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.


2 Answers 2


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 $Y$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Pierre PC
    Commented Jun 3, 2020 at 0:12
  • $\begingroup$ 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. $\endgroup$ Commented Jun 3, 2020 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.

  • $\begingroup$ 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. $\endgroup$ Commented Jun 2, 2020 at 10:33

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