ThmDex – An index of mathematical definitions, results, and conjectures.
Complex function convolution is homogeneous to degree one
Formulation 0
Let $M = (\mathbb{R}^D, \mathcal{L}, \mu)$ be a D1744: Lebesgue measure space such that
(i) $f, g : \mathbb{R}^D \to \mathbb{C}$ are each an D1921: Absolutely integrable function on $M$
Let $\lambda \in \mathbb{C}$ be a D1207: Complex number.
Then \begin{equation} (\lambda f) * g = \lambda (f * g) \end{equation}
Proofs
Proof 0
Let $M = (\mathbb{R}^D, \mathcal{L}, \mu)$ be a D1744: Lebesgue measure space such that
(i) $f, g : \mathbb{R}^D \to \mathbb{C}$ are each an D1921: Absolutely integrable function on $M$
Let $\lambda \in \mathbb{C}$ be a D1207: Complex number.
If $x \in \mathbb{R}^D$, then applying R1502: Complex-linearity of complex integral yields \begin{equation} \begin{split} ((\lambda f) * g) (x) & = \int_{\mathbb{R}^D} \lambda f(y) g(x - y) \, \mu(d y) \\ & = \lambda \int_{\mathbb{R}^D} f(y) g(x - y) \, \mu(d y) = \lambda (f * g)(x) \end{split} \end{equation} Since $x \in \mathbb{R}^D$ was arbitrary, this is true for all $x \in \mathbb{R}^D$ and the proof is complete. $\square$