European lookback put options uniformly continuous 
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I am currently reading Martingale optimal transport in the Skorokhod space by Yan Dolinsky and H. Mete Soner. 
The Skorokhod space used is the space inducted by the Skorokhod metric d defnied by,
$$ d(\omega,\widetilde{\omega}) = \inf_{\lambda \in \Lambda [0,T] }\sup_{t \in [0,T]}(|\omega(t)-\widetilde{\omega}(\lambda(t))|+|\lambda(t)-t|), $$
where $\Lambda [0,T]$ is the set of all strictly increasing onto functions
$\lambda:[0,T]\to[0,T]$.
At one point it is assumed that exotic options $$G:\mathbb{D}([0,T];\mathbb{R}^{d}) \to \mathbb{R}$$ are uniformly continuous and satisfy $|G(\mathbb{S})| \leq C(1+|\mathbb{S}_{T}|)$ (for some constant C). Where $\mathbb{D}([0,T];\mathbb{R}^{d})$ is the set of all $\mathbb{R}^{d}$-valued cadlag functions that are continuous at T and $\mathbb{S}_{0}=(1, \ldots ,1)$ for all $\mathbb{S} \in \mathbb{D}([0,T];\mathbb{R}^{d})$. Afterwards some examples for options G are given that statisfy this assumption. Namely for d=1, the lookback put option with fixed strike 
$$ G(\mathbb{S})=\left(K-\min\limits_{0 \leq t \leq T}\mathbb{S}_{t}\right)^{+} $$
and the lookback call options with floating price 
$$G(\mathbb{S})=\left(\mathbb{S}_{T}-\min\limits_{0 \leq t \leq T}\mathbb{S}_{t}\right)^{+}. $$ 
I understand why they satisfy the inequality assumption but why are these example options uniformly continuous? I tired to showing their uniform continuity by using the definition and and by finding fitting modulus of continuity. However with both approaches I got nowhere. 
 A: Let's show that the map $\omega \mapsto \inf_{0 \le t \le T} \omega(t)$ is $1$-Lipschitz with respect to Skorohod distance.  This should be enough to prove that both maps $G$ are uniformly continuous, since you already know that $\omega \mapsto \omega(T)$ is 1-Lipschitz.  Indeed, it follows that the first $G$ is 1-Lipschitz and the second G is 2-Lipschitz.
Let $\omega, \tilde{\omega}$ be given, and let $M = \inf_{0 \le t \le T} \omega(t)$, $\tilde{M}= \inf_{0 \le t \le T} \tilde{\omega}(t)$.  Let $\epsilon > 0$ be arbitrary.  Then there exists a strictly increasing onto $\lambda$ such that for all $t \in [0,T]$ we have
$$|\omega(t) - \tilde{\omega}(\lambda(t))| \le |\omega(t) - \tilde{\omega}(\lambda(t))| + |\lambda(t) - t| \le d(\omega,\tilde{\omega}) + \epsilon.$$
  We now note that for all $t$, we have
$$\begin{align*}\omega(t) &\ge \tilde{\omega}(\lambda(t)) - |\omega(t) - \tilde{\omega}(\lambda(t))|  \\
&\ge \tilde{M} - (d(\omega, \tilde{\omega}) + \epsilon).\end{align*}$$
Taking the inf over $t$, we have $M \ge \tilde{M} - (d(\omega, \tilde{\omega}) + \epsilon)$, or in other words, $\tilde{M} - M \le d(\omega, \tilde{\omega}) + \epsilon$.  By symmetry, we also have $M - \tilde{M} \le d(\omega, \tilde{\omega}) + \epsilon$.  Putting these together yields $|M - \tilde{M}| \le d(\omega, \tilde{\omega}) + \epsilon$, and since $\epsilon$ was arbitrary, we conclude $|M - \tilde{M}| \le d(\omega, \tilde{\omega})$.
