The Beurling lemma.

Theorem 1: Lebesgue measure version. Let φ ∈ L^{1}(ℝ), φ ⩾ 0. For every u > 0 define Y_{u}(t) = Y_{u}(φ; t) as the solution of the equation

∫_{ℝ} \frac{φ(s)ds}{(s − t)^{2} + Y_{u}(t)^{2}} = u

Then

∫_{ℝ} Y_{u}(t)dt ⩽ \frac{Const}{u} ∫_{ℝ} φ(t)dt.

Proof:

1) Some trivia (switching to the dyadic version).

Let D be any dyadic lattice. For I ∈ D define

f_{I} = f_{I} (φ) := \frac{1}{|I|^{2}} ∫_{I} φ(t)dt; P_{I} = P_{I} (φ) := ∑_{J∈D:I⊂J} f_{J} .

Note that f_{I} ⩽ |I|^{−2} ∫_{ℝ} φ(t)dt and therefore the series in the definition of P_{I} converges and P_{I} → 0 as |I| → ∞.

Define

Y_{u}^{D}(t) := sup{|I| : t ∈ I ∈ D, P_{I} ⩾ u}

(i.e. we use the dyadic Poisson kernel instead of the smooth one). Random dyadic lattice.

Let Ω be any probability space. Let x(ω) be a random variable uniformly distributed over (0, 1). Let ξ_{j}(ω) (j = 1, 2, …) be random variables such that ℙ\{ξ_{j} = 1\} = ℙ\{ξ_{j} = −1\} = \frac{1}{2}. Assume at last that x(ω) and ξ_{j}(ω) are independent.

Construct the random lattice D(ω) as follows.

1) Let I_{0}(ω) = [x(ω) − 1, x(ω)] ∈ D(ω). This uniquely determines all intervals in D(ω) of length 2^{k} with k ⩽ 0.

2) Intervals I_{k}(ω) ∈ D(ω) of length 2^{k} are determined inductively. If I_{k−1} ∈ D(ω) is already chosen, then (I_{k}(ω))_{+} = I_{k−1}(ω) if ξ_{k}(ω) = +1 and (I_{k}(ω))_{−} = I_{k−1}(ω) if ξ_{k}(ω) = −1 (as usual for an interval I ⊂ ℝ we denote by I_{−} and I_{+} its left and right halves respectively). Lemma. For sufficiently small λ > 0

𝔼 Y_{λu}^{D(ω)}(t) ⩾ λY_{u}(t)

for every t ∈ ℝ, u > 0. Proof:

Sublemma.

Typeset by AMS-TEX

Suppose that ξ is a random variable such that ξ(ω) ⩽ Cu for every ω ∈ Ω and 𝔼ξ ⩾ cu (0 < c ⩽ C).

Then

ℙ\{ξ ⩾ \frac{c}{2}u\} ⩾ \frac{c}{2C} .

Proof:

cu = 𝔼ξ = ∫_{Ω} ξ(ω)dℙ(ω) = ∫_{\{ξ⩾ \frac{c}{2} u\}} ⋯ + ∫_{\{ξ< \frac{c}{2} u\}} … ⩽ ℙ\{ξ ⩾ \frac{c}{2}u\}Cu + \frac{c}{2}u.

Now let a > 0. Fix t ∈ ℝ and define I(ω) as the interval in D(ω) such that t ∈ I(ω) and a ⩽ |I(ω)| < 2a. Note that |I(ω)| does not depend on ω: it is just the only integer power of 2 between a and 2a.

We have

P_{I(ω)} = ∑_{J∈D(ω):J⊃I(ω)} \frac{1}{|J|^{2}} ∫_{J} φ(s)ds = ∫_{ℝ} φ(s)(  ∑_{J∈D(ω):J⊃I(ω)} \frac{1}{|J|^{2}} χ_{J} (s))ds.

Note that

∑_{J∈D(ω):J⊃I(ω)} \frac{1}{|J|^{2}} χ_{J} (s) ⩽ \frac{C}{(s − t)^{2} + a^{2}}

for every ω with some absolute constant C > 0, and therefore

P_{I(ω)} ⩽ C ∫_{ℝ} \frac{φ(s)ds}{(s − t)^{2} + a^{2}} .

On the other hand,

𝔼P_{I(ω)} = ∫_{ℝ} φ(s)(∑_{j=0}^{∞} τ_{2^{j}|I(ω)|}(s − t))ds

where

τ_{A}(s) = \frac{1}{A^{2}} (1 − \frac{|s|}{A})_{+}

is the “triangle” kernel.

