# Compact sets in uniform norm

I was wondering whether an operator on the space $C$ of continuous real functions (defined on a closed interval $[a,b]$) i.e. $$C= \mathcal{C}([a,b];\mathbb{R}^d)$$ which is locally Lipschitz is also Lipschitz in sets as $$A:=\{ f \in C : || f ||_{\infty} \leq k \}$$ or $$B:=\{ f \in C : || f -g||_{\infty} \leq k \}$$ for some $k>0$ and some continuous function $g \in C$.

It's straightforward to see that locally Lipschitz implies Lipschitz on every compact set, but are those compact sets? I remember that Ascoli Arzelà was all about saying that in infinite dimension compactness is something harder to prove than close and bounded. Any intuitive ways to see whether those sets are compact?

P.S. I am using the sup norm (also called uniform norm, infinity norm or Chebyshev norm)

• Your bad: those sets are not compact, nor locally compact. Indeed closed balls are compact only in finite-dimensional topological vector spaces. – Crostul Jan 21 '18 at 18:16
• @Crostul, any idea on the fact whether they are Lipschitz? – ThePunisher Jan 21 '18 at 18:20

No, $A$ is not compact. I shall prove when $d=k=1$ and $[a,b]=[0,1]$; the general case is similar. Define $f_1\in A$ as$$f_1(x)=\begin{cases}4x&\text{ if }x\in\left[0,\frac14\right]\\1+4\left(\frac14-x\right)&\text{ if }x\in\left[\frac14,\frac12\right]\\0&\text{ otherwise.}\end{cases}$$Define $f_2\in A$ as$$f_2(x)=\begin{cases}8\left(x-\frac12\right)&\text{ if }x\in\left[\frac12,\frac58\right]\\1+8\left(\frac58-x\right)&\text{ if }x\in\left[\frac58,\frac34\right]\\0&\text{ otherwise}\end{cases}$$and so on. These functions are continuous and$$(\forall m,n\in\mathbb{N}):m\neq n\implies\|f_m-f_n\|_\infty=1.$$Therefore, no subsequence of the sequence $(f_n)_{n\in\mathbb N}$ converges, which would be impossible if $A$ was compact.