mathcal{L}\{\cos t\}=1$$

and from $\mathcal{L}\{(\sin t)^\prime\}=s\mathcal{L}\{\sin t\}-0$ we obtain

$$ s\mathcal{L}\{\sin t\}-\mathcal{L}\{\cos t\}=0$$

And the solution of these two gives

$$ \mathcal{L}\{\sin at\}=\dfrac{a}{s^2+a^2}\tag{4}$$

$$ \mathcal{L}\{\cos at\}=\dfrac{s}{s^2+a^2}\tag{5}$$

When I first adopted this approach decades ago in a differential equations class I believed I also had a proof that

$$ \mathcal{L}\{f(t)\}=\int_0^\infty e^{-st}f(t)\,dt$$

but I cannot recall it so I may have been mistaken.

Can anyone prove this usual definition of the Laplace transform from the two properties at the top of this post?

Answer 1

Here's a Heaviside-like heuristic proof, which makes some assumptions about smoothness and continuity (it's probably possible to tighten this up into a genuine proof, but this is a first approximation).

Suppose $f$ is an analytic function in $\mathbf{G}$, with a power series that converges everywhere, and $\mathcal{L}[f](s)$ finite in a neighbourhood of $s=0$ (there are plenty of these, such as sufficiently negative exponentials). Then, by expanding in a power series about the origin, we find $$ \mathcal{L}[xf](s) = -\mathcal{L}[f]'(s), $$ and further, induction gives $$ \mathcal{L}\left[ x^n f\right](s) = \left(-D\right)^n \mathcal{L}[f](s), $$ where $D=d/ds$.

Now, the important thing to know next is that for smooth functions, $$ e^{hD} F(s) = \sum_{n=0}^{\infty} \frac{h^n }{n!} D^n F(s) = \sum_{n=0}^{\infty} \frac{h^n }{n!} F^{(n)}(s) = f(s+h). $$

Then by linearity, $$ \mathcal{L}\left[ e^{-ax} f\right](s) = \mathcal{L}\left[ \sum_{n=0}^{\infty} (-a)^n\frac{x^n}{n!} f\right](s) = \sum_{n=0}^{\infty} \frac{(-a)^n}{n!} \mathcal{L}\left[x^n f\right](s) = \sum_{n=0}^{\infty} \frac{(-a)^n}{n!}(-D)^n \mathcal{L}[f](s) = e^{aD} \mathcal{L}[f](s) = \mathcal{L}[f](s+a) $$

It remains to determine $\mathcal{L}[f]$ at one point; the way we set things up, this may as well be $s=0$. The derivative property gives, if $\mathcal{L}[f](0)$ and $\mathcal{L}[f'](0)$ exist, $$ \mathcal{L}[f'](0) = -f(0). $$ There is, of course, another well-known linear operator that does this: $\int_0^{\infty} f'(x) \, dx = -f(0)$. It follows that where both operators are defined, they are equal, viz. $$ \mathcal{L}[f](s) = \int_0^{\infty} e^{-sx} f(x) \, dx, $$ although again, one needs to be careful about convergence and so on.