\magnification=\magstep1
\documentstyle{amsppt}
\loadmsbm
\NoBlackBoxes
\def\ttl#1{\par\bigskip\centerline{\bf #1. }\par\bigskip }
\def\st#1{ \par\bigskip\flushpar{\bf #1. }\medskip\par }
\def\C{\Bbb C}
\def\Z{\Bbb Z}
\def\R{\Bbb R}
\def\lb{\lambda}
\def\G{\Gamma}
\def\ep{\varepsilon}
\def\ph{\varphi}
\def\Sg{\Sigma}
\def\om{\omega}
\def\e{{\,e}}
\def\Dl{\Delta}
\def\Dt{\widetilde\Dl}
\def\ed{\overset{def}\to=}
\def\sm{\setminus}
\redefine\le{\leqslant}
\redefine\ge{\geqslant}
\ttl{Growth of entire functions with sparse spectra}
Let $f(z)=\sum_{k=0}^\infty c_k z^{\lb_k}$ be an entire function ($c_k\in
\C$, $0=\lb_0<\lb_1<\dots\in\Z$). Assume that $\sum_{k=1}^\infty\frac
1{\lb_k}<\infty$.
\st{Old conjecture}
There is no curve $\G$ joining $0$ and $\infty$ such that $f$ is bounded on
$\G$ (unless $f\equiv const$, of course).
\medskip
So far the only way to attack this problem was to show that for arbitrarily
large $R>0$,
$$
M_{\G,f}(R)\ed\max_{|z|\le R,z\in\G}|f(z)|\approx M_f(R)\ed
\max_{|z|\le R}|f(z)|
$$
(the idea that actually goes back to Littlewood and 1920's).
More precisely, the point was to show that
$$
\varlimsup_{R\to \infty}\frac{\log M_{\G,f}(R)}{\log M_f(R)}=1.\tag $*$
$$
This particular kind of interpretation is due to the fact that the
standard symmetry trick allows to reduce the case of an arbitrary curve $\G$
to the case $\G=\R_{+}$, which is much easier to handle.
Below we are going to squeeze from this approach everything it can give. The
result is the following
\st{Theorem}
(A) If the sequence $\{\lb_k\}$ is convex (i.e., if
$\lb_{k+1}-\lb_k\ge\lb_k-\lb_{k-1}$), then the condition
$\sum_{k=1}^\infty\frac
1{\lb_k}<\infty$
really implies ($*$). What is more, for every $\ep>0$, the set
\linebreak $\{R>1:\
M_{\G,f}(R)\le M_f(R)^{1-\ep}\}$ has finite logarithmic measure (i.e., the
measure of the set of $\log R$ is finite or, which is the same, the integral
of $1/R$ over this set converges).
(B) Under no regularity assumptions about the sequence $\{\lb_k\}$, the weakest
condition that still implies ($*$) is $\sum_{k=1}^\infty\frac
{\log\log\lb_k}{\lb_k}<\infty$ (or, which is the same, $\sum_{k=1}^\infty\frac
{\log\log k}{\lb_k}<\infty$).
\st{Proof}
Let
$f(z)=\sum_{k=0}^\infty c_kz^{\lb_k}$. Let $\mu(R)\ed\max_k(|c_k|R^{\lb_k})$ be
the central term for $f(z)$ when $|z|=R$, and let $j(R)$ be the central index,
i.e., the value of $k$ for which $\mu(R)$ is attained.
\st{Lemma 1}
$M_f(R)\le4\lb_{j(R)}^2\mu(R)$ for all $R>1$ except, maybe, a set of finite
logarithmic measure.
\newpage
\st{Proof}
Let $I_j=[a_j,b_j]$ be the interval of values of $\log R$ for which
$j(R)=j\ge 1$. \linebreak
If $\log R\in [a_j,b_j-\frac{1}{\lb_j^2}]$, then for every $z\in\C$ such that
$|z|=R$, we have
$$
|f(z)|=\biggl|\sum_{k=0}^\infty c_kz^{\lb_k}\biggr|
\le\sum_{k=0}^\infty |c_k|R^{\lb_k}=
\sum_{k=0}^{j-1} |c_k|R^{\lb_k}+\sum_{k=j}^\infty |c_k|R^{\lb_k}\ed\Sg_1+\Sg_2
$$
Since $|c_k|R^{\lb_k}\le|c_j|R^{\lb_j}$ for every $k$, we have $\Sg_1\le
j\mu(R)$.
