Consider the following one dimensional Young differential equation: $$ Y_t=\int_0^t Y_s dX_s,\quad t\in[0,1]. $$

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?

  • $\begingroup$ Just notice that your first inequality and the condition $Y\in Hol_\alpha$ forces $X\in Hol_\alpha$ as long as $Y$ is separated from $0$. $\endgroup$
    – fedja
    Jul 26, 2020 at 18:52
  • $\begingroup$ @fedja I agree and I also thought about using this fact, but unfortunately I don't see how it helps. It might be that $X$ is alpha Holder on any interval $[\epsilon,1]$ (with its Holder norm going to infinity as $\epsilon$ approaches $0$) and still not alpha Holder on $[0,1]$. Think about $\sqrt x$. Its Lipschitz on any interval $[\epsilon,1]$, but not Lipschitz on $[0,1]$. Maybe I am missing something here, but the proof definitely does not follow immediately from your comment. Do you have a formal proof or could you please expand your ideas? $\endgroup$
    – Oleg
    Jul 26, 2020 at 21:55
  • 2
    $\begingroup$ "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
  • $\begingroup$ @fedja thanks! that's indeed a great solution. If you would formally post is as a solution, I would be very happy to award you the promised bounty. $\endgroup$
    – Oleg
    Jul 28, 2020 at 16:00
  • $\begingroup$ I was hoping though that one can solve this problem by iterating the inequalities I wrote. My motivation is that i am stuck with a similar problem, but for PDEs. I want to show that dY/dt=\Delta Y+YdX has a unique zero solution (where now Y=Y(x,t)) under the same regularity assumptions. In this case one cannot write the solutions explicitly, but one can get very similar bounds to the ones presented in my question. $\endgroup$
    – Oleg
    Jul 28, 2020 at 16:11

1 Answer 1


For such small $\beta$, we need to use Rough path theory to make sense of the integral and so below I go over that. (Indeed for $\beta\in (\frac{1}{2},1]$, there is an ODE theory for Young integrals eg see theorem 2.5 in "Stochastic differential equations driven by fractional Brownian motions")

In "Rough Path Theory" by Andrew L. Allan (available free online), I found in Proposition 6.12 a solution in the spirit of what mentioned in the comments.

enter image description here

In the uniqueness, they do go over an estimates argument in terms of Holder norms that might be useful in your SPDE setting.

In your case instead of $1$, you will have $0$ and so the same argument will apply. Meaning that for general $y_{0}\in \mathbb{R}$ the solution to


can be similarly obtained by Ito's formula for rough paths (prop 6.11)

$$Y_{t}=y_{0}exp\left(X_{t}- \frac{1}{2}[\mathbf{X}]_{t}\right).$$

If you want me to add more details on the argument, please let me know.

For smaller $\beta$, the situation might break down. In here fractional Brownian Motion driven stochastic integrals, Hairer mentions some other approaches that might even apply to your SPDE setting (eg. regularity structures).


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