Little Level Theorem and Zero-Khinchin Inequality. (joint paper with Keith Ball).

Let ξ_{0} be any random variable with values in ℝ (everything works in ℝ^{n} too without any essential change in the proof). Let

F = {∑_{j=1}^{N} a_{j}ξ_{j} : N ⩾ 1; ∑_{j} a_{j}^{2} = 1; ξ_{j} are independent and equidistributed with ξ_{0}}.

For any ε > 0 define ε-concentration of a random variable ξ as

C_{ε}(ξ) = sup_{L∈ℝ} ℙ{|ξ − L| < ε}.

Then for any ξ ∈ F and any ε > 0

C_{ε}(ξ) ⩽ AC_{ε}(ξ_{0})

where A > 0 is some absolute constant (when you consider values in ℝ^{n}, this constant in our approach blows up as n → ∞). Note that ξ_{0} ∈ F and therefore the result is up to a certain extent unbeatable. Proof of the little level theorem. Let ξ = ∑_{j} a_{j}ξ_{j} ∈ F.

Consider 2 cases: Trivial case: there exists j_{0} such that a_{j_{0}} ⩾ \frac{1}{2}. Let Ω = Ω_{0} × Ω_{1} be the decomposition of the probability space such that ξ_{j_{0}} depends on ω_{0} ∈ Ω_{0} while all the other ξ_{j} depend on ω_{1} ∈ Ω_{1}. Let σ(ω_{1}) = ∑_{j≠j_{0}} a_{j}ξ_{j}.

For any fixed ω_{1} ∈ Ω_{1} we have

ℙ_{ω_{0}}\{|ξ − L| < ε\} = ℙ\{|ξ_{j_{0}} − \frac{L − σ(ω_{1})}{a_{j_{0}}} | < \frac{ε}{a_{j_{0}}} \} ⩽ C_{ε/a_{j_{0}}} (ξ_{0}) ⩽ C_{2ε}(ξ_{0}) ⩽ 2C_{ε}(ξ_{0}).

Therefore

ℙ{|ξ − L| < ε} = ∫_{Ω_{1}} ℙ_{ω_{0}}{|ξ − L| < ε}dℙ_{ω_{1}} ⩽ 2C_{ε}(ξ_{0})

as well and we are done. Interesting case: all a_{j} < \frac{1}{2}.

Key lemma. Let p : ℝ → [0, +∞); ∥p∥_{L^{1}} = 1; ∥p∥_{L^{∞}} = M. Let p̂(t) = ∫_{ℝ} p(x)e^{−2πitx}dx be the Fourier transform of p.

Then for every s ⩾ 2

∫_{ℝ} |p̂(t)|^{s}dt ⩽ \frac{AM}{\sqrt{s}} .

Typeset by AMS-TEX

Suppose that this lemma is proved. Let η_{j} be independent (and independent of ξ_{j}) random variables uniformly distributed over the interval (−ε, ε). Let

ξ′_{j} = ξ_{j} + η_{j};  η = ∑_{j} a_{j}η_{j};  ξ′ = ∑_{j} a_{j}ξ′_{j} = ξ + η.

Note that for any L ∈ ℝ

ℙ{|ξ′ − L| < 3ε} ⩾ ℙ{|ξ − L| < ε & |η| < 2ε} = ℙ{|ξ − L| < ε} · ℙ{|η| < 2ε}.

But 𝔼η^{2} ⩽ ε^{2} and therefore the second factor is at least 1/2. Thus

ℙ{|ξ − L| < ε} ⩽ 2ℙ{|ξ′ − L| < 3ε} ⩽ 2C_{3ε}(ξ′) ⩽ 6C_{ε}(ξ′).

So, it remains to prove that C_{ε}(ξ′) ⩽ AC_{ε}(ξ_{0}).

Note that the density p(x) of the distribution of every ξ′_{j} is the same (which is completely irrelevant) and satisfies

p(x) ⩾ 0;  ∥p∥_{L^{1}} = 1;  ∥p∥_{L^{∞}} ⩽ \frac{C_{ε}(ξ_{0})}{2ε}

(which is very important). For the density P (x) of the distribution of ξ′ we have the relation

P̂(t) = ∏_{j} p̂(a_{j}t).

Using the Holder inequality, we get

∥P ∥_{L^{∞}} ⩽ ∥P̂∥_{L^{1}} ⩽ ∏_{j}{∫_{ℝ} |p̂(a_{j}t)|^{1/a_{j}^{2}}dt}^{a_{j}^{2}}.

