Growth of entire functions with sparse spectra.

Let f(z) = ∑_{k=0}^{∞} c_{k}z^{λ_{k}} be an entire function (c_{k} ∈ ℂ, 0 = λ_{0} < λ_{1} < ⋯ ∈ ℤ). Assume that ∑_{k=1}^{∞} \frac{1}{λ_{k}} < ∞.

Old conjecture.

There is no curve Γ joining 0 and ∞ such that f is bounded on Γ (unless f ≡ const, of course).

So far the only way to attack this problem was to show that for arbitrarily large R > 0,

M_{Γ,f}(R) =^{def} max_{|z|⩽R,z∈Γ} |f(z)| ≈ M_{f}(R) =^{def} max_{|z|⩽R} |f(z)|

(the idea that actually goes back to Littlewood and 1920’s). More precisely, the point was to show that

(∗)  \overline{lim}_{R→∞} \frac{log M_{Γ,f}(R)}{log M_{f}(R)} = 1.

This particular kind of interpretation is due to the fact that the standard symmetry trick allows to reduce the case of an arbitrary curve Γ to the case Γ = 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

Theorem.

(A) If the sequence {λ_{k}} is convex (i.e., if λ_{k+1} − λ_{k} ⩾ λ_{k} − λ_{k−1}), then the condition ∑_{k=1}^{∞} \frac{1}{λ_{k}} < ∞ really implies (∗). What is more, for every ε > 0, the set {R > 1 : M_{Γ,f}(R) ⩽ M_{f}(R)^{1−ε}} 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 {λ_{k}}, the weakest condition that still implies (∗) is ∑_{k=1}^{∞} \frac{log log λ_{k}}{λ_{k}} < ∞ (or, which is the same, ∑_{k=1}^{∞} \frac{log log k}{λ_{k}} < ∞).

Proof.

Let f(z) = ∑_{k=0}^{∞} c_{k}z^{λ_{k}}. Let μ(R) =^{def} max_{k}(|c_{k}|R^{λ_{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 μ(R) is attained.

Lemma 1.

M_{f}(R) ⩽ 4λ_{j(R)}^{2}μ(R) for all R > 1 except, maybe, a set of finite logarithmic measure.

Proof.

Let I_{j} = [a_{j}, b_{j}] be the interval of values of log R for which j(R) = j ⩾ 1. If log R ∈ [a_{j}, b_{j} − \frac{1}{λ_{j}^{2}} ], then for every z ∈ ℂ such that |z| = R, we have

|f(z)| = |∑_{k=0}^{∞} c_{k}z^{λ_{k}} | ⩽ ∑_{k=0}^{∞} |c_{k}|R^{λ_{k}} = ∑_{k=0}^{j−1} |c_{k}|R^{λ_{k}} + ∑_{k=j}^{∞} |c_{k}|R^{λ_{k}} =^{def} Σ_{1} + Σ_{2}

Since |c_{k}|R^{λ_{k}} ⩽ |c_{j}|R^{λ_{j}} for every k, we have Σ_{1} ⩽ jμ(R).

For k ⩾ 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/λ_{j}^{2}})^{λ_{k}} ⩽ |c_{j}|(R e^{1/λ_{j}^{2}})^{λ_{j}} and therefore

|c_{k}|R^{λ_{k}} ⩽ |c_{j}|R^{λ_{k}} e^{− \frac{λ_{k}−λ_{j}}{λ^{2}_{j}}} ⩽ μ(R) e^{−_{\frac{k−j}{λ^{2}_{j}}}} .

Thus,

Σ_{2} ⩽ μ(R) ∑_{k=0}^{∞} exp(−\frac{k}{λ_{j}^{2}} ) ⩽ 2λ_{j}^{2}μ(R)

Finally we get M_{f}(R) ⩽ (j + 2λ_{j}^{2})μ(R) ⩽ 4λ_{j}^{2}μ(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}{λ_{j}^{2}} , b_{j}] is not greater than ∑_{k=1}^{∞} \frac{1}{λ_{j}^{2}} ⩽ ∑_{j=1}^{∞} \frac{1}{j^{2}} < ∞.

Now let us turn to the proof of the theorem. Consider first the case when the sequence {λ_{k}} is convex.

