For any irrational number
$\alpha $
define the Lagrange constant
$\mu \left(\alpha \right)$
by
$${\mu}^{1}\left(\alpha \right)=\underset{p\in \mathbb{Z},\phantom{\rule{0.3em}{0ex}}q\in \mathbb{N}}{liminf}\leftq\left(q\alpha p\right)\right.$$
The set of all values taken by
$\mu \left(\alpha \right)$
as
$\alpha $ varies is called the
Lagrange spectrum $\mathbb{L}$.
An irrational
$\alpha $
is called attainable if the inequality
$$\left\alpha \frac{p}{q}\right\le \frac{1}{\mu \left(\alpha \right){q}^{2}}$$
holds for infinitely many integers
$p$
and
$q$. We call a
real number
$\lambda \in \mathbb{L}$
admissible if there exists an irrational attainable
$\alpha $ such
that
$\mu \left(\alpha \right)=\lambda $.
In a previous paper we constructed an example of a nonadmissible element
in the Lagrange spectrum. In the present paper we give a necessary and
sufficient condition for admissibility of a Lagrange spectrum element. We
also give an example of an infinite sequence of left endpoints of gaps in
$\mathbb{L}$ which
are not admissible.