For $k\ge j$, we have (taking into account that $\log R$
does not lie too close to the
right end of $I_j$) $|c_k|(R\e^{1/\lb_j^2})^{\lb_k}\le
|c_j|(R\e^{1/\lb_j^2})^{\lb_j}$ and therefore
$$
|c_k|R^{\lb_k}\le|c_j|R^{\lb_k}\e^{-\frac{\lb_k-\lb_j}{\lb_j^2}}\le
\mu(R)\e^{-\frac{k-j}{\lb_j^2}}.
$$
Thus,
$$
\Sg_2\le\mu(R)\sum_{k=0}^\infty\exp(-\frac{k}{\lb_j^2})\le2\lb_j^2\mu(R)
$$
Finally we get $M_f(R)\le(j+2\lb_j^2)\mu(R)\le 4\lb_j^2\mu(R)$.
To finish the proof of the lemma, it remains only to note that the set of
$R>1$ for which $j(R)=0$ is bounded (if $f$ is not constant) and that
the measure of the union of the intervals $[b_j-\frac{1}{\lb_j^2},b_j]$
is not greater than
$\sum_{k=1}^\infty\frac{1}{\lb_j^2}\le\sum_{j=1}^\infty\frac{1}{j^2}<\infty$.
Now let us turn to the proof of the theorem.
Consider first the case when the sequence $\{\lb_k\}$ is convex.
Recall the formula for the distance $D$ in $L^2(0,1)$
from the function
$x^\lb$ to $span(x^{\lb_0},\dots,x^{\lb_m})$:
$$
D=\frac{1}{\sqrt{2\lb+1}}\prod_{k=0}^m\frac{|\lb-\lb_k|}{\lb+\lb_k+1}.
$$
A simple consequence of this formula is that for every polynomial
$P(x)=ax^\lb+\sum_{k=0}^ma_kx^{\lb_k}$, the inequality
$$
\max_{x\in[0,r]}|P(x)|\ge\frac{|a|r^\lb}{\sqrt{2\lb+1}}
\prod_{k=0}^m\frac{|\lb-\lb_k|}{\lb+\lb_k+1}
$$
holds.
\st{Lemma 2}
If $j(R)\ge1$ then $M_{\G,f}(R)\ge\e^{-15j(R)}\lb_{j(R)}^{-3}\mu(R)$
for every ray $\G=\{t\xi:t>0\}$.
\st{Proof}
Without loss of generality we may assume that $\G=\R_+$. Also we shall write
$j$ instead of $j(R)$ for convenience.
Write $f(z)=\sum_{k=0}^\infty c_kz^{\lb_k}=\sum_{k=0}^{2j}c_kz^{\lb_k}+
\sum_{k>2j} c_kz^{\lb_k}\ed P(z)+Q(z)$.
Let $\om>0$. We have
$$
M_{\G,f}(R)\ge\max_{x\in[0,Re^{-\om}]}|f(x)|\ge
\max_{x\in[0,Re^{-\om}]}|P(x)|-\max_{x\in[0,Re^{-\om}]}|Q(x)|.
$$
According to the consequence of the distance formula mentioned above,
$$
\max_{x\in[0,Re^{-\om}]}|P(x)|\ge|c_j|(R\e^{-\om})^{\lb_j}
\frac{1}{\sqrt{2\lb_j+1}}\prod\Sb 0\le k\le 2j\\
k\ne j\endSb\frac{|\lb_j-\lb_k|}{\lb_j+\lb_k+1}.
$$
The convexity of the sequence $\{\lb_k\}$ implies $\lb_m\le\frac mn\lb_n$ for
every $m2j}\e^{-\om\lb_k}\le\mu(R)
\frac{1}{1-\e^{-\om}}\e^{-\om\lb_{2j}}
\le(1+\frac{1}{\om})\e^{-2\om\lb_j}\mu(R)
$$
because (again, due to convexity) $\lb_{2j}\ge 2\lb_j$.
So, to get a nontrivial estimate for $M_{\G,f}(R)$ from below, we
should choose $\om>0$
satisfying
$$
\frac{1}{\lb_j}\e^{-6j}\e^{-\om\lb_j}\ge 2(1+\frac{1}{\om})\e^{-2\om\lb_j}
$$
Simple computation shows that $\om=\frac{1}{\lb_j}(8j+2\log\lb_j)$ suffices.
