Young's inequality for integral operators
In mathematical analysis, the Young's inequality for integral operators, is a bound on the operator norm of an integral operator in terms of norms of the kernel itself.
Statement
Assume that and are measurable spaces, is measurable and are such that . If
- for all
and
- for all
then [1]
Particular cases
Convolution kernel
If and , then the inequality becomes Young's convolution inequality.
gollark: Probably not people who violate ALL rules, but ones who violate *some subset* of them in interesting ways.
gollark: If you go out of your way to do exactly the opposite of what "rules" say, they have as much control over you as they do on someone who does exactly what the rules *do* say.
gollark: I'm glad you're making sure to violate norms in socially approved ways which signify you as "out there" or something.
gollark: > if you can convince them that their suffering benefits other people, then they'll happily submit to itI am not convinced that this is actually true of people, given any instance of "selfishness" etc. ever.
gollark: Yes, you can only make something optimize effectively for good if you can define what that is rigorously, and people haven't yet and wouldn't agree on it.
See also
Notes
- Theorem 0.3.1 in: C. D. Sogge, Fourier integral in classical analysis, Cambridge University Press, 1993. ISBN 0-521-43464-5
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