Convolution of Measurable Functions is Bilinear
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Theorem
Let $\BB^n$ be the Borel $\sigma$-algebra on $\R^n$.
Let $f, f', g, g': \R^n \to \R$ be $\BB^n$-measurable functions.
Then for all $\lambda \in \R$:
- $\paren {\lambda f + f'} * g = \lambda \paren {f * g} + f' * g$
- $f * \paren {\lambda g + g'} = \lambda \paren {f * g} + f * g'$
provided the convolutions in these expressions exist.
That is, convolution $*$ is a bilinear operation.
Proof
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Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $14.5 \ \text{(i)}$