Recall the formula for the distance D in L^{2}(0, 1) from the function x^{λ} to span(x^{λ_{0}}, …, x^{λ_{m}}):

D = \frac{1}{\sqrt{2λ + 1}} ∏_{k=0}^{m} \frac{|λ − λ_{k}|}{λ + λ_{k} + 1}.

A simple consequence of this formula is that for every polynomial P (x) = ax^{λ} + ∑_{k=0}^{m} a_{k}x^{λ_{k}}, the inequality

max_{x∈[0,r]} |P (x)| ⩾ \frac{|a|r^{λ}}{\sqrt{2λ + 1}} ∏_{k=0}^{m} \frac{|λ − λ_{k}|}{λ + λ_{k} + 1}

holds.

Lemma 2.

If j(R) ⩾ 1 then M_{Γ,f}(R) ⩾ e^{−15j(R)}λ_{j(R)}^{−3}μ(R) for every ray Γ = {tξ : t > 0}.

Proof.

Without loss of generality we may assume that Γ = ℝ_{+}. Also we shall write j instead of j(R) for convenience.

Write f(z) = ∑_{k=0}^{∞} c_{k}z^{λ_{k}} = ∑_{k=0}^{2j} c_{k}z^{λ_{k}} + ∑_{k>2j} c_{k}z^{λ_{k}} =^{def} P (z) + Q(z).

Let ω > 0. We have

M_{Γ,f}(R) ⩾ max_{x∈[0,Re^{−ω}]} |f(x)| ⩾ max_{x∈[0,Re^{−ω}]} |P (x)| − max_{x∈[0,Re^{−ω}]} |Q(x)|.

According to the consequence of the distance formula mentioned above,

max_{x∈[0,Re^{−ω}]} |P (x)| ⩾ |c_{j}|(R e^{−ω})^{λ_{j}} \frac{1}{\sqrt{2λ_{j} + 1}} ∏_{\begin{matrix}0⩽k⩽2j \\ k≠j \end{matrix}} \frac{|λ_{j} − λ_{k}|}{λ_{j} + λ_{k} + 1}.

The convexity of the sequence \{λ_{k}\} implies λ_{m} ⩽ \frac{m}{n}λ_{n} for every m < n and thereby \frac{|λ_{n}−λ_{m}|}{λ_{n}+λ_{m}+1} ⩾ \frac{|m−n|}{m+n+1} for every m, n ⩾ 0.

Thus

∏_{\begin{matrix}0⩽k⩽2j \\ k≠j \end{matrix}} \frac{|λ_{j} − λ_{k}|}{λ_{j} + λ_{k} + 1} ⩾ ∏_{\begin{matrix}0⩽k⩽2j \\ k≠j \end{matrix}} \frac{|j − k|}{j + k + 1} ⩾ \frac{(j!)^{2}}{(2j + 1)^{2j}} ⩾ e^{−5j}, say.

Together with the trivial estimate \frac{1}{\sqrt{2λ_{j}+1}} ⩾ \frac{1}{2λ_{j}}, this yields

max_{x∈[0,Re^{−ω}]} |P (x)| ⩾ |c_{j}|(R e^{−ω})^{λ_{j}} \frac{1}{2λ_{j}} e^{−5j} ⩾ \frac{1}{λ_{j}} e^{−6j} e^{−ωλ_{j}}μ(R).

As to Q(x), we have |c_{k}|(R e^{−ω})^{λ_{k}} = |c_{k}|R^{λ_{k}} e^{−ωλ_{k}} ⩽ |c_{j}|R^{λ_{j}} e^{−ωλ_{k}} = μ(R) e^{−ωλ_{k}}. Thus, for x ∈ [0, R e^{−ω}],

|Q(x)| ⩽ μ(R) ∑_{k>2j} e^{−ωλ_{k}} ⩽ μ(R)\frac{1}{1 − e^{−ω}} e^{−ωλ_{2j}} ⩽ (1 + \frac{1}{ω}) e^{−2ωλ_{j}}μ(R)

because (again, due to convexity) λ_{2j} ⩾ 2λ_{j}.

