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Let $M$ be the positive borel measures on a hausdorff topological space $X$, which are finite on compacts sets $--$ i.e. the real cone of radon measures.

I am given a definition of a derivative of two radon measures, say $\mu, \eta \in M$, as follows.

Let

$D^+_\mu \eta (x) := \left\{ \begin{array}{1 1} \limsup \limits_{r \rightarrow 0} \frac{ \eta(\overline{B_r(0)}) }{ \mu(\overline{B_r(0)}) } & \mu(\overline{B_r(0)}) > 0 \forall r > 0\\ \infty & \mu(\overline{B_r(0)}) = 0 \forall r > 0 \end{array} \right. $

and

$D^-_\mu \eta (x) := \left\{ \begin{array}{1 1} \liminf \limits_{r \rightarrow 0} \frac{ \eta(\overline{B_r(0)}) }{ \mu(\overline{B_r(0)}) } & \mu(\overline{B_r(0)}) > 0 \forall r > 0\\ \infty & \mu(\overline{B_r(0)}) = 0 \forall r > 0 \end{array} \right. $

and we write in case of equality

$D_\mu \eta (x) := D^+_\mu \eta (x) = D^-_\mu \eta (x)$

I could accept this ad-hoc definition "as is". Nevertheless, I know there is a canonical topology on the linear space of bounded radon measures, so is probably on the cone of positive (not necessarily bounded) radon measures (Furthermore, I guess the thoughs pass over the linear space of signed Radon measures)

Hence, I would like to obtain a good intuition how the definition of derivative of above relates to the canonical definition (if applicable) of derivative on infinite-dimensional vector spaces. (I suppose there is such a connection). Can give a explanation or tell me a good tight resource to look this up?

Thank you.

  • 0
    What do you mean by *derivative on infinite-dimensional vector space*?2011-01-14
  • 0
    Compare the Frechet-derivative or Gateaux-derivative. Of course, the question of convergence of the difference quotient is more subtle in the infinite-dimensional case.2011-01-14
  • 0
    There's something funky about your definition. What if $\mu(B_r(0)) = 0$ for all $0 < r < r_0$? Neither case of your definition apply.2011-01-14
  • 0
    Of course, the for all quantifier in that case is to be regarded as "for all $r$ sufficiently small$. As the measure is positive, there is either a $\mu$-negligible ball or not.2011-01-16

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