# Almost everywhere differentiability for a class of functions on $\mathbb{R}^2$

A while ago, I came across the following problem, which I was not able to resolve one way or the other.

Let $f,g\colon\mathbb{R}^2\to\mathbb{R}$ be continuous functions such that $f(t,x)$ and $g(t,x)$ are Lipschitz continuous with respect to $x$ and increasing in $t$. Is it necessarily true that the partial derivative $f_x=\partial f/\partial x$ exists almost everywhere with respect to the measure $\iint\cdot\,d_tg(t,x)dx$?

The notation $\int\cdot\,d_tg(t,x)$ refers to the Lebesgue-Stieltjes integral with respect to $g(t,x)$, for each fixed $x$. At first sight, this looks like a basic real-analysis problem for which it should be possible to either prove true or construct a simple counterexample. However, after playing around with this for a while, I was none the wiser.

The reason for the interest in this question comes from attempts to construct a particular decomposition of stochastic processes. Supposing that $X$ is a continuous martingale, it can be useful to decompose $f(t,X_t)$ into a martingale term and a 'drift' component, \begin{align} f(t,X_t)=\int_0^tf_x(s,X_s)\,dX_s+A_t.&&{\rm(1)} \end{align} For this expression to be well-defined, it is necessary that $f_x$ exists outside of a null set. That is, $\int1_{\{f_x(s,X_s)\rm\ doesn't\ exist\}}dX_s$ should be zero. However, letting $g(t,x)=\mathbb{E}[\vert X_t-x\vert]$, it can be shown that the expected value of the square of this integral is $\iint1_{\{f_x\rm\ doesn't\ exist\}}d_tg(t,x)dx$. So a positive answer to my question would mean that decomposition (1) is well-defined. Furthermore, it can then be shown that $A$ has zero quadratic variation, so that $f(t,X_t)$ is what is sometimes referred to as a Dirichlet process (although this term does also have a completely different meaning). Decomposition (1) was used to prove some results about the distribution of martingales, but I also published it as a stand-alone paper (here). However, not knowing an answer to the question above, this did require imposing the annoying additional condition that the left and right hand derivatives of $f(t,x)$ with respect to $x$ exist everywhere.

Some further points are in order.

• Lebesgue's differentiation theorem tells us that the partial derivative $f_x$ exists almost everywhere with respect to the Lebesgue measure. That does not help here though, because $\iint\cdot d_tg(t,x)dx$ need not be absolutely continuous. In fact, even restricting to the case where $g(t,x)$ is convex in $x$, it is possible to construct nontrivial examples where $\iint\cdot d_tg(t,x)dx$ is supported on the graph $\{(t,\gamma(t))\colon t\in\mathbb{R}\}$ of a (discontinuous) function $\gamma\colon\mathbb{R}\to\mathbb{R}$.
• If the left and right hand partial derivatives of $f(t,x)$ with respect to $x$ exist everywhere, then the question has a positive answer.
• If the left and right hand partial derivatives of $g(t,x)$ with respect to $x$ exist everywhere, then $f_x$ exists in a weak sense. Given any bounded measurable $\theta\colon\mathbb{R}^2\to\mathbb{R}$ with compact support,$$\iint h^{-1}(f(t,x+h)-f(t,x))\theta d_tgdx\to\iint f_x\theta d_tgdx$$as $h\to0$.
• Rather than monotonicity in $t$, I am really interested in the weaker condition that $\iint\cdot\vert d_tf(t,x)\vert dx$ is locally finite ($\int\cdot\vert d_tf(t,x)\vert$ is the variation of $\int\cdot d_tf(t,x)$), and similarly for $g$. If I find a positive answer to this, then I think I would update my papers on arXiv to reflect this.
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