So, to get a nontrivial estimate for M_{Γ,f}(R) from below, we should choose ω > 0 satisfying

\frac{1}{λ_{j}} e^{−6j} e^{−ωλ_{j}} ⩾ 2(1 + \frac{1}{ω}) e^{−2ωλ_{j}}

Simple computation shows that ω = \frac{1}{λ_{j}} (8j + 2 log λ_{j}) suffices. This yields the estimate

M_{Γ,f}(R) ⩾ \frac{1}{2}λ_{j}^{−3} e^{−14j}μ(R) ⩾ λ_{j}^{−3} e^{−15j}μ(R)

desired.

The symmetry trick allows to derive from Lemma 2 the estimate

M_{Γ,f}(R) ⩾ e^{−30j(R)}λ_{j(R)}^{−6}μ(R)^{2}M_{f}(R)^{−1}

for an arbitrary curve Γ. Together with the estimate from Lemma 1, it yields the inequality

M_{Γ,f}(R) ⩾ \frac{1}{4} e^{−30j(R)}λ_{j(R)}^{−8}μ(R)

valid for all R > 1 except, maybe, a set of finite logarithmic measure.

Assume now that M_{Γ,f}(R) ⩽ M_{f}(R)^{1−ε} and that R doesn’t belong to the exceptional set. Then we have

\frac{1}{4} e^{−30j(R)}λ_{j(R)}^{−8}μ(R) ⩽ (4λ_{j}(R)^{2}μ(R))^{1−ε}

and therefore

(∗∗)  log μ(R) ⩽ A(ε)(j(R) + log λ_{j(R)})

Let’s show now that the set of all R > 1 satisfying (∗∗) has finite logarithmic measure. Let N(ℓ) be the number of the elements λ_{k} that are less than 2^{ℓ}. The condition ∑_{k=1}^{∞} \frac{1}{λ_{k}} < ∞ is equivalent to ∑_{ℓ⩾1} \frac{N(ℓ)}{2^{ℓ}} < ∞. Note that the slope of log μ(R) against log R is λ_{j(R)}. Therefore the logarithmic measure of the set of “good” (in the sense of Lemma 1) R for which (∗∗) is satisfied, λ_{j(R)} ∈ [2^{ℓ−1}, 2^{ℓ}), j(R) ⩾ 1 and μ(R) > 1 (the last two conditions are not very essential since they are fulfilled automatically for all sufficiently large R), doesn’t exceed A(ε)2^{−(ℓ−1)}(j(R) + log λ_{j(R)}) ⩽ 2A(ε)2^{−ℓ}(N(ℓ) + ℓ) because j(R) ⩽ N(ℓ) and log λ_{j(R)} ⩽ ℓ.

Thus, the logarithmic measure of the set of “large” R for which (∗∗) holds doesn’t exceed 2A(ε) ∑_{ℓ⩾1} \frac{N(ℓ)+ℓ}{2^{ℓ}} < ∞, proving the statement (A) of the theorem.

Let’s now show that the condition ∑_{k⩾2} \frac{log log λ_{k}}{λ_{k}} < ∞ 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 ∈ (0, R) such that m_{f}(r) =^{def} min_{|z|=r} |f(z)| ⩾ M_{f}(R)^{1−ε}. (From the proof of (A), it follows only that for most R the connected component of the open set {z : |f(z)| < M_{f}(R)^{1−ε}} containing the origin lies in some disk of radius strictly less then R and therefore the origin can be surrounded by some simple closed Jordan curve on which f is everywhere large but there is no warranty that this curve is a circumference centered at the origin).

The core of the reasoning below is the following

Lemma 3 (factorization of lacunary polynomials).

Let P (z) = ∑_{k=0}^{m} c_{k}z^{λ_{k}}, 0 ⩽ j ⩽ m, d_{k} =^{def} |λ_{k} − λ_{j}|. Then

P (z) = c_{j}z^{λ_{j}} ∏_{\begin{matrix}0⩽k⩽m \\ k≠j \end{matrix}} φ_{k}(z)^{−1}

where

φ_{k}(z) =^{def} \{ \begin{matrix}\frac{1}{d_{k}} ∑_{ℓ=1}^{D_{k}} \frac{z}{z − ζ_{kℓ}} if λ_{k} < λ_{j}; \\ \frac{1}{d_{k}} (D_{k} − ∑_{ℓ=1}^{D_{k}} \frac{z}{z − ζ_{kℓ}}) if λ_{k} > λ_{j}, \end{matrix}

and D_{k} ⩽ 2d_{k} for every k.

Proof.

