I am reading a paper (the 2015 paper by A. Falkowski and L. Slominski Stochastic Differential Equation with Constraints Driven by Processes with Bounded $p-$variation, page 353, proof of the Lemma 3.1) for my master degree thesis.


we fix a continuous function $$ f:\Bbb R^d\to\Bbb R^d $$ such that satisfies the linear growth condition $$ |f(x)|\le L(1+|x|),\;\;\;x\in\Bbb R^d\;. $$ Let $1<p\le2$ and take $$ g:\Bbb R^d\to\mathcal M_d(\Bbb R) $$ $\alpha$-Holder continous function, where $p-1<\alpha\le1$ i.e. $$ ||g(x)-g(y)||\le C_{\alpha}|x-y|^{\alpha},\;\;x,y\in\Bbb R^d $$ where $||A||:=\sup\{|Ax|\;:\;|x|=1\}$ is the usual matrix norm and $|\cdot|$ is the euclidean norm in $\Bbb R^d$.

Fix then $a:\Bbb R_{\ge0}\to\Bbb R$ and $z:\Bbb R_{\ge0}\to\Bbb R^d$ right continous functions which admit finite left limit (RCFLL for short) such that \begin{align*} V_1(a)_{[0,T]}&:=\sup_{\pi[0,T]}\sum_{j=1}^n|a_{t_j}-a_{t_{j-1}}|<+\infty\\ V_p(z)_{[0,T]}&:=\sup_{\pi[0,T]}\left[\sum_{j=1}^n|z_{t_j}-z_{t_{j-1}}|^p\right]^{1/p}<+\infty\\ \end{align*} where obviously $\pi[0,T]$ denotes the generic subdivision of the closed interval $[0,T]$.

Fix then $l:\Bbb R_{\ge0}\to\Bbb R^d$ another RCFLL function and consider the following integral equation $$ x_t=x_0+\int_0^tf(x_{s-})\,da_s+\int_0^tg(x_{s-})\,dz_s+k_t $$ whose unknown are the RCFLL functions $x,k:\Bbb R_{\ge0}\to\Bbb R^d$, with constraint given by $x_0\ge l_0$ (inequality taken componentwise); both integral are Riemann-Stieltjes ones.


the Lemma of the paper I refer to, says what follows: suppose there exists $b>0$ such that $$ \max\left[V_1(a)_{[0,T]},\;V_p(z)_{[0,T]},\;\sup_{t\le T}|l_t|\right]\le b $$ then there exists $\bar C$ depending ONLY on $d, p, \alpha, L, g(0), x_0, b$ such that, if $(x,k)$ is a solution of the integral equation above, then $$ \bar V_p(x)_{[0,T]}:=|x_0|+V_p(x)_{[0,T]} \le\bar C\;. $$

I will skip most of detail of the proof, going directly to the core of the problem (otherwise the post would be the longest ever!).

I am able to prove the following inequality: \begin{align*} \bar V_p(x)_{[0,t]} &\le (d+1)\left[|x_0|+LV_1(a)_{[0,t]}(1+V_p(x)_{[0,t]})+DV_p(z)_{[0,t]}(1+V_p(x)_{[0,t]})\right]+db\\ &= (d+1)\left[|x_0|+(LV_1(a)_{[0,t]}+DV_p(z)_{[0,t]})(1+V_p(x)_{[0,t]})\right]+db \end{align*} where $D$ is another absolute constant. Here $t\le T$.

Starting from here, we define $$ t_1:=\inf\left\{t>0\;:\;LV_1(a)_{[0,t]}>\frac1{4(d+1)}\;\;\mbox{or}\;\;DV_p(z)_{[0,t]}>\frac1{4(d+1)}\right\}\wedge T $$ from which immediately $$ \bar V_p(x)_{[0,t_1[}\le(d+1)|x_0|+\frac12(1+V_p(x)_{[0,t_1[})+db $$ and thus $$ \bar V_p(x)_{[0,t_1[}\le2(d+1)|x_0|+1+2db\;\;. $$ Next we accept that $$ |\Delta x_{t_1}|=|x_{t_1}-x_{t_1-}|\le(L(1+|x_{t_1-}|)+C_{\alpha}|x_{t_1}|^{\alpha}+||g(0)||+2)b. $$

Then the authors write there exists $C_1,C_2>0$ depending only on $d, p, \alpha, L, g(0), b$, such that $$ \bar V_p(x)_{[0,t_1]}\le C_1+C_2|x_0|. $$ WHY?! I know that $\bar V_p(x)_{[0,t_1]}\le\bar V_p(x)_{[0,t_1[}+|\Delta x_{t_1}|$ but controlling $\Delta x_{t_1}$, terms depending on $x$ appear!

Then they set $$ t_k:=\inf\left\{t>t_{k-1}\;:\;LV_1(a)_{[t_{k-1},t_k]}>\frac1{4(d+1)}\;\;\mbox{or}\;\;DV_p(z)_{[t_{k-1},t_k]}>\frac1{4(d+1)}\right\}\wedge T $$ thus for the same constants $$ V_p(x)_{[t_{k-1},t_k]}\le C_1+C_2|x_{t_{k-1}}|\le C_1+C_2\bar V_p(x)_{[0,t_{k-1}]} $$ Set now $m:=\sup\{k\;:\;t_k\le T\}$ and accept that $m$ is finite and absolute (in the sense it doesn't depend on $(x,k)$ but only on the constants already written).

How can I reach the conclusion from here? I am really confused: how can I get rid of the terms depending on $x$ in order to get my absolute estimate?

Many thanks for any suggestion!

  • $\begingroup$ There are a lot of trivial edits to this question; let's bring that to an end, please. Each edit bumps a question to the top of the page and knocks another one off the page; this when taken too far is not fair. $\endgroup$ – Todd Trimble Jul 24 '17 at 17:43

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