Scaling Property of Dirac Delta Function

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Theorem

Let $\map \delta t$ be the Dirac delta function.

Let $a$ be a non zero constant real number.

Then:

$\map \delta {a t} = \dfrac {\map \delta t} {\size a}$


Proof

The equation can be rearranged as:

$\size a \map \delta {a t} = \map \delta t$

We will check the definition of Dirac delta function in turn.


Definition of Dirac delta function:

$\paren 1:\map \delta t = \begin{cases}

+\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}$

$\paren 2:\ds \int_{-\infty}^{+\infty} \map \delta t \rd t = 1$


$\paren 1:$

\(\ds \size a \map \delta {a t}\) \(=\) \(\ds \begin{cases} \paren {\size a} \paren {+\infty} & : a t = 0 \\

\paren {\size a} 0 & : \text{otherwise} \end{cases}\)

Definition of Dirac Delta Function
\(\ds \leadstoandfrom \ \ \) \(\ds \size a \map \delta {a t}\) \(=\) \(\ds \begin{cases}

+\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}\)

simplifying


$\paren 2:$

The proof of this part will be split into two parts, one for positive $a$ and one for negative $a$.


For $a > 0$:

\(\ds \int_{-\infty}^{+\infty} \size a \map \delta {a t} \rd t\) \(=\) \(\ds \int_{-\infty}^{+\infty} \size a \map \delta t \dfrac {\rd t} a\) Substitute $t \mapsto \dfrac t a$
\(\ds \) \(=\) \(\ds \dfrac {\size a} a \int_{-\infty}^{+\infty} \map \delta t \rd t\) Simplifying
\(\ds \) \(=\) \(\ds \dfrac a a \int_{-\infty}^{+\infty} \map \delta t \rd t\) $a > 0$
\(\ds \) \(=\) \(\ds 1\) Definition of Dirac Delta Function


$\Box$


For $a < 0$:

\(\ds \int_{-\infty}^{+\infty} \size a \map \delta {a t} \rd t\) \(=\) \(\ds \int_{+\infty}^{-\infty} \size a \map \delta t \dfrac {\rd t} a\) Substitute $t \mapsto \dfrac t a$
\(\ds \) \(=\) \(\ds \dfrac {\size a} a \int_{+\infty}^{-\infty} \map \delta t \rd t\) Simplifying
\(\ds \) \(=\) \(\ds \dfrac {-\size a} a \int_{-\infty}^{+\infty} \map \delta t \rd t\) Reversal of Limits of Definite Integral
\(\ds \) \(=\) \(\ds \dfrac a a \int_{-\infty}^{+\infty} \map \delta t \rd t\) $a < 0$
\(\ds \) \(=\) \(\ds 1\) Definition of Dirac Delta Function

$\Box$


Therefore, by definition, $\size a \map \delta {a t} = \map \delta t$.

The result follows after rearrangement.


$\blacksquare$


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