Consider the following one dimensional Young differential equation: \begin{align*} &Y_t=\int_0^t Y_s dX_s,\quad t\in[0,1];\\ &Y_0=0. \end{align*}

Here the driving process $X$ is a bounded functions $[0,1]\to\mathbb{R}$, which is $\beta$-Holder with $\beta<1/2$.

If $Y$ is an $\alpha$-Holder function, $\alpha>1-\beta$, then this equation is well defined (because the integral then becomes just the Young integral).

Question: how to prove that the only solution to this equation in the class of $\alpha$-Holder functions is $Y\equiv0$?

Warning: note that $\beta<1/2$! If $\beta>1/2$, then this result is standard, but what to do if $\beta<1/2$?

Failed solution attempt

Denote by $[X]_{\beta,[0,T]}$, $[Y]_{\alpha,[0,T]}$ the corresponding Holder norms of $X$ and $Y$ on the interval $[0,T]$, respectively. Then the standard inequality for the Young integral implies $$ |Y_t-Y_s-Y_s(X_t-X_s)|\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} (t-s)^{\alpha+\beta},\quad s,t\in[0,T]. $$ This in turn leads $$ |Y_t-Y_s|\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} (t-s)^{\alpha+\beta}+|Y_s|\,|X_t-X_s|, $$ and thus $$ [Y]_{\beta,[0,T]}\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} T^{\alpha}+[X]_{\beta,[0,T]}\sup_{r\in[0,T]}|Y_r|. $$

However, because $\beta<\alpha$, the last inequality gives us nothing (we are estimating a smaller norm by a larger norm). The iteration over $T$ also seems hopeless. So what to do?

  • 3
    $\begingroup$ "Maybe I am missing something here, but the proof definitely does not follow immediately from your comment." It does. Let $t>0$. Choose small $\varepsilon>0$ and consider the last moment $t_1\in[0,t]$ when $|Y_t|\le\varepsilon$. Then $|Y_s|\ge\varepsilon$ on $[t_1,t]$, so $X\in Hol_\alpha([t_1,t])$ and then $Y_s=Y_{t_1}\exp(X_s-X_{t_1})$ for $s\in[t_1,t]$. In particular $|Y_t|\le\varepsilon\exp(2\|X\|_{C([0,t])})$. But $\varepsilon>0$ is arbitrarily small! $\endgroup$
    – fedja
    Jul 27, 2020 at 12:50
  • 1
    $\begingroup$ @ThomasKojar What do you mean I cannot impose a different regularity? Of course I can, this is the whole point of the question. If we know in advance that $Y$ is very regular, show that $Y$ is identically zero (or provide a counter-example). $\endgroup$
    – Oleg
    Jan 30, 2023 at 21:48
  • 1
    $\begingroup$ @ThomasKojar Let me rephrase my question. We are given two processes. $Y$ which is very regular and $X$ which is less regular. It is known that $dY_t=Y_td X_t$ (in the usual ODE, not RDE sense, all the integrals are well-defined Young integrals). Question: is it true that $Y\equiv0$? $\endgroup$
    – Oleg
    Jan 30, 2023 at 21:51
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    $\begingroup$ In other words: I am asking for uniqueness only among $C^\alpha$ solutions (not among all solutions) for which we can always define this as a Young integral. Is it more clear now? $\endgroup$
    – Oleg
    Jan 30, 2023 at 23:34
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    $\begingroup$ @ThomasKojar I really do not understand what you do not understand in my question. We are given two processes $Y$ and $X$, $Y$ is regular $X$ is not. They solve a certain equation. Question: does this mean that $Y=0$ or there is another regular process $\widetilde Y$ such that $d\widetilde Y=\widetilde Y dX$. You are asking a different question, but which part in my question you do not understand? I am talking here about standard ODEs not RDEs. You are trying to answer a different question. $\endgroup$
    – Oleg
    Jan 30, 2023 at 23:39

1 Answer 1


We will follow the Lemma 8.10. (Rough Gronwall) and Proposition 8.12 from "a course in rough paths" but modify them for this particular setting of studying

$$Y_{t}=Y_{0}+\int^{t}_{0} Y dX,$$

for $Y\in C^{\alpha},X\in C^{\beta}$ where $\alpha+\beta>1$ but $\beta<\frac{1}{2}$ and $\alpha>\frac{1}{2}$.

The Lemma 8.10. (Rough Gronwall) comes verbatim with no changes

Assume $Y\in C([0,1])$, $r<1$ and $$|Y_{s,t}|\leq M ||Y||_{\infty;[s,t]}|t-s|^{r},(*)$$ for $0\leq s< t\leq 1$. Then there exists $c=c_{r}<\infty$ such that $$||Y||_{\infty;[0,1]}\leq cexp(cM^{1/r})|Y_{0}|.$$

So it suffices to prove (*). From a slight variation of the proof of Proposition 8.12 we have the bound

$$|Y_{s,t}|=|\int_{s}^{t}Y_{r}dX_{r}|\leq c||Y||_{\infty;[s,t]}||X||_{\beta;[s,t]}|t-s|^{\beta}\leq c||Y||_{\infty;[s,t]}|t-s|^{\beta},$$ where we used that $\beta<\frac{1}{2}\leq \alpha$ and so $||Y||_{\alpha;[s,t]}\leq ||Y||_{\beta;[s,t]}$. So the result follows from here.


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