This yields the estimate
$$
M_{\G,f}(R)\ge\frac 12\lb_j^{-3}\e^{-14j}\mu(R)\ge\lb_j^{-3}\e^{-15j}\mu(R)
$$
desired.
\medskip
The symmetry trick allows to derive from Lemma 2 the estimate
$$
M_{\G,f}(R)\ge\e^{-30j(R)}\lb_{j(R)}^{-6}\mu(R)^2M_f(R)^{-1}
$$
for an arbitrary curve $\G$. Together with the estimate from Lemma 1,
it yields the inequality
$$
M_{\G,f}(R)\ge\frac 14\e^{-30j(R)}\lb_{j(R)}^{-8}\mu(R)
$$
valid for all $R>1$ except, maybe, a set of finite logarithmic measure.
Assume now that $M_{\G,f}(R)\le M_f(R)^{1-\ep}$ and that $R$ doesn't belong
to the exceptional set. Then we have
$$
\frac 14\e^{-30j(R)}\lb_{j(R)}^{-8}\mu(R)\le(4\lb_j(R)^2\mu(R))^{1-\ep}
$$
and therefore
$$
\log\mu(R)\le A(\ep)(j(R)+\log\lb_{j(R)}) \tag $**$
$$
Let's show now that the set of all $R>1$ satisfying ($**$) has finite
logarithmic measure. Let $N(\ell)$ be the number of the elements $\lb_k$
that are less than $2^\ell$. The condition $\sum_{k=1}^\infty
\frac{1}{\lb_k}<\infty$ is equivalent to $\sum_{\ell\ge1}\frac{N(\ell)}
{2^\ell}<\infty$. Note that the slope of $\log\mu(R)$ against $\log R$
is $\lb_{j(R)}$. Therefore the logarithmic measure of the set of ``good'' (in
the sense of Lemma 1) $R$ for which
($**$) is satisfied, $\lb_{j(R)}\in[2^{\ell-1},2^\ell)$,\
$j(R)\ge1$ and $\mu(R)>1$
(the last two conditions are not very essential since they are fulfilled
automatically for all sufficiently large $R$), doesn't exceed
$A(\ep)2^{-(\ell-1)}(j(R)+\log\lb_{j(R)})\le 2A(\ep)2^{-\ell}(N(\ell)+\ell)$
because $j(R)\le N(\ell)$ and $\log\lb_{j(R)}\le\ell$.
Thus, the logarithmic measure of the set of ``large'' $R$ for which ($**$)
holds doesn't exceed $2A(\ep)\sum_{\ell\ge1}\frac{N(\ell)+\ell}{2^\ell}
<\infty$, proving the statement (A) of the theorem.
\bigskip
Let's now show that the condition $\sum_{k\ge2}\frac{\log\log\lb_k}
{\lb_k}<\infty$ implies ($*$) without any regularity assumptions. We might
just repeat the proof of (A) but I'd like to demonstrate a different
technique that will yield a little bit stronger result.
Namely, we shall show that
for all $R>1$ except, maybe, a set of finite logarithmic measure,
one can find
$r\in(0,R)$ such that $m_f(r)\ed\min_{|z|=r}|f(z)|\ge M_f(R)^{1-\ep}$.
(From the proof of (A), it follows only that for most $R$ the connected
component of the open set $\{z:|f(z)|\lb_j,
\endaligned
\right.
$$
and $D_k\le 2d_k$ for every $k$.
\st{Proof}
If $d_m\le d_0$ put $Q(z)\ed z^{-\lb_0}P(z)$ and $P_1(z)\ed
\frac{1}{d_0}z^{\lb_0+1}\frac{d}{dz}(z^{-\lb_0}P(z))$.
Note that $P_1(z)=\sum_{k=0}^{m}c_k\frac{\lb_k-\lb_0}{d_0}z^{\lb_k}$ and
therefore the term $c_0z^{\lb_0}$ disappears while the coefficient at
$z^{\lb_j}$ doesn't change. Besides,
$$
\frac{d_0P_1(z)}{P(z)}=\frac{zQ'(z)}{Q(z)}=
\sum_{\ell=1}^{D_0}\frac{z}{z-\zeta_{0\ell}}
$$
where $\zeta_{0\ell}$ are the roots of the polynomial $Q(z)$ of degree
$D_0=\lb_m-\lb_0=d_0+d_m\le 2d_0$.
