Theorem: Convolution Theorem

Let $\GF \in \set {\R, \C}$. 
Let $f: \R \to \GF$ and $g: \R \to \GF$ be functions.
Let their Laplace transforms $\laptrans {\map f t} = \map F s$ and $\laptrans {\map g t} = \map G s$ exist.
Then:
:$\map F s \map G s = \ds \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\int_0^t \map f u \map g {t - u} \rd u}
      | r = \int_{t \mathop = 0}^\infty e^{-s t} \paren {\int_{u \mathop = 0}^t \map f u \map g {t - u} \rd u} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \int_{t \mathop = 0}^\infty \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t
      | c = 
}}
{{eqn | r = \lim_{M \mathop \to \infty} \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t
      | c = 
}} 
{{eqn | n = 1
      | r = \lim_{M \mathop \to \infty} s_M
      | c = 
}} 
{{end-eqn}}
where $s_M$ is defined to be:
:$\ds \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$
The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below:
:400px
Setting $t - u = v$, that is $t = u + v$, the shaded region above is transformed into the region $\mathscr R_{u v}$ the $u v$ plane:
:400px
Thus:
{{begin-eqn}}
{{eqn | l = s_M
      | r = \iint_{\mathscr R_{t u} } e^{-s t} \map f u \map g {t - u} \rd u \rd t
      | c = from $(1)$ above
}}
{{eqn | n = 2
      | r = \iint_{\mathscr R_{u v} } e^{-s \paren {u + v} } \map f u \map g v \size {\dfrac {\map \partial {u, t} } {\map \partial {u, v} } } \rd u \rd v
      | c = Change of Variables Theorem (Multivariable Calculus)
}}
{{end-eqn}}
where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation:
{{begin-eqn}}
{{eqn | l = J
      | r = \dfrac {\map \partial {u, t} } {\map \partial {u, v} }
      | c = 
}}
{{eqn | r = \begin{vmatrix} \dfrac {\partial u} {\partial u} & \dfrac {\partial u} {\partial v} \\ \dfrac {\partial t} {\partial u} & \dfrac {\partial t} {\partial v} \end{vmatrix}
      | c = 
}}
{{eqn | r = \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}
      | c = 
}} 
{{end-eqn}}
Thus the {{RHS}} of $(2)$ is:
:$(3): \quad \ds s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$
Let $\map K {u, v}$ be the function defined as:
:$\map K {u, v} = \begin{cases} e^{-s \paren {u + v} } \map f u \map g v & : u + v \le M \\ 0 & : u + v > M \end{cases}$
This function is defined over the square region in the diagram below:
:400px
but is zero over the lighter shaded portion.
Now we can write $(3)$ as:
{{begin-eqn}}
{{eqn | l = s_M
      | r = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^M \map K {u, v} \rd u \rd v
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{M \mathop \to \infty} s_M
      | r = \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty \map K {u, v} \rd u \rd v
      | c = 
}}
{{eqn | r = \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v
      | c = 
}}
{{eqn | r = \paren {\int_{v \mathop = 0}^\infty e^{-s u} \map f u \rd u} \paren {\int_{v \mathop = 0}^\infty e^{-s v} \map g u \rd v}
      | c = 
}}
{{eqn | r = \map F s \map G s
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}