I am working on the following SDE (but we will dealing only with deterministic object: $\omega\in\Omega$ is fixed): \begin{equation}\label{sde}%sde x_t=\underbrace{\xi_0+\int_0^tb(s,x_s)\,ds+\int_0^t\sigma(s,x_s)\,dW_s}_{=:z_t}+y_t \end{equation} where $y_t=\sup_{0\le s\le t}(z_s)^{-}$ is the regulator term which ensures that the positivity constraint is respected and the stochastic integral is a Young integral.
Let us suppose everything real valued.
The following hypothesis are considered: $\xi_0>0$ is fixed, \begin{align*} &b:[0,L]\times\Bbb R\to\Bbb R\;\;\;\;,\\ &\sigma:[0,L]\times\Bbb R\to\Bbb R \end{align*} are measurable and bounded functions which satisfy \begin{equation}\label{hyp1}%hyp1 |b(t,x)-b(t,y)|\le K_0|x-y|\;\;\forall x,y\in\Bbb R,\;\forall t\in[0,L] \end{equation} \begin{equation}\label{hyp2}%hyp2 |\sigma(t,x)-\sigma(t,y)|\le K_0|x-y|\;\;\forall x,y\in\Bbb R,\;\forall t\in[0,L] \end{equation} \begin{equation}\label{hyp3}%hyp3 |\sigma(t,x)-\sigma(s,x)|\le K_0|t-s|^{\nu}\;\;\forall x\in\Bbb R,\;\forall s,t\in[0,L] \end{equation} where $\nu\in]\frac12,1]$ and $K_0>0$.
Hence, both $b$ and $\sigma$ are Lipschitz in space, moreover $\sigma$ is $\nu$-Holder continous in time.
Next we take a fractional Brownian motion $W$ of Hurst parameter $H$ that is, $W\in\mathcal C^{H-\varepsilon}([0,L],\Bbb R)$ for every $\varepsilon>0$, with $1/2<H\le\nu$.
We will consider an arbitrary fixed trajectory of this FBM (i.e. $\omega$ is fixed and we mean $W=W(\omega)$).
Then it is proved that for every fixed $\lambda\in]\frac12,H[$, the equation at the beginning admits a solution $x$ such that, a.s. $x(\omega)\in\mathcal C^{\lambda}([0,L],\Bbb R)$.
We summarize here the relations between the parameters considered above, in order to be clear: $$ \frac12<\lambda<H\le\nu\le1\;\;. $$
I would like to prove uniqueness for this SDE.
In the following we will consider $x^{(1)},x^{(2)}$ two $\lambda$-solutions of the SDE on $[0,L]$, writing $z^{(i)},y^{(i)}$ with the obvious meaning: $x^{(i)}=z^{(i)}+y^{(i)}\;\;i=1,2$.
We will take $0\le T\le L$, which will be chosen conveniently later.
I will skip the details (if someone is interested, tell me this!) in order not to annoy you, going directly to the core of the problem: setting $$ H_T:=\|x^{(1)}-x^{(2)}\|_{\infty,[0,T]} $$ I was able to prove that $$ H_T\le2K_0H_TT+\eta_TT^{\lambda+H-\varepsilon} $$ where $$ \eta_T:=K_0\|W\|_{H-\varepsilon,[0,T]}\left(\|x^{(1)}\|_{\infty,[0,T]}+\|x^{(2)}\|_{\infty,[0,T]}+2T^{1-\lambda}\right) $$ which is a bounded non-negative increasing function of $T$.
Now from this I got $$ \frac{H_T}{\eta_TT^{\lambda+H-\varepsilon}}\le\frac1{1-2K_0T} $$ and thus, passing to the $\limsup_{T\to0+}$ we get \begin{align*} 1 &=\limsup_{T\to0+}\frac1{1-2K_0T}\\ &\ge\limsup_{T\to0+}\frac{H_T}{\eta_TT^{\lambda+H-\varepsilon}}\\ &\ge\limsup_{T\to0+}\frac{|x_T^{(1)}-x_T^{(2)}|}{\eta_TT^{\lambda+H-\varepsilon}} \end{align*} and setting $f_t:=x_t^{(1)}-x_t^{(2)}$ we have that $f_0=0$ and $f\in\mathcal C^{\lambda}[0,L]$, thus the last term can be rewritten as $$ \limsup_{T\to0+}\frac{|f_T-f_0|}{|T-0|}\frac1{\eta_TT^{\alpha}}=(*). $$ where $\alpha:=\lambda+H-\varepsilon-1>0$.
Now if $\limsup_{T\to0+}\frac{|f_T-f_0|}{|T-0|}>0$ we'd get $(*)=+\infty$ thus we would have reached a contradiction, which would allow to get uniqueness.
The other case is $\limsup_{T\to0+}\frac{|f_T-f_0|}{|T-0|}=0$, from which we would have $\exists\lim_{T\to0+}\frac{|f_T-f_0|}{|T-0|}=0$, and thus, $f$ would be differentiable at 0 on the right. But here I'm stuck: how can we exclude this last limsup is zero?
