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For folks wondering about applications of this theorem: it is a key building block in the theory of reproducing kernel Hilbert spaces (RKHS), which in turn are
by ajtulloch 5y ago
For folks wondering about applications of this theorem: it is a key building block in the theory of reproducing kernel Hilbert spaces (RKHS), which in turn are the building block of kernel support vector machines (kernel SVMs), which are widely used in machine learning applications.
The "kernel trick" from kernel SVMs only works because of the existence and uniqueness result from the RRT on the underlying Hilbert space.
- hilber_traum 5y agoThere are several results called the Riesz representation theorem. The article is about representing continuous linear functionals on a space of continuous functions as signed measures (or Riemann-Stieltjes integrals). This has lots of applications in ergodic theory or representation theory (e.g. disintegration of measures). This result is essentially unrelated to the result characterizing continuous linear functionals on Hilbert spaces. It is also much more difficult to prove (the result on Hilbert spaces is rather simple).
- contravariant 5y agoIf 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.