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?

regularprocess $\widetilde Y$ 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$11more comments