If d_{m} ⩽ d_{0} put Q(z) =^{def} z^{−λ_{0}}P (z) and P_{1}(z) =^{def} \frac{1}{d_{0}} z^{λ_{0}+1} \frac{d}{dz} (z^{−λ_{0}}P (z)).

Note that P_{1}(z) = ∑_{k=0}^{m} c_{k} \frac{λ_{k}−λ_{0}}{d_{0}} z^{λ_{k}} and therefore the term c_{0}z^{λ_{0}} disappears while the coefficient at z^{λ_{j}} doesn’t change. Besides,

\frac{d_{0}P_{1}(z)}{P (z)} = \frac{zQ′(z)}{Q(z)} = ∑_{ℓ=1}^{D_{0}} \frac{z}{z − ζ_{0ℓ}}

where ζ_{0ℓ} are the roots of the polynomial Q(z) of degree D_{0} = λ_{m} −λ_{0} = d_{0} +d_{m} ⩽ 2d_{0}.

If d_{m} > d_{0} put Q(z) =^{def} z^{−λ_{m}}P (z) and P_{1}(z) =^{def} \frac{1}{d_{m}} z^{λ_{m}+1} \frac{d}{dz} (z^{−λ_{m}}P (z)).

Everything remains the same except Q(z) is now a polynomial of degree D_{m} ⩽ 2d_{m} in 1/z, not in z, and thereby

\frac{d_{m}P_{1}(z)}{P (z)} = \frac{zQ′(z)}{Q(z)} = ∑_{ℓ=1}^{D_{0}} \frac{\frac{1}{z}}{\frac{1}{z} − \frac{1}{ζ_{mℓ}}} = − ∑_{ℓ=1}^{D_{m}} \frac{ζ_{mℓ}}{z − ζ_{mℓ}} = D_{m} − ∑_{ℓ=1}^{D_{m}} \frac{z}{z − ζ_{mℓ}}

where 1/ζ_{mℓ} 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.

Now we want to show that the functions φ_{k}(z) are small on most circumferences centered at the origin. Note that

|φ_{k}(z)| ⩽ 2(1 + \frac{1}{D_{k}} ∑_{ℓ=1}^{D_{k}} \frac{|z|}{|z − ζ_{kℓ}|}) ⩽ 2(1 + \frac{1}{D_{k}} ∑_{ℓ=1}^{D_{k}} \frac{|z|}{| |z| − |ζ_{kℓ}| |}).

Lemma 3.

Let φ(x) = \frac{1}{D} ∑_{ℓ=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 : φ(x) > A\} doesn’t exceed \frac{9+4 log D}{A}.

Proof.

Let E_{ℓ} =^{def} \{x > 0 : |x − x_{ℓ}| ⩽ \frac{3}{A}x_{ℓ}\}. We have

\frac{x}{|x − x_{k}|} = \frac{x}{|x − x_{k}|}χ_{E_{ℓ}} + \frac{x}{|x − x_{k}|}χ_{ℝ_{+}\E_{ℓ}} =^{def} η_{ℓ}(x) + θ_{ℓ}(x)

Note that for every ℓ we have θ_{ℓ}(x) ⩽ \frac{A}{3}(1+ \frac{3}{A}) = \frac{A}{3} +1 ⩽ \frac{A}{2}. Thus, \frac{1}{D} ∑_{ℓ=1}^{D} θ_{ℓ}(x) ⩽

\frac{A}{2}. Let F_{ℓ} =^{def} \{x > 0 : |x − x_{ℓ}| ⩽ \frac{3}{AD} x_{ℓ}\}, F =^{def} ⋃_{ℓ} F_{ℓ}. The logarithmic measure

of F doesn’t exceed D log \frac{AD+3}{AD−3} ⩽ D \frac{9}{AD} = \frac{9}{A}. To finish the proof of Lemma 3 it remains only to note that for every ℓ

∫_{ℝ_{+}\F} η_{ℓ}(x)\frac{dx}{x} ⩽ ∫_{ℝ_{+}\F_{ℓ}} η_{ℓ}(x)\frac{dx}{x} = 2 ∫ \frac{3}{A} ; \frac{3}{AD} \frac{dx}{x} = 2 log D.

