Compactness of set of Lipschitz functions $f:D\rightarrow D$ Denote by $D:=\{z\in \mathbb{R}^2\text{ such that } ||z||\le 1\}$ the unitary closed disk: it is a metric space with the induced Euclidean metric of $\mathbb{R}^2$.
Fix any $k>0$ and denote by $Lip_k(D)$ the set of continuous functions $f:D\rightarrow D$ with Lipschitz constant $k$.
On $Lip_k(D)$ consider any of the following topologies:


*

*the compact open topology

*the topology of pointwise convergence

*the $L^\infty$-topology 
Question 1: Is $Lip_k(D)$ compact with respect to any of the previous topologies? Can you motivate your answer or point me our a reference?
I am sorry if I can not show my attempt, but I do not really know how to prove that, given any sequence $\{f_n\}\subset Lip_k(D)$, there must always be a converging subsequence $\{f_{n_k}\}$ with limit $f\in Lip_k(D)$.
Question 2: If $Lip_k(D)$ is not compact, then will it change if we consider $Lip_k^0(D)$, the set of $k$-Lipschitz functions which coincide with the identity on $\partial D$?
 A: First lets deal with the case of the $L^\infty$-topology. Here it is natural to think of $\operatorname{Lip}_k(D)$ as a subspace of $C(D)$, the space of continuous functions on $D$ equipped with the supremum norm.
We will use the Arzela-Ascoli Theorem which tells us that $F \subset C(D)$ is relatively compact if and only if it is a bounded set in the sup norm and it is equicontinuous. It is easy to see that $\operatorname{Lip}_k(D)$ is equicontinuous and since $f \in \operatorname{Lip}_k(D)$ implies $f(D) \subset D$, $\operatorname{Lip}_k(D)$ is bounded in the supremum norm with bound $1$. Hence $\operatorname{Lip}_k(D)$ is relatively compact. 
It's an easy exercise to see that $\operatorname{Lip}_k(D)$ is closed in $C(D)$ for the supremum norm and hence is compact.
Similarly $\operatorname{Lip}_k^0(D)$ is a bounded, equicontinuous and closed subset of $C(D)$ and so is compact.
Now $D$ is compact, so the compact open topology coincides with the $L^\infty$-topology. So we are only left to consider the topology of pointwise convergence.
This is straightforward since the $L^\infty$-topology is strictly finer than the topology of pointwise convergence and hence any compact subset of $\operatorname{Lip}_k(D)$ (or even of $C(D)$) for the $L^\infty$-topology is also compact in the topology of pointwise convergence.