If $d_m> d_0$ put $Q(z)\ed z^{-\lb_m}P(z)$ and $P_1(z)\ed
\frac{1}{d_m}z^{\lb_m+1}\frac{d}{dz}(z^{-\lb_m}P(z))$.
Everything remains the same except $Q(z)$ is now a polynomial of degree
$D_m\le 2d_m$ in $1/z$, not in $z$, and thereby
$$
\frac{d_mP_1(z)}{P(z)}=\frac{zQ'(z)}{Q(z)}=
\sum_{\ell=1}^{D_0}\frac{\frac 1z}{\frac 1z-\frac 1{\zeta_{m\ell}}}=
-\sum_{\ell=1}^{D_m}\frac{\zeta_{m\ell}}{z-\zeta_{m\ell}}=
D_m-\sum_{\ell=1}^{D_m}\frac{z}{z-\zeta_{m\ell}}
$$
where $1/{\zeta_{m\ell}}$ are the roots of the polynomial $Q(1/z)$.
Constructing $P_2(z)$ from $P_1(z)$ in the same way and so on, we get the
representation desired.
\medskip
Now we want to show that the functions $\ph_k(z)$ are small on most
circumferences centered at the origin. Note that
$$
|\ph_k(z)|\le
2\biggl(1+\frac{1}{D_k}\sum_{\ell=1}^{D_k}\frac{|z|}{|z-\zeta_{k\ell}|}
\biggr)\le
2\biggl(1+\frac{1}{D_k}\sum_{\ell=1}^{D_k}\frac{|z|}{|\,|z|-|\zeta_{k\ell}|\,|}
\biggr).
$$
\st{Lemma 3}
Let $\ph(x)=\frac{1}{D}\sum_{\ell=1}^D\frac{x}{|x-x_k|}$\ \ $(x,x_k>0)$.
Then for every $A>6$, say, the logarithmic measure of the set
$\{x>0:\ph(x)>A\}$ doesn't exceed $\frac{9+4\log D}{A}$.
\newpage
\st{Proof}
Let $E_\ell\ed \{x>0:|x-x_\ell|\le\frac 3A x_\ell\}$. We have
$$
\frac{x}{|x-x_k|}=\frac{x}{|x-x_k|}\chi_{E_\ell}+
\frac{x}{|x-x_k|}\chi_{\R_+\sm E_\ell}\ed\eta_\ell(x)+\theta_\ell(x)
$$
Note that for every $\ell$ we have $\theta_\ell(x)\le \frac A3(1+\frac 3A)=
\frac A3+1\le\frac A2$. Thus, $\frac{1}{D}\sum_{\ell=1}^D
\theta_\ell(x)\le\frac A2$. Let $F_\ell\ed\{x>0:|x-x_\ell|\le\frac{3}{AD}
x_\ell\}$,
$F\ed\bigcup_\ell F_\ell$. The logarithmic measure of $F$ doesn't exceed
$D\log\frac{AD+3}{AD-3}\le D\frac 9{AD}=\frac 9A$. To finish the proof of
Lemma 3 it remains only to note that for every $\ell$
$$
\int_{\R_+\sm F}\eta_\ell(x)\frac{dx}{x}\le
\int_{\R_+\sm F_\ell}\eta_\ell(x)\frac{dx}{x}=
2\int_{\frac 3{AD}}^{\frac 3A}\frac{dx}{x}=2\log D.
$$
It follows from Lemma 3 that for every $A\ge 1$,
the logarithmic measure of the set
$$
\{r:\text{ there exists }z\text{ for which } |z|=r\text { and }
|\ph_k(z)|>4(9+4\log D_k)A\}
$$
is less than $\frac 1A$. Though weak type estimates are difficult to add,
they are easy to multiply and we conclude that every set $S\subset \R_+$ of
logarithmic measure $\om<1$ contains a point $r$ such that
$$
\prod\Sb 0\le k\le m\\k\ne
j\endSb\ph_k(z)^{-1}\ge\left(\frac{\om}{4(9+4\log\lb_m)e}\right)^{m}
$$
(we have used the trivial estimate $D_k\le\lb_m$).
Now everything is ready to prove that the condition
$\sum\frac{\log\log\lb_k}{\lb_k}<\infty$ implies ($*$).