It follows from Lemma 3 that for every A ⩾ 1, the logarithmic measure of the set

{r : there exists z for which |z| = r and |φ_{k}(z)| > 4(9 + 4 log D_{k})A}

is less than \frac{1}{A}. Though weak type estimates are difficult to add, they are easy to multiply and we conclude that every set S ⊂ ℝ_{+} of logarithmic measure ω < 1 contains a point r such that

∏_{\begin{matrix}0⩽k⩽m \\ k≠j \end{matrix}} φ_{k}(z)^{−1} ⩾ (\frac{ω}{4(9 + 4 log λ_{m})e})^{m}

(we have used the trivial estimate D_{k} ⩽ λ_{m}).

Now everything is ready to prove that the condition ∑ \frac{log log λ_{k}}{λ_{k}} < ∞ implies (∗). Let j = j(R) ⩾ 1. Assume that 2^{ℓ−1} ⩽ λ_{j} < 2^{ℓ}. Again we are going to use the

decomposition f(z) = P (z) + Q(z) where

P (z) =^{def} ∑_{k:λ_{k}<2^{ℓ+2}} c_{k}z^{λ_{k}}, Q(z) =^{def} ∑_{k:λ_{k}⩾2^{ℓ+2}} c_{k}z^{λ_{k}}.

As before, we observe that for every |z| ⩽ Re^{−ω},

|Q(z)| ⩽ (1 + \frac{1}{ω})e^{−2^{ℓ+2}ω}μ(R)

while there exists r ∈ [Re^{−2ω}, Re^{−ω}] such that for every |z| = r,

|P (z)| ⩾ (\frac{ω}{4(9 + 4 log(2^{ℓ+2}))e})^{N(ℓ+2)} e^{−2^{ℓ+1}ω}μ(R) ⩾ (\frac{ω}{1000ℓ})^{N(ℓ+2)} e^{−2^{ℓ+1}ω}μ(R).

Again, to get a nontrivial result, we must choose ω < 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

ω = 2^{−ℓ}N(ℓ + 2) [log(1000ℓ) + log(\frac{2^{ℓ}}{N(ℓ + 2)}) + ℓ + 3]

(since the condition ∑_{k=2}^{∞} \frac{log log λ_{k}}{λ_{k}} < ∞ implies that 2^{−ℓ}N(ℓ) log ℓ → 0 as ℓ → ∞, we really have ω < 1 for large ℓ).

Proceeding as before, we find that to get the result desired, it is enough to check that each of the series ∑_{ℓ} 2^{−ℓ}N(ℓ) log ℓ and ∑_{ℓ} 2^{−ℓ}N(ℓ) log(\frac{2^{ℓ}}{N(ℓ)}) converges. But the convergence of the first series is just a reformulation of the condition ∑_{k} \frac{log log λ_{k}}{λ_{k}} < ∞ assumed, while the second series can be split into two parts:

the sum over ℓ for which \frac{N(ℓ)}{2^{ℓ}} > \frac{1}{ℓ^{2}} , which is actually the same as the first series, and the sum over ℓ for which \frac{N(ℓ)}{2^{ℓ}} ⩽ \frac{1}{ℓ^{2}} , which is dominated by ∑_{ℓ} \frac{2 log ℓ}{ℓ^{2}} < ∞.

It remains only to show that no weaker condition of the same type is sufficient to get (∗) without additional regularity assumptions.

Let ℓ ∈ ℤ, ℓ > 0. Put

Δ(ℓ) = log ( max_{j:λ_{j}∈[2^{ℓ},2^{ℓ+1})} ∏_{\begin{matrix}k:λ_{k}∈[2^{ℓ},2^{ℓ+1}) \\ k≠j \end{matrix}} \frac{λ_{k} + λ_{j}}{|λ_{k} − λ_{j}|})

If the number of elements λ_{k} ∈ [2^{ℓ}, 2^{ℓ+1}) is less than 2, we put Δ(ℓ) = 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, Δ(ℓ) > 0.

The statement (A) of the theorem can be generalized in the following way without any essential change in the proof:

Refined version of (A).

If ∑_{ℓ} \frac{Δ(ℓ)}{2^{ℓ}} < ∞, then \overline{lim}_{R→∞} \frac{log M_{Γ,f} (R)}{log M_{f} (R)} = 1.

