Is the weak derivative a Radon-Nikodym derivative?

Question

It seems reasonable to me to conjecture that the weak derivative is a special case of Radon-Nikodym derivative between measures. Recall that (missing some technical details):

• Weak derivative: given a function $f:I=(a,b) \rightarrow \mathbb{R}$, $f'$ is called its weak derivative if for every $\varphi \in C^\infty_c(I)$ it holds that $ \int_I f'\varphi dx = - \int_I f \varphi' dx $
• Radon-Nikodym derivative: given two measures $\mu, \nu$ (on the same $\sigma-$algebra), if $\nu \ll \mu$ ($\nu$ absoultely continuous w.r.t $\mu$), then there exists a unique $f \in L^1(\mu)$ such that $ \nu(E) = \int_E f d\mu $. Such $f$ is called the Radon-Nikodym derivative of $\nu$ w.r.t $\mu$: $f = \frac{d\nu}{d\mu}$.
• Lebesgue-Stieltjes measure: given a right-continuous, monotone-increasing function $f$, we can define a measure (called the Lebesgue-Stieltjes measure of $f$) $LS_f((a,b]) = f(b) - f(a)$ (and then we have to extend it to all measurable sets as a Radon measure).

My question is: does something like the following hold true?

given $f \in W^{1,1}(I)$ ($f$ Lebesgue integrable, with weak derivative Lebesgue integrable), its weak derivative $f'$ coincides with the Radon-Nikodym derivative of the Lebesgue-Stieltjes measure of $f$ w.r.t. the Lebesgue measure:

$$f' = \frac{d (LS_f)_{ac}}{dx}$$

where $\nu_{ac}$ indicates the absolutely continuous part of $\nu$ w.r.t. the Lebesgue measure.

This seems to me to be true from the Fundamental Theorem of Calculus for $W^{1,1}$, but I have never seen it stated like that.

I would say it looks interesting as a formulation because it would regard the FTC in a more geeral setting.

Answer 1

There is indeed a connection there, but your require a bit more regularity. You actually need $f\in W^{1,\infty}_{loc}(\mathbb{R})$, i.e. $f$ has to be locally Lipschitz. Then by Rademachers Theorem (see e.g. https://en.wikipedia.org/wiki/Rademacher%27s_theorem) $f$ has a classical derivative almost everwhere. This and monotonicity of $f$ is enough to show, that $LS_f$ is a Radon measure and absolutely continuous w.r.t. Lebesgue measure $\mathcal{L}^1$. Since we are working in $\mathbb{R}$ we can apply the differentiation theorem for Radon measures, see e.g. Leon Simons book about geometric measure theory. It states, that the Radon-Nikodym derivative can be calculated as follows: $$\frac{ d(LS_f)}{d\mathcal{L}^1}(x) = \lim_{r\rightarrow 0} \frac{LS_f(B_r(x))}{\mathcal{L}^1(B_r(x))} = \lim_{r\rightarrow 0}\frac{f(x+r)-f(x-r)}{2r} = f'(x)\ \mathcal{L}^1\mbox{-a.e.}$$ The last equality is by Rademachers Theorem.

EDIT: I did some research and I do not think the added regularity is really needed. For this you have to show, that $f\in W^{1,1}$ is enough for $f$ to be absolutely continuous, see e.g. https://www.math.ucdavis.edu/~hunter/m218a_09/ch3A.pdf Theorem 3.57

This is indeed true, because $f'\in L^1$ and we therefore have $$\forall \varepsilon > 0\ \exists \delta>0:\ \mbox{for all measurable }A\subseteq\mathbb{R}\mbox{ with }\mathcal{L}^1(A)<\delta$$ we have $\int_A|f'|\, dx < \varepsilon$. This is true by e.g. Vitalis convergence theorem. So let us check absolute continuity for $f$: We have to check the following: For all $\varepsilon>0$ exists a $\delta>0$ such that for all $-\infty < a_1 < b_1 < a_2 < \ldots< a_n< b_n< \infty$ with $$\sum_{k=1}^n b_k-a_k < \delta,\mbox{ we have }\sum_{k=1}^nf(b_k) -f(a_k) < \varepsilon.$$ Hence we choose $\delta$ as above for the integral estimate. Then $$\mathcal{L}^1(\bigcup_{k=1}^n [a_k,b_k]) = \sum_{k=1}^n b_k-a_k < \delta$$ and therefore $$\sum_{k=1}^n f(b_k)-f(a_k) = \sum_{k=1}^n\int_{a_k}^{b_k}f'(x)\, dx \leq \int_{\bigcup_{k=1}^n[a_k,b_k]}|f'(x)|\, dx < \varepsilon.$$