# Intuition behind Gubinelli derivative

I apologise for the confusion of the following sentences. I'm lazy to give more information about Rough path theory as Is a fairly broad subject.

On page 14 of "A Course on Rough Paths With an Introduction to Regularity Structures" by Peter K. Friz & Martin Hairer has written:

For $$\alpha \in (1/ 3; 1 /2]$$, define the space of $$\alpha$$-Hölder rough paths (over V ), in symbols $$\mathcal C^{\alpha} ([0,T]; V )$$, as those pairs $$(X; \mathbb X) =: \mathbf{X}$$ such that $$||X||_{\alpha}:= \sup_{ s\neq t \in [0;T ]} \frac{|X_{s,t}|}{|t-s|^{\alpha}} < \infty , \quad ||X||_{2\alpha}:=\sup_{ s\neq t \in [0;T ]} \frac{|\mathbb X_{s,t}|}{|t-s|^{\alpha}} < \infty ,$$ and such that the algebraic Chen relation ( is satisfed.

And on page 56 it hase written: Given a path $$X \in \mathcal C^{\alpha}([0, T ]; V )$$, we say that $$Y \in \mathcal C^{\alpha}([0, T ]; \hat{W} )$$, is controlled by $$X$$ if there exists $$Y' \in \mathcal C^{\alpha}([0, T ]; \mathcal L(V , \hat{W})$$, so that the remainder term $$R^Y$$ given implicitly through the relation $$Y_{s,t }= Y_{s0} X_{s,t} + R_{s,t}^Y ,$$ satisfes $$||R^Y||_{ 2 \alpha}< 1$$. This defines the space of controlled rough paths, $$(Y, Y') \in \mathcal D_X^{2α}([0, T ]; \hat{W })$$: Although $$Y'$$ is not, in general, uniquely determined from Y. We call any such $$Y'$$ the Gubinelli derivative of $$Y$$ (with respect to $$X$$). Here, $$R_{s,t}^Y$$ takes values in $$\hat{W}$$, and the norm $$|| \cdot||$$.

Question : What is the intuition behind this idea of the Gubinelli derivative?

Any help is appreciated with thanks in advance.

In a way it is very much like a usual derivative. Recall first that for a regular function $$Y$$, its derivative $$Y'_s$$ at a point $$s$$ is the (unique) number such that $$Y_{t,s}=Y'_s(t-s)+ R_{s,t},$$ where $$R_{s,t}\to0$$ faster than linearly. If $$Y$$ is twice differentiable, then $$R_{s,t}\lesssim |t-s|^2$$. That is, as a function of $$t$$, $$Y_t$$ "looks like" the linear function $$Y_s+Y'_s(t-s)$$, in the neighborhood of $$s$$.

Now simply replace the linear function by $$X$$. So we impose $$Y_{t,s}=Y'_sX_{t,s}+R_{s,t}$$ with the remainder $$R_{s,t}\to0$$ faster than the first term, that is, faster than $$|t-s|^\alpha$$ (The condition $$R_{s,t}\lesssim|t-s|^{2\alpha}$$ from Friz-Hairer corresponds to the twice differentiable scenario in the previous case). Then as a function of $$t$$, $$Y_t$$ "looks like" the path $$Y_s+Y'_sX_{s,t}$$. This is great news for integration: we can of course integrate $$Y_s$$ against $$dX_t$$ (since as a function of $$t$$ it is just constant), and we can also integrate $$X_{s,t}$$ against $$dX_t$$ (by the definition of a rough path).

Actually, I wouldn't focus so much on assigning a meaning to $$Y'$$ itself, but rather focus on what the existence of a $$Y'$$ means for $$Y$$.

• I like this answer very much. With this in mind, in many situations the approximation $Y_{t,s}\simeq Y'_sX_{t,s}$ gives a good idea of what $Y'$ should be: intuitively, if $Y=f(X)$, then $Y_{t,s}\simeq f'(X_s)X_{t,s}$; if $Y=\int A\mathrm dX$ for some process $A$, then $Y_{t,s}\simeq A_s\cdot X_{t,s}$; etc. – Pierre PC Jun 3 '20 at 0:12
• Morality: this derivative is none other than the usual derivative modulo the regularities as it has just been pointed out in the commentary of Pierre PC. – Furdzik Jun 3 '20 at 10:36

We want to define $$\int_0^T f(X_s) dX_s$$ for smooth bounded $$f$$ with bounded derivatives of all orders. Using linearity and a partition $${t_k}$$ of $$[0,T]$$, we have

\begin{align*}\int_0^T f(X_s) dX_s&=\sum_k\int_{t_k}^{t_{k+1}}f(X_s) dX_s\\&=\sum_k\int_{t_k}^{t_{k+1}}f(X_{t_k})+f'(X_{t_k})(X_s-X_{t_k})+O(|s-t_k|^{2\alpha})dX_s\\&=\sum_k f(X_{t_k})(X_{t_{k+1}}-X_{t_k})+f'(X_{t_k})\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s+O(|t_{k+1}-t_k|^{3\alpha})\end{align*}

As $$3\alpha>1$$ the third term goes to zero as the mesh size goes to $$0$$. The first term is just a Riemann integral. The second term is the "rough path" term. $$f'(X_{t_k})$$ is the Gubinelli derivative and $$\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s$$ is your area process.

• That is for a smooth function $f$ where the Gubinelli derivative is intuitively seen as a velocity or tangent vector at a point $X_{t_k}$, but what about function with some regularity. Thanks for your answer. – Furdzik Jun 2 '20 at 10:33