# Convolution Theorem/Proof 1

## Theorem

Let $\mathbb F \in \set {\R, \C}$.

Let $f: \R \to \F$ and $g: \R \to \F$ be functions.

Let their Laplace transforms $\laptrans {\map f t} = \map F s$ and $\laptrans {\map g t} = \map G s$ exist.

Then:

$\map F s \map G s = \displaystyle \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$

## Proof

 $\displaystyle \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$ $=$ $\displaystyle \int_{t \mathop = 0}^\infty e^{-s t} \paren {\int_{u \mathop = 0}^t \map f u \map g {t - u} \rd u} \rd t$ Definition of Laplace Transform $\displaystyle$ $=$ $\displaystyle \int_{t \mathop = 0}^\infty \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$ $\displaystyle$ $=$ $\displaystyle \lim_{M \mathop \to \infty} \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$ $\text {(1)}: \quad$ $\displaystyle$ $=$ $\displaystyle \lim_{M \mathop \to \infty} s_M$

where $s_M$ is defined to be:

$\displaystyle \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$

The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below: 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: Thus:

 $\displaystyle s_M$ $=$ $\displaystyle \iint_{\mathscr R_{t u} } e^{-s t} \map f u \map g {t - u} \rd u \rd t$ from $(1)$ above $\text {(2)}: \quad$ $\displaystyle$ $=$ $\displaystyle \iint_{\mathscr R_{u v} } e^{-s \paren {u + v} } \map f u \map g v \size {\dfrac {\map \partial {u, t} } {\map \partial {u, v} } } \rd u \rd v$

where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation:

 $\displaystyle J$ $=$ $\displaystyle \dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ $\displaystyle$ $=$ $\displaystyle \begin{vmatrix} \dfrac {\partial u} {\partial u} & \dfrac {\partial u} {\partial v} \\ \dfrac {\partial t} {\partial u} & \dfrac {\partial t} {\partial v} \end{vmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}$

Thus the right hand side of $(2)$ is:

$(3): \quad \displaystyle s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$

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: but is zero over the lighter shaded portion.

Now we can write $(3)$ as:

 $\displaystyle s_M$ $=$ $\displaystyle \int_{v \mathop = 0}^M \int_{u \mathop = 0}^M \map K {u, v} \rd u \rd v$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{M \mathop \to \infty} s_M$ $=$ $\displaystyle \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty \map K {u, v} \rd u \rd v$ $\displaystyle$ $=$ $\displaystyle \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$ $\displaystyle$ $=$ $\displaystyle \paren {\int_{v \mathop = 0}^\infty e^{-s u} \map f u \rd u} \paren {\int_{v \mathop = 0}^\infty e^{-s v} \map g u \rd v}$ $\displaystyle$ $=$ $\displaystyle \map F s \map G s$

Hence the result.

$\blacksquare$