Since

∑_{j=0}^{∞} τ_{2^{j}|I(ω)|}(s) ⩾ \frac{c}{s^{2} + a^{2}} ,

with some small c > 0, we get

𝔼P_{I(ω)} ⩾ c ∫_{ℝ} \frac{φ(s)ds}{(s − t)^{2} + a^{2}} .

Hence, due to the sublemma,

ℙ\{P_{I(ω)} ⩾ \frac{c}{2} ∫_{ℝ} \frac{φ(s)ds}{(s − t)^{2} + a^{2}} \} ⩾ \frac{c}{2C} > 0.

Put now a = Y_{u}(t). Then we conclude that

ℙ\{P_{I(ω)} ⩾ \frac{c}{2}u\} ⩾ \frac{c}{2C} .

But P_{I(ω)} ⩾ \frac{c}{2}u ⟺ Y_{\frac{c}{2} u}^{D(ω)}(t) ⩾ |I(ω)| ⩾ a = Y_{u}(t). Therefore we get

𝔼Y_{\frac{c}{2} u}^{D(ω)}(t) ⩾ \frac{c}{2C} Y_{u}(t)

and it remains to put λ = min\{\frac{c}{2}, \frac{c}{2C} \}.

This lemma shows that it is enough to prove that for every dyadic grid D

∫_{ℝ} Y_{u}^{D}(t)dt ⩽ \frac{Const}{u} ∫_{ℝ} φ(t)dt.

2) Translation into the Bellman function language. Since clearly Y_{Au}^{D}(Aφ; t) = Y_{u}^{D}(φ; t), we may restrict ourselves to the case u = 1. Let I^{0} ∈ D be so large that |I^{0}|^{−2} ∫_{ℝ} φ(t)dt << 1 (<< means “much less than”). Then, of course, also f_{I^{0}} = |I^{0}|^{−2} ∫_{I^{0}} φ(t)dt << 1 and even

P_{I^{0}} = ∑_{J∈D:J⊃I^{0}} f_{J} ⩽ (∫_{ℝ} φ(t)dt)( ∑_{J∈D:J⊃I^{0}} |J|^{−2}) = \frac{4}{3}|I^{0}|^{−2} ∫_{ℝ} φ(t)dt << 1.

The “averages” f_{I} and P_{I} will be very important for us. Let us look at their dynamics. It is given by three following relations:

f_{I} = \frac{f_{I_{+}} + f_{I_{−}}}{4} ;  P_{I_{+}} = P_{I} + f_{I_{+}} ,  P_{I_{−}} = P_{I} + f_{I_{−}}  (*)

Fix an interval I ∈ D and define the Bellman function corresponding to the Beurling lemma as

B(f, P) = sup\{\frac{1}{|I|^{2}} ∫_{I} Y_{1}^{D}(φ; t)dt : f_{I} (φ) = f; P_{I} (φ) = P \}.

It is easy to check that the normalization is chosen right, i.e. B(f, P) does not depend on the choice of I.

3) Heuristic: properties of the Bellman function.

Domain: Clearly,

P_{I} = ∑_{J∈D:J⊃I} |J|^{−2} ∫_{J} φ(t)dt ⩾ ∑_{J∈D:J⊃I} |J|^{−2} ∫_{I} φ(t)dt = \frac{4}{3}|I|^{−2} ∫_{I} φ(t)dt = \frac{4}{3}f_{I} .

Also, we can hope to have Y_{1}^{D}(t) under control on the interval I only if P_{I} < 1 (then we are sure that for every t ∈ I the maximal interval J ∋ t for which P_{J} ⩾ 1 is a subinterval of I). So, the whole description of the domain is

0 ⩽ \frac{4}{3}f ⩽ P < 1.

Range: When P_{I} < 1 we clearly have Y_{1}^{D}(t) ⩽ |I| for every t ∈ I. Thus we get the “trivial” estimate

B(f, P) ⩽ 1.

We will call a pair f = f_{I} , P = P_{I} terminal if there is a possibility that for one of the intervals I_{+} and I_{−} the corresponding pair lies outside the domain. Trivial computation shows that a pair (f, P) is terminal if and only if P + 4f ⩾ 1. Note that for all terminal pairs

B(f, P) ⩾ \frac{1}{4}.

Indeed, consider the function φ such that φ ≡ 2f|I| on I_{−}; φ ≡ 0 on I_{+} (this automatically gives f_{I} (φ) = f) and φ is defined outside I in some way to get P_{I} (φ) = P (it is an excercise for the reader to show that it is possible, provided that P ⩾ \frac{4}{3}f). Then P_{I_{−}} (φ) = P + 4f ⩾ 1 and thereby Y_{1}^{D}(t) = |I_{−}| = |I|/2 for every t ∈ I_{−}. This results in B(f, P) ⩾ \frac{1}{4} immediately.

