In Axler's Linear Algebra Done Right we have the theorem

6.42: (Riesz Representation Theorem) Suppose $V$ is a finite dimensional inner product space and $\phi$ is a linear functional on $V$. Then there is a unique vector $u \in V$ such that $$\phi(v) = \langle v, u\rangle$$ for every $v \in V$

I'm currently learning measure theory and have came across Radon-Nikodym

(Radon-Nikodym) Consider a measurable space $(X,\mathcal{M})$ on which two $\sigma$-finite signed measures $\mu,\nu$ are defined such that $\nu << \mu$ ($\nu$ is absolutely continuous with respect to $\mu$) then there is a $\mu$-integrable function $f: X \to \mathbb{R}$ such that $$\nu(E) = \int_E f d\mu$$ for every $E \in \mathcal{M}$ and any other function $g$ satisfying this is equal to $f$ almost everywhere with respect to $\mu$.

These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $f$ "acts" like the vector $u$ in Riesz, and the signed measure $\nu$ acts like the linear functional $\phi$.

I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $\sigma$-algebra, $\mathcal{M}$ on $V$ so that $(V, \mathcal{M})$ is a measurable space. This has to be a very particular $\sigma$-algebra so that somehow the integral can be reduced to the inner product on $V$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.

So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?

  • $\begingroup$ if $\phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $\lambda(E)=\phi(\chi_E)$, if $\phi$ is positive we can (generalize after) show that $\lambda$ is a measure and if the set $E$ has measure $\mu(E)=0$ than $\lambda(E)=0$ because $\phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $\phi(f)=\int fg d\mu$ for some $g\in L^q$ $\endgroup$ – Robson Nov 18 '18 at 22:20

I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem (for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).

  • $\begingroup$ I think this is a good answer though. $\endgroup$ – user370967 Jan 30 '19 at 15:52
  • $\begingroup$ Although not the exact thing I was looking for, this is a very good answer. Thank you. $\endgroup$ – carsandpulsars Feb 4 '19 at 4:10
  • $\begingroup$ Any idea what the original von Neumann reference for this proof is? $\endgroup$ – aduh Jan 11 at 22:05
  • $\begingroup$ @aduh I think it might be: Neumann, J.V., 1940. On rings of operators. III. Annals of Mathematics, pp.94-161. <doi.org/10.2307/1968823>. But paywalled, so I'm not sure $\endgroup$ – UserNamesAreHard Jan 14 at 14:12

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