Convolution of Measures is Bilinear
Jump to navigation
Jump to search
Theorem
Let $\mu, \mu', \nu$ and $\nu'$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.
Then for all $\lambda \in \R$:
- $\paren {\lambda \mu + \mu'} * \nu = \lambda \paren {\mu * \nu} + \mu' * \nu$
- $\mu * \paren {\lambda \nu + \nu'} = \lambda \paren {\mu * \nu} + \mu * \nu'$
where $*$ denotes convolution.
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): $14.5 \ \text{(i)}$