I am reading Differential inclusions by J.-P. Aubin, A. Cellina. Here is Lemma 1.4.1 from the book:
Suppose $\{x_k\}$ is a bounded sequence in a Hilbert space $H$. We associated the bounded sequence a function $\phi$ by defining \begin{align*} \phi(y) = \limsup_{k \to \infty} \|x_k - y\|^2 = \inf_{k \ge 1} \sup_{m \ge k} \| x_m - y\|^2. \end{align*} Then $\phi$ is locally Lipschitzean, strictly convex and satisfies $\lim_{\|y\| \to \infty} \phi(y) = \infty$.
In the proof to this lemma, the first sentence is 'It is obvious the function is Lipschitzean.' Well it is not so obvious to me. I would show the function is strictly convex, and convex function is locally Lipschitzean in the interior of domain. What am I missing here? Is the Lipschitzean property so 'obvious'?
I realized it should be just straightforward application of triangle inequality and using boundedness of the sequence to separate a sum.
Let's consider a related function $$\psi (y) = \limsup_{k \to \infty} \|x_k - y\| $$ By the triangle inequality, $|\psi(y)-\psi(z)| \le \|y-z\|$, meaning the function $\psi$ is globally Lipschitz with constant equal to $1$. (This is an instance of a general fact: the infimum, or supremum, of any family of $L$-Lipschitz functions is $L$-Lipschitz, as long as the constant $L$ is fixed.)
The function $\phi = \psi^2$ is only Lipschitz on bounded sets. Indeed, consider the ball $B = \{y : \|y\|\le R\}$. Then $|\psi(y)|\le R+\sup_k\|x_k\|$ on $B_R$, so $\psi$ is bounded on $B_R$. Also, the squaring map $t\mapsto t^2$ is Lipschitz on any bounded subset of $\mathbb{R}$; and the composition of Lipschitz maps is Lipschitz.