Little and Great Picard consequences of Nevanlinna theory

A nonconstant rational function $P\mathrm{/}Q$, with coprime polynomials and $d=\max (\mathrm{deg}P,\mathrm{deg}Q)\ge 1$, omits at most one sphere value. If $\mathrm{deg}Q<d$, then $\mathrm{deg}(P-aQ)=d$ for every finite $a$, so [F4] makes every finite value attained; $\mathrm{\infty }$ is omitted only when $Q$ is constant. If $\mathrm{deg}Q=d$, then $Q$ has a root, so $\mathrm{\infty }$ is attained. When $\mathrm{deg}P=d$, the polynomial $P-aQ$ has degree $d$ except possibly for the single value $a$ that cancels its leading term. When $\mathrm{deg}P<d$, every $a\mathrm{\ne }0$ gives a polynomial $P-aQ$ of degree $d$, and $a=0$ is attained unless $P$ is a nonzero constant. Thus at most one finite value can be omitted in this case.

Let $f$ be meromorphic on $0<\mathrm{\mid }z-z_0\mathrm{\mid }<r^{\ast }$ with an isolated essential singularity. If three distinct sphere values are omitted on $0<\mathrm{\mid }z-z_0\mathrm{\mid }<\rho$ for some $\rho <r^{\ast }$, then $G(w):=f(z_0+\rho \mathrm{/}w)$ is meromorphic on $\mathrm{\mid }w\mathrm{\mid }>1$ and omits those values there. Apply [F3] with $R=1$ and $R_1=2$: $G$ extends meromorphically across infinity. Inversion then extends $f$ meromorphically across $z_0$, a contradiction. Thus no three distinct values are omitted on any punctured neighbourhood.

Suppose a nonconstant meromorphic plane function $h$ omits three distinct sphere values. By step 1.1 it is transcendental. [F1] gives $T(r,h)\le C({\log }^{+}T(r,h)+\log r)$ outside a set $E$ of finite linear measure, while [F2] makes the right side $o(T(r,h))$ as $r\to \mathrm{\infty }$. The complement of $E$ is unbounded, so this inequality is impossible at sufficiently large $r\mathrm{\notin }E$. Hence a nonconstant meromorphic plane function omits at most two sphere values.

If three distinct sphere values each had only finitely many preimages in some punctured neighbourhood, choose one radius smaller than all three neighbourhood radii. Their combined preimage set inside it is finite. Choose a still smaller radius below the distance from $z_0$ to every point of that finite set; if the set is empty, any smaller radius works. All three values are then omitted on that smaller punctured disc, contrary to step 1.2. Hence at most two sphere values fail to occur infinitely often in every punctured neighbourhood.

An entire function omits $\mathrm{\infty }$, so step 2.1 leaves at most one omitted finite value. A function holomorphic on the punctured disc also omits $\mathrm{\infty }$, so step 2.2 leaves at most one finite value that fails to occur infinitely often near the puncture.