For pairs which are far from terminal (i.e. when f, P << 1) we should have much better estimate like

B(f, P) ⩽ C(f + P)

(our belief that the theorem is true formalized: consider a function supported by I and write down what the theorem means for it). Key property: If (f, P), (f_{+}, P_{+}), (f_{−}, P_{−}) belong to the domain and satisfy the dynamics relations (∗), i.e.

f = \frac{f_{+} + f_{−}}{4} ;  P_{+} = P + f_{+},  P_{−} = P + f_{−},

then

B(f, P) ⩾ \frac{1}{4}\{B(f_{+}, P_{+}) + B(f_{−}, P_{−})\}.

Note that if the pair (f, P) is not terminal, then the pairs (f_{±}, P_{±}) belong to the domain automatically:

P_{±} = f_{±} + P ⩽ P + f_{+} + f_{−} = P + 4f < 1.

Now everything is ready to run the machine backward.

4) Running the machine backward. Note that this is the point from which the formal proof begins.

Suppose we have any function B(f, P) defined in the region

0 ⩽ \frac{4}{3}f ⩽ P < 1

and satisfying

1) B(f, P) ⩾ \frac{1}{4} for every terminal pair (f, P); 2) B(f, P) ⩽ C(f + P);

3) B(f, P) ⩾ \frac{1}{4}\{B(f_{−}, P_{−}) + B(f_{+}, P_{+})\} whenever all three pairs belong to the domain and satisfy the dynamics relations (∗).

Then the theorem is proved. Indeed, let us take a very large interval I^{0} as above (so, for sure, f_{I^{0}} , P_{I^{0}} << 1). Draw a binary tree in the following way: start from I^{0} and go down stopping at terminal intervals (i.e. at intervals J ∈ D for which P_{J} + 4f_{J} ⩾ 1). Put the quantity |J|^{2}B(f_{J} , P_{J}) at the vertex corresponding to a dyadic interval J ⊂ I^{0}. Then (due to the key property (3)) the value at any vertex is at least the sum of two values at the vertices right below it.

If the function φ is separated from 0 on I (which we can always assume without loss of generality), then the whole tree is finite and the terminal intervals form a splitting of I^{0}. Let us denote the family of all terminal intervals by T. We have

∫_{I^{0}} Y_{1}^{D}(t)dt = ∑_{J∈T} ∫_{J} Y_{1}^{D}(t)dt ⩽ ∑_{J∈T} |J|^{2} ⩽ 4 ∑_{J∈T} |J|^{2}B(f_{J} , P_{J})

⩽ 4|I^{0}|^{2}B(f_{I^{0}} , P_{I^{0}} ) ⩽ 4C|I^{0}|^{2}(f_{I^{0}} + P_{I^{0}} ) ⩽ 4C(1 + \frac{4}{3}) ∫_{ℝ} φ(t)dt.

Taking I^{0} larger and larger, we see that the integral on the left can be replaced by the integral over ℝ and we are done.

Note by the way that we only needed the property (2) for very-very small f and P and therefore we need to check it only for f, P < \frac{1}{2}, say. The Bellman type function: construction. It remains only to find a Bellman type function satisfying 1-3 above. Note that the true Bellman function actually depends only on the ratio \frac{f}{1−P}. If the reader cannot understand why, it does not matter: at this stage we may do whatever we want without any explanation, so let us just say that we will look for a function B(f, P) of the kind

B(f, P) = Φ(\frac{f}{1 − P} ).

The function Φ should be defined on the whole ray [0, +∞). Let three pairs (f, P), (f_{±}, P_{±}) satisfy the dynamics relations (∗). Let

x_{−} = \frac{f_{−}}{1 − P} , x_{+} = \frac{f_{+}}{1 − P} .

All the three pairs belong to the domain (if and) only if x_{±} < 1, and the key property can be rewritten as

Φ(\frac{x_{+} + x_{−}}{4} ) ⩾ \frac{1}{4}\{Φ(\frac{x_{−}}{1 − x_{−}} ) + Φ(\frac{x_{+}}{1 − x_{+}} )\}

whenever 0 ⩽ x_{±} < 1 (because \frac{f_{+}}{1−P_{+}} = \frac{f_{+}}{1−P} (\frac{1−P −f_{+}}{1−P} )^{−1} = \frac{x_{+}}{1−x_{+}} ). Also the conditions (1) and (2) are equivalent to

2′) Φ(x) ⩾ \frac{1}{4} when x ⩾ \frac{1}{4} and

3′) Φ(x) ⩽ Cx for small x

(recall that we need (2) only for f, P < \frac{1}{2}, say and under this condition \frac{f}{1−P} ⩽ 2f).