Let $j=j(R)\ge1$. Assume that $2^{\ell-1}\le\lb_j<2^\ell$. Again we are
going to use the decomposition $f(z)=P(z)+Q(z)$ where
$$
P(z)\ed\sum_{k:\lb_k<2^{\ell+2}}c_kz^{\lb_k}
,\qquad\quad Q(z)\ed\sum_{k:\lb_k\ge 2^{\ell+2}}c_kz^{\lb_k}.
$$
As before, we observe that for every $|z|\le Re^{-\om}$,
$$
|Q(z)|\le(1+\frac 1\om)e^{-2^{\ell+2}\om}\mu(R)
$$
while there exists $r\in[Re^{-2\om},Re^{-\om}]$ such that for every $|z|=r$,
$$
|P(z)|\ge
\left(\frac{\om}{4(9+4\log(2^{\ell+2}))e}\right)^{N({\ell+2})}e^{-2^{\ell+1}\om}
\mu(R)
\ge \left(\frac{\om}{1000\ell}\right)^{N({\ell+2})}e^{-2^{\ell+1}\om}\mu(R).
$$
Again, to get a nontrivial result, we must choose $\om<1$ for which the estimate
for $P(z)$ from below is at least twice as large as
the estimate for $Q(z)$ from above.
Solving the corresponding inequality, we see that it is possible to take
$$
\om=2^{-\ell}N(\ell+2)\left[\log(1000\ell)+\log\Bigl(
\frac{2^\ell}{N(\ell+2)}\Bigr)+\ell+3\right]
$$
(since the condition $\sum_{k=2}^{\infty}\frac{\log\log\lb_k}{\lb_k}<\infty$
implies that $2^{-\ell}N(\ell)\log\ell\to 0$ as $\ell\to\infty$, we really
have $\om<1$ for large $\ell$).
Proceeding as before, we find that to get the result desired, it is enough to
check that each of the series $\sum_{\ell}2^{-\ell}N(\ell)\log\ell$ and
$\sum_\ell2^{-\ell}N(\ell)\log(\frac{2^\ell}{N(\ell)})$ converges. But the
convergence of the first series is just a reformulation of the condition
$\sum_k\frac{\log\log\lb_k}{\lb_k}<\infty$ assumed, while the second series can
be split into two parts: the sum over $\ell$ for which
$\frac{N(\ell)}{2^\ell}>\frac{1}{\ell^2}$, which is actually the same as the
first series, and the sum over $\ell$ for which
$\frac{N(\ell)}{2^\ell}\le\frac{1}{\ell^2}$, which is dominated by
$\sum_\ell\frac{2\log\ell}{\ell^2}<\infty$.
\bigskip
It remains only to show that no weaker condition of the same type is
sufficient to get ($*$) without additional regularity assumptions.
Let $\ell\in\Z$, $\ell>0$. Put
$$
\Dl(\ell)=\log\left(\max_{j:\lb_j\in [2^\ell,2^{\ell+1})}
\prod\Sb k:\lb_k\in [2^\ell,2^{\ell+1}) \\ k\ne
j\endSb\frac{\lb_k+\lb_j}{|\lb_k-\lb_j|}\right)
$$
If the number of elements $\lb_k\in [2^\ell,2^{\ell+1})$ is less than $2$,
we put $\Dl(\ell)=0$ by definition (which is in good accordance with the
common agreement that the product of elements of the empty set is 1).
Otherwise,
obviously, $\Dl(\ell)>0$.
The statement (A) of the theorem can be generalized in the following way
without any essential change in the proof:
\st{Refined version of (A)}
If $\sum_{\ell}\frac{\Dl(\ell)}{2^\ell}<\infty$, then
$\varlimsup_{R\to\infty}\frac{\log M_{\G,f}(R)}{\log M_f(R)}=1$.
\medskip
Now we are going to show that if
$\sum_{\ell}\frac{\Dl(\ell)}{2^\ell}=\infty$, then there exists an entire
function $f(z)=\sum_kc_kz^{\lb_k}$ for which $\lim_{R\to\infty}\frac{\log
|f(R)|}{\log M_f(R)}=0$.
Consider all polynomials $P(x)=\sum_{k:\lb_k\in[2^\ell,2^{\ell+1})}a_k
z^{\lb_k}$
satisfying $|P(x)|\le\max(1,x^{2^{\ell+2}})$. Let $A_\ell$ be the maximal
possible coefficient of such a polynomial, $\Dt(\ell)\ed\log A_\ell$.
