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If you can represent a continuous linear functional as the inner product with the Riesz representative then doesn't this also define a signed measure? It kind o
by contravariant 5y ago
If you can represent a continuous linear functional as the inner product with the Riesz representative then doesn't this also define a signed measure? It kind of seems like one of those theorems should imply the other to me, or is there some subtle aspect I'm missing?
- turminal 5y agoIt's the same thing really. It's just that Riesz first proved it for the special case and it was then generalized to Hilbert spaces. It's such a huge generalization that it causes a lot of confusion.
- ogogmad 5y agoC([0,1]) is not a Hilbert space but a Banach space. Every Hilbert space is a Banach space, but not vice versa. The version of the theorem for Hilbert spaces is indeed a lot easier to prove than the one given in the article.
- turminal 5y agoAnd that just proves my point about confusion :) Thanks for the correction.