Let's call $\mathcal D(\Bbb R_{\ge0},\Bbb R^d)$ the Skohrokod space, i.e. the space of the functions $f:\Bbb R_{\ge0}\to\Bbb R^d$ continous on the right which admit limit on the left.
Let's fix $y,l\in \mathcal D(\Bbb R_{\ge0},\Bbb R^d)$ s.t. $y_0\ge l_0$ (when we write inequalities we mean them componentwise).
Then it is well known (Skohrokod problem) that there exist $x,k\in \mathcal D(\Bbb R_{\ge0},\Bbb R^d)$ which solve the following \begin{align*} &\bullet \;\;\; x_t=y_t+k_t\ge l_t\;\;\;\forall t\ge0\\ &\bullet\;\;\; k_0=0 \;\;\mbox{, the components of}\;\; k=(k^{(1)},\dots,k^{(d)}),\;\;k^{(i)}\;\;\\ &\mbox{are nondecreasing functions, s.t.}\\ &\;\;\;\;\;\;\;\;\;\;\;\int_0^t(x_s^{(i)}-l_s^{(i)})\,dk_s^{(i)}=0\;\;\;\;\forall t\ge0,\;\; i=1,\dots,d \end{align*}
It can be proven that $k_t=\sup_{s\le t}(y_s-l_s)^{-}\;$.
One last thing: fixed $p\ge1$, we define the $p-$variation norm of $f\in \mathcal D(\Bbb R_{\ge0},\Bbb R^d)$ as $$ \bar V_p(f)_T:=|f_0|+\left[\sup_{\pi}\sum_{j=1}^n|f_{t_{j}}-f_{t_{j-1}}|^p\right]^{1/p} $$
where $\pi=\{0=t_0<t_1<\dots<t_n=T\}$ runs through all the partitions of the interval $[0,T]$.
Ok, here comes my problem: I have to prove the following inequality
$$ \bar V_p(k)_T\le d\sup_{s\le T}|y_s|+d\sup_{s\le T}|l_s|\;\;. $$
Here's my try:
\begin{align*} \bar V_p(k)_T =&\sup_{\pi}\left[\sum_{j=1}^n|k_{t_j}-k_{j-1}|^p\right]^{1/p}+\underbrace{|k_0|}_{=0}\\ \le&\sup_{\pi}\sum_{j=1}^n\left|k_{t_j}-k_{t_{j-1}}\right|\\ =&\sup_{\pi}\sum_{j=1}^n\left[\sum_{i=1}^d(k_{t_j}^{(i)}-k_{t_{j-1}}^{(i)})^2\right]^{1/2}\\ \le&\sup_{\pi}\sum_{j=1}^n\left[\sum_{i=1}^d(k_{t_j}^{(i)}-k_{t_{j-1}}^{(i)})\right]\\ =&\sum_{i=1}^d\left[\sup_{\pi}(k_{t_1}^{(i)}-k_{t_0}^{(i)}+k_{t_2}^{(i)}-k_{t_1}^{(i)}+\cdots+k_{t_n}^{(i)}-k_{t_{n-1}}^{(i)})\right]\\ =&\sum_{i=1}^d1\cdot k_T^{(i)}\\ \mbox{(Cauchy-Schwarz)}\;\;\;\le&\left(\sum_{i=1}^d1^2\right)^{1/2}\cdot\left(\sum_{i=1}^d(k_T^{(i)})^2\right)^{1/2}\\ =&d^{1/2}\left(\sum_{i=1}^d(\sup_{s\le T}(y_s^{(i)}-l_s^{(i)})^{-})^2\right)^{1/2}\\ \le&d\left(\sum_{i=1}^d(\sup_{s\le T}|y_s^{(i)}-l_s^{(i)}|)^2\right)^{1/2}\\ \le&d\left(\sum_{i=1}^d(\sup_{s\le T}|y_s^{(i)}|+\sup_{s\le T}|l_s^{(i)}|)^2\right)^{1/2}\\ \mbox{(Minkowsky)}\;\;\le&d\left(\sum_{i=1}^d(\sup_{s\le T}|y_s^{(i)}|)^2\right)^{1/2}+d\left(\sum_{i=1}^d(\sup_{s\le T}|l_s^{(i)}|)^2\right)^{1/2}\\ %=&d\left|\sup_{s\le T}|y_s|\right|+d\left|\sup_{s\le T}|l_s|\right| \end{align*} thus in every sum, the $\sup$ is taken for every summand; instead, in the original inequality, the $\sup$ is taken for the norm of the vector, which gives a stronger inequality than the one written above, since I guess that
$$ \sup_{s\le T}|y_s|=\sup_{s\le T}\left(\sum_{i=1}^d(y_s^{(i)})^2\right)^{1/2}\;\;. $$ I don't think the authors of the paper wrote
$$ \sup_{s\le T}|y_s| $$ wanting to meaning exactly $$ \left(\sum_{i=1}^d(\sup_{s\le T}|y_s^{(i)}|)^2\right)^{1/2}\;\;. $$