Our first aim is to show that
$$
\Dt(\ell)\ge\frac{\Dl(\ell)}{2}.
$$
\st{Lemma 4}
Assume that a function $F(z)$ is
analytic in the strip $0<\Re z<1$ and continuous up
to the boundary. Let $\zeta,\zeta_1,\dots,\zeta_m\in[\frac 14,\frac 12]$. If
$|F(z)|\le1$ in the whole strip and $F(\zeta_k)=0$ for $k=1,\dots,m$, then
$$
|F(\zeta)|\le\biggl(\prod_{k=1}^m\frac{|\zeta-\zeta_k|}{\zeta+
\zeta_k}\biggr)^{\frac 12}.
$$
\st{Proof}
It is clear that
$$
|F(\zeta)|\le\prod_k\frac{\sin\frac\pi 2|\zeta-\zeta_k|}{\sin
\frac\pi 2(\zeta+\zeta_k)}
$$
(the Blaschke product for the strip).
Note now that for every $x\in[0,1]$, we have \linebreak
$ x\le\sin\frac\pi 2 x\le
\frac\pi 2 x$ and
therefore
$$
\frac{\sin\frac\pi 2|\zeta-\zeta_k|}{\sin
\frac\pi 2(\zeta+\zeta_k)}\le\frac \pi 2
\frac{|\zeta-\zeta_k|}{\zeta+\zeta_k}\le\left(
\frac{|\zeta-\zeta_k|}{\zeta+\zeta_k}
\right)^{\frac12}.
$$
(The last inequality is due to the fact that
$\frac{|\zeta-\zeta_k|}{\zeta+\zeta_k}\le\frac13$ while $\frac\pi 2\le\sqrt
3$).
Now the inequality $\Dt(\ell)\ge\frac{\Dl(\ell)}{2}$ follows by standard
duality argument from the fact that for every (signed) measure $\mu$
satisfying $\int_0^1 1 d|\mu|(x)+\int_1^\infty x^{2^{\ell+2}}d|\mu|(x)\le1$
and $\int_0^\infty x^{\lb_k}d\mu(x)=0$ for $k\ne j$,
the function $F(z)\ed\int_0^\infty x^{2^{\ell+2}z}d\mu(x)$ satisfies the
conditions of Lemma 4 (with $\zeta=2^{-(\ell+2)}\lb_j$ etc.).
So now we have the condition $\sum_\ell\frac{\Dt(\ell)}{2^\ell}=\infty$.
The sequence $\{\Dt(\ell)\}$ may have very irregular behavior. Our next
aim will be to choose a subsequence $\{\ell_k\}$ such that
$\Dt(\ell_{k+1})\ge2\Dt(\ell_k)$ for every $k$,\ \
$\ell_{k+1}-\ell_{k}\to\infty$ as $k\to\infty$, and still
$\sum_k\frac{\Dt(\ell_k)}{2^{\ell_k}}=\infty$.
Note that for every $\ell_0$,
$$
\sum\Sb\ell\ge\ell_0 \\ \Dt(\ell)\le 2\Dt(\ell_0)\endSb
\frac{\Dt(\ell)}{2^\ell}\le 4\frac{\Dt(\ell_0)}{2^{\ell_0}}.
$$
Thus we can construct the sequence $\{\ell_k\}$ by induction taking for every
step the least possible $\ell_k$ for which $\Dt(\ell_k)\ge2\Dt(\ell_{k-1})$.
This sequence will satisfy our first regularity condition.
Now it remains to notice that for every $N>0$, $k_0>0$, at least one
of the series $\sum_{m\ge0}\frac{\Dt(\ell_{k_{0}+Nm+d})}{2^{k_0+Nm+d}}$
($d=1,\dots,N$) diverges and therefore starting from any $k_0$, we can
pull out a finite subsequence $\{\ell_{k_m}\}$
of the sequence $\{\ell_k\}$ such that $\ell_{k_{m+1}}-\ell_{k_m}\ge
k_{m+1}-k_m\ge N$ and
$\sum_{m}\frac{\Dt(\ell_{k_m})}{2^{k_m}}>1$, say.
Then we may take $N+1$ instead of $N$ and so on.
Dropping several first terms from the sequence $\{\ell_k\}$, we may also
assume that $\sum_ke^{-\frac{\Dt(\ell_k)}{2}}\le 1$.
Now everything is ready to construct the ``bad'' function $f(z)$.
