8
$\begingroup$

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?

$\endgroup$
18
  • 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 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 at 21:51
  • 2
    $\begingroup$ 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 at 23:34
  • 2
    $\begingroup$ @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 at 23:39

1 Answer 1

0
$\begingroup$

With existing technology of Young differential equations for $x_{t}\in C^{\gamma}$, for the Picard iteration to go through one can prove existence/uniquence over $C^{\beta}$-solution space with $\gamma>\beta>1-\gamma$. For example, in "Stochastic differential equations driven by fractional Brownian motions" they show existence and uniqueness over the solution space $C^{\beta}$ with $\gamma>\beta>1-\gamma$. So one cannot use the existing Young technology for $\gamma<1/2$ since $\beta>1-\gamma>1/2>\gamma$. enter image description here

But one can use RDEs see Proposition 6.12 "Rough Path Theory Lecture Notes Andrew L. Allan" where the author has to use RDEs in order to solve the equation $Y_t=\int Y_s dX_s$.

Having said that, here is an argument showing that only $Y\equiv 0$ can solve the above DE

$$Y_t=\int_{0}^{t} Y_s dX_s. $$

Suppose otherwise that there is some nonconstant solution $Y_{s}\in C^{\beta}$ and $x_{s}\in C^{\gamma}$ with $\beta+\gamma>1$ and $\gamma<1/2$. Using the lemma 2.3

enter image description here

we have

$$\frac{Y_t-Y_s}{(t-s)^{\beta}}=Y(s)\frac{X_t-X_s}{(t-s)^{\beta}}+\frac{\sum\sum...}{(t-s)^{\beta}}.$$

In the proof of prop 2.2, they show that the second term with the sums is bounded by $c (t-s)^{\gamma}(t-s)^{\beta}$ and so

$$\frac{Y_t-Y_s}{(t-s)^{\beta}}=Y(s)\frac{X_t-X_s}{(t-s)^{\beta}}+O((t-s)^{\gamma}).$$

However, since $\gamma<1/2$, that means $\beta>1/2>\gamma$ and so the first term on the RHS diverges whereas the LHS is finite, which is a contradiction. So we can only have zero solutions $Y\equiv 0$.

$\endgroup$
2
  • $\begingroup$ The answer is totally unrelated to the question. $\endgroup$
    – user479223
    Jan 30 at 12:51
  • 2
    $\begingroup$ @user479223 ok I updated the answer. Thank you for your help. $\endgroup$ Jan 31 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy