Let $f: [0;1] \rightarrow \mathbb{R}$ be a Lipschitz continuous function, with Lipschitz constant $L>0$ and $X$ be a random variable uniformly distributed on $[0;1]$.
Let $X_n=2^{-n}\left\lfloor2^nX\right\rfloor,Y_n=2^n(f(X_n+\frac{1}{2^n})-f(X_n)),
\mathcal{F}_n=\sigma(X_0,…,X_n)$ and $\mathcal{F}_{\infty}=\bigcap_{n \in \mathbb{N}}\sigma(\bigcup_{k \geq n}\sigma(X_k)).$
We can prove that $\mathcal{F}_n=\sigma(X_n),\mathcal{F}_{\infty}=\sigma(X)$ and that $(Y_n)_n$ is a martingale for the filtration $(\mathcal{F}_n)_n$ such that $|Y_n| \leq L,$ then $(Y_n)_n$ converges a.s and in $L^1$ to a random variable Y such that there exists a bounded measurable function $h:[0;1] \rightarrow \mathbb{R}$ such that $Y=h(X)$ a.s.
We have $$Y_n=E[Y|\mathcal{F}_n]=E[h(X)|X_n]=2^n\int_{X_n}^{X_n+\frac{1}{2^n}}h(x)dx$$ so
$$\forall x \in [0;1],f(x)=f(0)+\int_0^xh(y)dy$$
If the function $f$ was from $[0;1]^d$ to $\mathbb{R}$, how should we define $X_n$ and $Y_n$? We should take $X$ uniformly distributed on $[0,1]^d$?
Best Answer
A function $f:\mathbb{R}^d\rightarrow \mathbb{R}$ is differentiable at the point $X$ iff there exists a function $\displaystyle J:\mathbb{R}^d\rightarrow \mathbb{R}$ such that $\lim_{h\rightarrow 0}\frac{f(X+h)-f(X)-J(h)}{||h||}=0$.
This is what I guess might happen in the $d=2$ case:
Given $X=(x,y)\in [0,1]^2$, define $X_{n}=(x_n,y_n)=2^{2n}(\lfloor2^{-n}x\rfloor,\lfloor 2^{-n}y\rfloor)$.
We want to find a function $J=(J_1,J_2)$ such that, at almost all points $X=(x,y)$, $J=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$.
Define $\displaystyle Y_{1,n}=2^{2n}\int_{y_n}^{y_n+2^{-n}} \left(f(x_n+2^{-n},s)-f(x_n,s) \right)\mathrm{d}s$ and $\displaystyle Y_{2,n}=2^{2n}\int_{x_n}^{x_n+2^{-n}}\left( f(t,y_n+2^{-n})-f(t,y_n)\right)\mathrm{d}t$.
The main motivation for this definition of $Y_{1,n}$ is the fact that, if $f$ could be assumed to be differentiable, then we would have $ \displaystyle 2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}}\frac{\partial f}{\partial x} \mathrm{d}s\mathrm{d}t=2^{2n}\int_{y_n}^{y_n+2^{-n}}\left(f(x_n+2^{-n})-f(x_n)\right)\mathrm{d}t$
As before, $(Y_{1,n},Y_{2,n})$ converges to a vector $Y=(Y_1,Y_2)$ and there exists a bounded measurable $J$ such that $Y=J(X)$ a.s.
We have
$\displaystyle Y_{1,n}(X)=\mathbb{E}(Y_1|X_{n})=2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}} J_1(s,t)\mathrm{d}s \mathrm{d}t$ $\displaystyle Y_{2,n}(X)=\mathbb{E}(Y_2|X_{n})=2^{2n}\int_{x_n}^{x_n+2^{-n}}\int_{y_n}^{y_n+2^{-n}} J_2(s,t) \mathrm{d}s \mathrm{d}t$
If $L$ is a smooth curve going from $0$ to $X$, then $\displaystyle f(X)=f(0)+\int_L J\cdot\mathrm{d}L$.