Now we are going to show that if ∑_{ℓ} \frac{Δ(ℓ)}{2^{ℓ}} = ∞, then there exists an entire function f(z) = ∑_{k} c_{k}z^{λ_{k}} for which lim_{R→∞} \frac{log |f(R)|}{log M_{f} (R)} = 0.

Consider all polynomials P (x) = ∑_{k:λ_{k}∈[2^{ℓ},2^{ℓ+1})} a_{k}z^{λ_{k}} satisfying |P (x)| ⩽ max(1, x^{2^{ℓ+2}}).

Let A_{ℓ} be the maximal possible coefficient of such a polynomial, Δ̃(ℓ) =^{def} log A_{ℓ}. Our first aim is to show that

Δ̃(ℓ) ⩾ \frac{Δ(ℓ)}{2}.

Lemma 4.

Assume that a function F (z) is analytic in the strip 0 < ℜz < 1 and continuous up to the boundary. Let ζ, ζ_{1}, …, ζ_{m} ∈ [\frac{1}{4}, \frac{1}{2}]. If |F (z)| ⩽ 1 in the whole strip and F (ζ_{k}) = 0 for k = 1, …, m, then

|F (ζ)| ⩽ (∏_{k=1}^{m} \frac{|ζ − ζ_{k}|}{ζ + ζ_{k}})^{\frac{1}{2}} .

Proof.

It is clear that

|F (ζ)| ⩽ ∏_{k} \frac{sin \frac{π}{2} |ζ − ζ_{k}|}{sin \frac{π}{2} (ζ + ζ_{k})}

(the Blaschke product for the strip). Note now that for every x ∈ [0, 1], we have x ⩽ sin \frac{π}{2} x ⩽ \frac{π}{2} x and therefore

\frac{sin \frac{π}{2} |ζ − ζ_{k}|}{sin \frac{π}{2} (ζ + ζ_{k})} ⩽ \frac{π}{2} \frac{|ζ − ζ_{k}|}{ζ + ζ_{k}} ⩽ (\frac{|ζ − ζ_{k}|}{ζ + ζ_{k}})^{\frac{1}{2}} .

(The last inequality is due to the fact that \frac{|ζ−ζ_{k}|}{ζ+ζ_{k}} ⩽ \frac{1}{3} while \frac{π}{2} ⩽ \sqrt{3}).

Now the inequality Δ̃(ℓ) ⩾ \frac{Δ(ℓ)}{2} follows by standard duality argument from the fact that for every (signed) measure μ satisfying ∫_{0}^{1} 1d|μ|(x) + ∫_{1}^{∞} x^{2^{ℓ+2}}d|μ|(x) ⩽ 1

and ∫_{0}^{∞} x^{λ_{k}}dμ(x) = 0 for k ≠ j, the function F (z) =^{def} ∫_{0}^{∞} x^{2^{ℓ+2}z}dμ(x) satisfies the conditions of Lemma 4 (with ζ = 2^{−(ℓ+2)}λ_{j} etc.).

So now we have the condition ∑_{ℓ} \frac{_{Δ}(ℓ)}{2^{ℓ}} = ∞. The sequence \{Δ̃(ℓ)\} may have very irregular behavior. Our next aim will be to choose a subsequence {ℓ_{k}} such that

Δ̃(ℓ_{k+1}) ⩾ 2Δ̃(ℓ_{k}) for every k, ℓ_{k+1} − ℓ_{k} → ∞ as k → ∞, and still ∑_{k} \frac{_{Δ}(ℓ_{k})}{2^{ℓ_{k}}} = ∞.

Note that for every ℓ_{0},

∑_{\begin{matrix}ℓ⩾ℓ_{0} \\ _{Δ}(ℓ)⩽2_{Δ}(ℓ_{0}) \end{matrix}} \frac{Δ̃(ℓ)}{2^{ℓ}} ⩽ 4\frac{Δ̃(ℓ_{0})}{2^{ℓ_{0}}}.