Let $\{R_k\}$ be the sequence of positive numbers defined by $R_1=1$, $\log
R_{k}-\log R_{k-1}=\frac12\frac{\Dt(\ell_k)-\Dt(\ell_{k-1})}
{2^{\ell_k+1}}$ for
$k\ge2$. Note that the condition
$\sum_k\frac{\Dt(\ell_k)}{2^{\ell_k}}=\infty$ implies that $R_k\to\infty$ as
$k\to\infty$.
Let $P_k(x)$ be the polynomial at which $\Dt(\ell_k)$ is attained.
Put
$$
f(z)=\sum_ke^{-\frac{\Dt(\ell_k)}{2}}P_k\Bigl(\frac{z}{R_k}\Bigr).
$$
The convergence of the series can be easily justified if we note that
the condition
\linebreak
$|P_k(x)|\le\max(1,x^{2^{\ell_k+2}})$ implies (by subharmonicity of
$\log|P_k(z)|$) that $|P_k(z)|\le e^{A2^{\ell_k+2}}$ for every $|z|=1$ with some
absolute constant $A$ (which is just the maximal value over the unit
circumference of the harmonic continuation of the function $u(x)=\log^{+}x$
from $R_+$ to $\Bbb C\sm\R_+$).
As $P(z)$ has zero of multiplicity $2^{\ell_k}$ at the origin, the Schwartz
lemma yields that $|P(z)|\le1$ for every $|z|\le e^{-4A}$, which is enough to
get the convergence of the series defining $f(z)$ for every $z$.
Let $\psi(x)$ be the piecewise linear function that takes the values
$\frac{\Dt(\ell_k)}{2}$ at the points
$\log R_k$ and has the slope $2^{\ell_k+1}$
between $\log R_{k-1}$ and $\log R_k$.
Note that the existence of a coefficient $a_j$ in $P_k$ for which $|a_j|\ge
e^{\Dt(\ell_k)}$ implies that the central term of the function $f(z)$
$$
\mu(R)\ge\left(\frac{R}{R_k}\right)^{\lb_j}e^{\frac{\Dt(\ell_k)}{2}}\ge
\left(\frac{R}{R_k}\right)^{2^{\ell_k+1}}e^{\frac{\Dt(\ell_k)}{2}}
$$
and therefore
$$
\log\mu(R)\ge\frac{\Dt(\ell_k)}{2}+2^{\ell_k+1}(\log R-\log R_k)
=\psi(\log R)
$$
for every $R\in [R_{k-1},R_k]$.
Thus, $\log\mu(R)\ge\psi(\log R)$ for every $R\ge R_1$.
On the other hand, since $\sum_ke^{-\frac{\Dt(\ell_k)}{2}}\le1$,
we have $|f(R)|\le\max_k|P_k(\frac{R}{R_k})|$ and thereby
$
\log|f(R)|\le\max_k\log|P_k(\frac{R}{R_k})|
$.
But
$$
\log\left|P_k\Bigl(\frac{R}{R_k}\Bigr)\right|
\le\left\{\aligned &0\text{\qquad\ \ \ when }R\le
R_k;
\\
&2^{\ell_k+2}(\log R-\log R_k)\text{ when }R\ge R_k.
\endaligned\right.
$$
Note now that for all $x\ge\log R_k$ the slope of $\psi(x)$ is at least
$2^{\ell_{k+1}}$ and therefore \linebreak
$\log|P_k(\frac{R}{R_k})|\le
2^{\ell_k+2-\ell_{k+1}}\psi(\log R)$.
Since the difference $\ell_{k+1}-\ell_k\to\infty$ as $k\to\infty$, and since
for
every fixed $k$, we have $\lim_{R\to\infty}\frac{2^{\ell_k+2}(\log R-\log
R_k)}{\psi(\log R)}=0$ (because the slope of $\psi(x)$ tends to infinity
as $x\to\infty$), we may conclude that
$$
\lim_{R\to\infty}\frac{\log|f(R)|}{\mu(R)}=0,
$$
which is even (a very little bit) stronger than we needed.
Consider now a condition $\sum_kg(\lb_k)<\infty$ with some decreasing
function $g(x)$ for which $xg(x)$ increases. It is quite an easy exercise to
show that this condition implies the convergence of the series
$\sum\frac{\Dl(\ell)}{2^\ell}$ (which really governs the situation) if
and only if
$g(x)\ge const\log\log x$, finishing the proof of (B).
\bye