MathOverflow Asked on December 13, 2021
Consider the following one dimensional Young differential equation:
$$
Y_t=int_0^t Y_s dX_s,quad tin[0,1].
$$
Here the driving process $X$ is a bounded functions $[0,1]tomathbb{R}$, 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 $Yequiv0$?
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
$$
|Y_t-Y_s-Y_s(X_t-X_s)|le C[X]_{beta,[0,T]}[Y]_{alpha,[0,T]} (t-s)^{alpha+beta},quad s,tin[0,T].
$$
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
$$
[Y]_{beta,[0,T]}le C[X]_{beta,[0,T]}[Y]_{alpha,[0,T]} T^{alpha}+[X]_{beta,[0,T]}sup_{rin[0,T]}|Y_r|.
$$
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?
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP