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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.


$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. $


$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.

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What do you mean by derivative on infinite-dimensional vector space? – AD. Jan 14 '11 at 5:46
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. – shuhalo Jan 14 '11 at 20:21
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. – Willie Wong Jan 14 '11 at 20:50
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. – shuhalo Jan 16 '11 at 1:41

I think there is some misunderstanding here. While it is possible to have derivatives of functions defined between Banach spaces (say) (e.g. with the concept of Frechet- or Gateaux-derivative), the derivative of a Radon measure is a derivative of an object in such a space. Hence, the derivative of a Radon measure is somewhat more related to the distributional derivative of a function or the weak derivative.

To get an impression of the derivative you are talking about, the following example may be helpful: Let $\mu$ be the one dimensional Lebesgue measure on $[0,1]$. For a integrable and positive function $f:[0,1] \to \mathbf{R}$ and define a measure $\nu$ by $\nu(A) = \int_A f\,d\mu$. Now, $D_\mu \eta$ can be calculated explicitely and is related to the weak derivative of $f$. (Actually I do not know what precisely is meant by $B_r^-(0)$ and hence, I could not work out the details.)

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$\bar{B_r(x)}$ denotes just the closure \bar{B_r(x)}. Corrected, didn't know how to do until now. – shuhalo Jan 14 '11 at 20:40
@Martin this is a little late, but if you happen to read this. Try \overline{B_r(x)} $\Bigl(\,\overline{B_r(x)}\,\Bigr)$ for the closure of the $r$-ball around $x$ and \overline{B}_r(x) $\Bigl(\,\overline B_r(x)\,\Bigr)$ for the closed $r$-ball around $x$. If you're writing a $\mathrm{\LaTeX}$ paper, you can even use the \widebar suggested on TeX.SX:… – kahen Nov 25 '12 at 17:25

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