[proofplan] We prove that the supremum-norm functional on $\ell^\infty(T)$ is Lipschitz, hence continuous and Borel measurable. This lets us apply the Continuous Mapping Theorem to the [weak convergence](/page/Weak%20Convergence) $Z_n \xrightarrow{d} Z$ in $\ell^\infty(T)$. The image random variables under this functional are exactly $\sup_{t \in T}|Z_n(t)|$ and $\sup_{t \in T}|Z(t)|$. [/proofplan] [step:Define the supremum functional on $\ell^\infty(T)$] Define the map $\Phi: \ell^\infty(T) \to \mathbb{R}$ as follows: for each $z \in \ell^\infty(T)$, set \begin{align*} \Phi(z) := \sup_{t \in T} |z(t)|. \end{align*} This map is well-defined because every $z \in \ell^\infty(T)$ is bounded, so $\sup_{t \in T}|z(t)| < \infty$. [/step] [step:Prove that the supremum functional is Lipschitz] Let $z,w \in \ell^\infty(T)$. For each $t \in T$, applying the ordinary triangle inequality in $\mathbb{R}$ to $z(t)=w(t)+(z(t)-w(t))$ gives \begin{align*} |z(t)| \le |w(t)| + |z(t)-w(t)| \le |w(t)| + \|z-w\|_{\ell^\infty(T)}. \end{align*} Taking the supremum over $t \in T$ yields \begin{align*} \Phi(z) \le \Phi(w) + \|z-w\|_{\ell^\infty(T)}. \end{align*} Interchanging $z$ and $w$ gives \begin{align*} \Phi(w) \le \Phi(z) + \|z-w\|_{\ell^\infty(T)}. \end{align*} Combining these two inequalities, \begin{align*} |\Phi(z)-\Phi(w)| \le \|z-w\|_{\ell^\infty(T)}. \end{align*} Thus $\Phi$ is $1$-Lipschitz, and therefore continuous on $\ell^\infty(T)$. [guided] We need a deterministic continuity statement before applying the probabilistic convergence theorem. The relevant deterministic map is $\Phi: \ell^\infty(T) \to \mathbb{R}$, defined for each $z \in \ell^\infty(T)$ by \begin{align*} \Phi(z) := \sup_{t \in T} |z(t)|. \end{align*} To prove continuity, it is enough to prove a Lipschitz estimate. Fix $z,w \in \ell^\infty(T)$. For each $t \in T$, applying the ordinary triangle inequality in $\mathbb{R}$ to $z(t)=w(t)+(z(t)-w(t))$ gives \begin{align*} |z(t)| \le |w(t)| + |z(t)-w(t)|. \end{align*} Since $z-w \in \ell^\infty(T)$, the definition of the supremum norm gives \begin{align*} |z(t)-w(t)| \le \sup_{s \in T}|z(s)-w(s)| = \|z-w\|_{\ell^\infty(T)}. \end{align*} Therefore, for every $t \in T$, \begin{align*} |z(t)| \le |w(t)| + \|z-w\|_{\ell^\infty(T)}. \end{align*} Taking the supremum over $t \in T$ on the left-hand side and using that the added term $\|z-w\|_{\ell^\infty(T)}$ is independent of $t$, we obtain \begin{align*} \Phi(z) = \sup_{t \in T}|z(t)| \le \sup_{t \in T}|w(t)| + \|z-w\|_{\ell^\infty(T)} = \Phi(w) + \|z-w\|_{\ell^\infty(T)}. \end{align*} Repeating the same argument with $z$ and $w$ exchanged gives \begin{align*} \Phi(w) \le \Phi(z) + \|z-w\|_{\ell^\infty(T)}. \end{align*} The [first inequality](/theorems/2897) says $\Phi(z)-\Phi(w) \le \|z-w\|_{\ell^\infty(T)}$, and the second says $\Phi(w)-\Phi(z) \le \|z-w\|_{\ell^\infty(T)}$. Combining them gives \begin{align*} |\Phi(z)-\Phi(w)| \le \|z-w\|_{\ell^\infty(T)}. \end{align*} Thus $\Phi$ is $1$-Lipschitz. Every Lipschitz map between metric spaces is continuous, so $\Phi$ is continuous at every point of $\ell^\infty(T)$. [/guided] [/step] [step:Apply the continuous mapping theorem to $\Phi(Z_n)$] Since $\Phi: \ell^\infty(T) \to \mathbb{R}$ is continuous, it is Borel measurable. The hypothesis gives $Z_n \xrightarrow{d} Z$ as $\ell^\infty(T)$-valued random elements. Therefore, by the [Continuous Mapping Theorem](/theorems/1847), applied to the continuous map $\Phi$, we have \begin{align*} \Phi(Z_n) \xrightarrow{d} \Phi(Z) \end{align*} in $\mathbb{R}$. [/step] [step:Identify the image random variables with the desired suprema] By the definition of $\Phi$, for each $n \in \mathbb{N}$, \begin{align*} \Phi(Z_n) = \sup_{t \in T}|Z_n(t)|, \end{align*} and also \begin{align*} \Phi(Z) = \sup_{t \in T}|Z(t)|. \end{align*} Substituting these identities into $\Phi(Z_n) \xrightarrow{d} \Phi(Z)$ gives \begin{align*} \sup_{t \in T}|Z_n(t)| \xrightarrow{d} \sup_{t \in T}|Z(t)|. \end{align*} This is the desired convergence in distribution. [/step]