Again, the best thing the reader can do at this point is to put the text aside and to find such a function by himself.

Found ? Then compare to how I did it.

Note that any concave function Φ(x) satisfying (2′), (3′) and

Φ(\frac{x}{1 − x}) ⩽ 2Φ(\frac{x}{2}) (0 ⩽ x < 1)

fits (taking x_{−} = x_{+} = x, we see that this condition is also necessary). Indeed, then we have

\frac{1}{4}\{Φ(\frac{x_{−}}{1 − x_{−}} ) + Φ(\frac{x_{+}}{1 − x_{−}} )\} ⩽ \frac{1}{2}\{Φ(\frac{x_{−}}{2} ) + Φ(\frac{x_{+}}{2} )\} ⩽ Φ(\frac{x_{−} + x_{+}}{4} )

(the last inequality is due to concavity). Note now that the function

Φ(x) = \frac{2x}{2x + 1}

satisfies the identity

Φ(\frac{x}{1 − x}) = 2Φ(\frac{x}{2}) (0 ⩽ x < 1),

is concave, behaves like 2x near 0 and, at last, is not less than \frac{1}{3} for all x ⩾ \frac{1}{4}. Moral: The Beurling lemma is just the function \frac{2x}{2x+1} and it is being proved again and again by calculus students when drawing the graph of this function as a homework. This is not a joke, but a bitter truth of this world: no one understands what he is really doing. Theorem 2: (Beurling’s lemma in the \frac{dt}{1+t^{2}} -measure form) Let φ(t) ⩾ 0;

∫_{ℝ} \frac{φ(t)dt}{1 + t^{2}} = μ < 1.

Define Y_{1}(t) exactly as before. Then

∫_{ℝ} \frac{Y_{1}(t)}{1 + t^{2}} dt ⩽ \frac{Cμ}{(1 − μ)^{3}} .

Proof: Let A ⩾ 1. Note that for any s, t ∈ ℝ satisfying |t| ⩽ A < |s| we have

(s − t)^{2} ⩾ (|s| − A)^{2} = (1 − \frac{A}{|s|})s^{2} + A^{2}

Let η < 1. Then for |s| > \frac{A}{1−η} we can conclude from here that

(s − t)^{2} ⩾ η(1 + s^{2})

(as before we assume that |t| ⩽ A). Now for any y ⩾ 0 we have

∫_{\{|s|> \frac{A}{1−η} \}} \frac{φ(s)ds}{(s − t)^{2} + y^{2}} ⩽ η^{−1} ∫ \frac{φ(s)ds}{1 + s^{2}} = η^{−1}μ.

Let η = \frac{1+μ}{2}, φ_{A} = φχ_{[− \frac{A}{1−η} , \frac{A}{1−η}]} Then from the definition of Y_{1}(t) it follows

immediately that

∫_{[− \frac{A}{1−η} , \frac{A}{1−η}]} \frac{φ(s)ds}{(s − t)^{2} + Y_{1}(φ; t)^{2}} ⩾ 1 − η^{−1}μ = \frac{1 − μ}{1 + μ} ⩾ \frac{1 − μ}{2} ,

i.e.

Y_{1}(φ; t) ⩽ Y_{\frac{1−μ}{2}} (φ_{A}; t) for every t ∈ [−A, A] .

This results in

∫_{[−A,A]} Y_{1}(φ; t)dt ⩽ ∫_{ℝ} Y_{\frac{1−μ}{2}} (φ_{A}; t)dt ⩽ \frac{C}{1 − μ} ∫_{ℝ} φ_{A}(t)dt

and thereby

∫_{ℝ} \frac{Y_{1}(t)dt}{1 + t^{2}} ⩽ 2 ∫_{1}^{∞}(∫_{[−A,A]} Y_{1}(t)dt)\frac{dA}{A^{3}}

⩽ \frac{2C}{1 − μ} ∫_{1}^{∞}(∫_{ℝ} φ_{A}(t)dt)\frac{dA}{A^{3}} ⩽ \frac{C′}{(1 − η)^{3}} ∫_{ℝ} \frac{φ(t)dt}{1 + t^{2}} .

Kargaev’s version of the Beurling lemma: Let φ ⩾ 0, ∫_{ℝ} \frac{φ(t)dt}{1+t^{2}} < +∞. Then for every u > 0

∫_{ℝ} \frac{Y_{u}(t)dt}{Y\frac{u}{3} (0)^{2} + t^{2}} ⩽ C

where C is some absolute constant. Proof: If u = 1, Y\frac{u}{3} = 1, then it is just the Beurling lemma in the weird form with μ = \frac{1}{3}. The general result follows immediately from invariance of the problem under dilations and multiplication by constants.