Following Doob, we say that a function
f
(
z
)
f(z)
analytic in the unit disk U has the property
K
(
ρ
)
K(\rho )\,
if
f
(
0
)
=
0
f(0)\, = \,0
and for some
arc
A
\operatorname {arc} \,A\,
on the unit circle whose measure
|
A
|
⩾
2
ρ
>
0
\left | A \right |\, \geqslant \,2\rho \, > \,0
,
\[
lim
inf
i
→
∞
|
f
(
P
i
)
|
⩾
1
where
P
i
→
P
∈
A
and
P
i
∈
U
.
\lim \,\inf \limits _{i \to \infty } \,\left | {f({P_i})} \right |\, \geqslant \,1\,{\text {where}}\,{P_i}\, \to \,P\, \in \,A\,{\text {and}}\,{P_i}\, \in \,U.
\]
We recently have solved a problem of Doob by showing that there is an integer
N
(
ρ
)
N(\rho )
such that no function with the property
K
(
ρ
)
K(\rho )
can satisfy
\[
(
1
−
|
z
|
)
|
f
n
′
(
z
)
|
⩽
1
/
n
for
z
∈
U
,
where
n
>
N
(
ρ
)
.
(1\, - \,\left | z \right |)\left | {{f_n}’ (z)} \right |\, \leqslant \,1/n\,{\text {for}}\,z\, \in \,U,\,{\text {where}}\,n\, > \,N(\rho ).
\]
The function
\[
f
n
(
z
)
=
1
+
(
1
−
z
n
)
/
n
2
,
{f_n}(z)\, = \,1\, + \,(1\, - \,{z^n})/{n^2},
\]
shows that the condition
f
n
(
0
)
=
0
{f_n}(0)\, = \,0
is necessary and cannot be replaced by
f
n
(
0
)
=
r
e
i
α
{f_n}(0)\, = \,r{e^{i\alpha }}
, for
r
>
1
r\, > \,1
. Naturally, we may ask whether this can be replaced by
f
n
(
0
)
=
r
e
i
α
{f_n}(0)\, = \,r{e^{i\alpha }}
, for
r
>
1
r\, > \,1
? The answer turns out to be yes, when
n
>
N
(
r
,
ρ
)
n\, > \,N\,(r,\,\rho )
, where
\[
N
(
r
,
ρ
)
≑
(
1
/
(
1
−
r
)
)
log
(
1
/
(
1
−
cos
ρ
)
)
.
N(r,\,\rho )\,\doteqdot \,(1/(1\, - \,r))\log (1/(1\, - \,\cos \rho )).
\]
.