Convolution of Measurable Function and Measure is Bilinear
Jump to navigation
Jump to search
Theorem
Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.
Let $f, f': \R^n \to \R$ be $\BB^n$-measurable functions.
Then for all $\lambda \in \R$:
- $\paren {\lambda f + f'} * \mu = \lambda \paren {f * \mu} + f' * \mu$
- $f * \paren {\lambda \mu + \nu} = \lambda \paren {f * \mu} + f * \nu$
provided the convolutions in these expressions exist.
That is, convolution $*$ is a bilinear operation.
Proof
This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $14.5 \ \text{(i)}$