Convolution Theorem/Proof 1

Proof
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:


 * ConvolutionTheorem1.png

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:


 * ConvolutionTheorem2.png

Thus:

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

Thus the 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:


 * ConvolutionTheorem3.png

but is zero over the lighter shaded portion.

Now we can write $(3)$ as:

Hence the result.