Let $X$ be the Banach space of the continuous functions $x:[0,1]\longrightarrow \mathbb{R}$, and conside the set
$$ C:=\{x\in X: 0=x(0\leq x(t)\leq x(1)=1,\textrm{ for all } t\in[0,1]\} . $$
Also, recall that the Hausdorff measure of noncompactness of a bounded subset of $X$, put $B$, is defined as
$$ \beta(B):=\inf\{\varepsilon>0: B\subset \cup_{i=1}^{n}U(x_{i},\varepsilon),\textrm{ for some }x_{1},\ldots,x_{n}\in X\}, $$ where $U(x_{i},\varepsilon)$ denotes the closed ball centered at $x_{i}$ and radisu $\varepsilon$.
I have read, without proof, that $\beta(C)=1/2$. My question is the following: There is $r>0$ such that $\beta(B)\geq r$ for each non-precompact $B\subset C$?
Thanks in advance for your suggestions.
I think the answer is no: Let $\alpha > 0$, set $y(t):=t$ $(t \in [0,1])$ (then $y \in C$) and $$ B:=C\cap U(y,\alpha)= \{x \in C: |x(t)-t| \le \alpha ~ (t \in [0,1])\}. $$ Then $B$ is closed, not compact (hence $\beta(B) > 0$) but $\beta(B) \le \alpha$ since $U(y,\alpha)$ covers $B$. A simple way to prove that $B$ is not compact is to construct a pointwise convergent sequence in $B$ with pointwise limit function beeing discontinuous at some $t_0 \in (0,1)$.