Thus we can construct the sequence {ℓ_{k}} by induction taking for every step the least possible ℓ_{k} for which Δ̃(ℓ_{k}) ⩾ 2Δ̃(ℓ_{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 ∑_{m⩾0} \frac{_{Δ}(ℓ_{k_{0}+Nm+d})}{2^{k_{0}+Nm+d}} (d = 1, …, N) diverges and therefore starting from any k_{0}, we can pull out a finite subsequence {ℓ_{k_{m}}} of the sequence {ℓ_{k}} such

that ℓ_{k_{m+1}} − ℓ_{k_{m}} ⩾ k_{m+1} − k_{m} ⩾ N and ∑_{m} \frac{_{Δ}(ℓ_{km})}{2^{km}} > 1, say. Then we may take

N + 1 instead of N and so on. Dropping several first terms from the sequence {ℓ_{k}},

we may also assume that ∑_{k} e^{− \frac{_{Δ}(ℓ_{k})}{2}} ⩽ 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} = \frac{1}{2} \frac{_{Δ}(ℓ_{k})−_{Δ}(ℓ_{k−1})}{2^{ℓ_{k}+1}}

for k ⩾ 2. Note that the condition ∑_{k} \frac{_{Δ}(ℓ_{k})}{2^{ℓ_{k}}} = ∞ implies that R_{k} → ∞ as k → ∞. Let P_{k}(x) be the polynomial at which Δ̃(ℓ_{k}) is attained. Put

f(z) = ∑_{k} e^{− \frac{_{Δ}(ℓ_{k})}{2}} P_{k}(\frac{z}{R_{k}} ).

The convergence of the series can be easily justified if we note that the condition |P_{k}(x)| ⩽ max(1, x^{2^{ℓ_{k}+2}}) implies (by subharmonicity of log |P_{k}(z)|) that |P_{k}(z)| ⩽ e^{A2^{ℓ_{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 ℂ \ ℝ_{+}). As P (z) has zero of multiplicity 2^{ℓ_{k}} at the origin, the Schwartz lemma yields that |P (z)| ⩽ 1 for every |z| ⩽ e^{−4A}, which is enough to get the convergence of the series defining f(z) for every z.

Let ψ(x) be the piecewise linear function that takes the values \frac{_{Δ}(ℓ_{k})}{2} at the points log R_{k} and has the slope 2^{ℓ_{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}| ⩾ e^{_{Δ}(ℓ_{k})} implies that the central term of the function f(z)

μ(R) ⩾ (\frac{R}{R_{k}} )^{λ_{j}} e^{\frac{_{Δ}(ℓ_{k})}{2}} ⩾ (\frac{R}{R_{k}} )^{2^{ℓ_{k}+1}} e^{\frac{_{Δ}(ℓ_{k})}{2}}

and therefore

log μ(R) ⩾ \frac{Δ̃(ℓ_{k})}{2} + 2^{ℓ_{k}+1}(log R − log R_{k}) = ψ(log R)

for every R ∈ [R_{k−1}, R_{k}]. Thus, log μ(R) ⩾ ψ(log R) for every R ⩾ R_{1}.

On the other hand, since ∑_{k} e^{− \frac{_{Δ}(ℓ_{k})}{2}} ⩽ 1, we have |f(R)| ⩽ max_{k} |P_{k}(\frac{R}{R_{k}})| and thereby log |f(R)| ⩽ max_{k} log |P_{k}(\frac{R}{R_{k}})|. But

log |P_{k}(\frac{R}{R_{k}} )| ⩽ \{ \begin{matrix}0 when R ⩽ R_{k}; \\ 2^{ℓ_{k}+2}(log R − log R_{k}) when R ⩾ R_{k}. \end{matrix}

Note now that for all x ⩾ log R_{k} the slope of ψ(x) is at least 2^{ℓ_{k+1}} and therefore log |P_{k}(\frac{R}{R_{k}})| ⩽ 2^{ℓ_{k}+2−ℓ_{k+1}}ψ(log R).

Since the difference ℓ_{k+1} − ℓ_{k} → ∞ as k → ∞, and since for every fixed k, we have lim_{R→∞} \frac{2^{ℓ_{k}+2}(log R−log R_{k})}{ψ(log R)} = 0 (because the slope of ψ(x) tends to infinity as x → ∞), we may conclude that

lim_{R→∞} \frac{log |f(R)|}{μ(R)} = 0,

which is even (a very little bit) stronger than we needed.

Consider now a condition ∑_{k} g(λ_{k}) < ∞ 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 ∑ \frac{Δ(ℓ)}{2^{ℓ}} (which really governs the situation) if and only if g(x) ⩾ const log log x, finishing the proof of (B).