But (according to the key lemma) for every j

∫_{ℝ} |p̂(a_{j}t)|^{1/a_{j}^{2}}dt = \frac{1}{a_{j}} ∫_{ℝ} |p̂(t)|^{1/a_{j}^{2}}dt ⩽ \frac{1}{a_{j}} A\frac{C_{ε}(ξ_{0})}{2ε}a_{j} = A\frac{C_{ε}(ξ_{0})}{2ε}.

Hence

∥P ∥_{L^{∞}} ⩽ A\frac{C_{ε}(ξ_{0})}{2ε}

as well and thereby for any L ∈ ℝ

ℙ\{|ξ′ − L| < ε\} = ∫_{L−ε}^{L+ε} P (t)dt ⩽ 2εA\frac{C_{ε}(ξ_{0})}{2ε} = AC_{ε}(ξ_{0}),

finishing the proof. Proof of the key lemma.

Distribution inequality:

Let δ ⩽ \frac{1}{2}. Then

μ{t ∈ ℝ : |p̂(t)| ⩾ 1 − δ^{2}} ⩽ 64Mδ.

(μ is the Lebesgue measure on ℝ). Proof of the distribution inequality: Let p_{−}(x) = p(−x). Consider the function f = p_{−} ∗ p. We clearly have

f ⩾ 0;  ∥f∥_{L^{1}} = 1;  ∥f∥_{L^{∞}} ⩽ M.

Also f̂(t) = |p(t)|^{2} and therefore f̂(t) ⩾ (1−δ^{2})^{2} ⩾ 1−2δ^{2} whenever |p̂(t)| ⩾ 1−δ^{2}. The main advantage of f(x) compared to p(x) is that it is even and therefore

f̂(t) = ∫_{ℝ} f(x) cos(2πtx)dx = ∫_{ℝ} f(x)(1−2 sin^{2}(πtx))dx = 1−2 ∫_{ℝ} f(t) sin^{2}(πtx) dx.

Assume first that δ = \frac{1}{2m} where m ∈ ℕ. Then

f̂(t) ⩾ 1 − 2δ^{2} ⟺ ∫_{ℝ} f(t) sin^{2}(πtx) dx ⩽ δ^{2} ⟹

⟹ ∫_{ℝ} f(t) sin^{2}(π \frac{t}{2δ} x) dx ⩽ \frac{1}{4} ⟺ f̂(\frac{t}{2δ} ) ⩾ 1 − 2\frac{1}{4} = \frac{1}{2}

(we used the inequality | sin(my)| ⩽ m| sin y|) and thereby

μ{t ∈ ℝ : |p̂(t)| ⩾ 1 − δ^{2}} ⩽ μ{t ∈ ℝ : f̂(t) ⩾ 1 − 2δ^{2}} ⩽

2δμ\{t ∈ ℝ : f̂(t) ⩾ \frac{1}{4}\} ⩽ 32δ∥f̂∥_{L^{2}}^{2} = 32δ∥f∥_{L^{2}}^{2} ⩽ 32Mδ.

To get the result for an arbitrary δ < \frac{1}{2}, just note that we always can find a fraction \frac{1}{2m} between δ and min\{2δ, \frac{1}{2}\}.

Now it is easy to finish the proof of the key lemma. We have

∫_{ℝ} |p̂(t)|^{s}dt = ∫_{\{t:|p̂(t)|⩾ \frac{3}{4} \}} |p̂(t)|^{s}dt + ∫_{\{t:|p̂(t)|< \frac{3}{4} \}} |p̂(t)|^{s}dt = I_{1} + I_{2}.

Due to the distribution inequality,

I_{1} ⩽ 64M ∫0^{\frac{1}{2}} (1 − δ^{2})^{s}dδ ⩽ 64M ∫_{0}^{∞} e^{−sδ^{2}}dδ = \frac{A′M}{\sqrt{s}} .

At last,

I_{2} ⩽ ∫_{ℝ} |p̂(t)|^{2}(\frac{3}{4})^{s−2}dt = ∫_{ℝ} |p(t)|^{2}(\frac{3}{4})^{s−2}dt ⩽ (\frac{3}{4})^{s−2}M ⩽ \frac{A′′M}{\sqrt{s}}

for all s ⩾ 2, provided that A′′ is large enough.

That’s all. We worked under minimal assumptions about the density p(t) of the random variable ξ_{0} (it could be just any measure). If you assume just a little bit more, namely, that

p ∈ L^{1+γ} with some γ > 0,

everything gets completely trivial because then for any ξ = ∑_{j} a_{j}ξ_{j} ∈ F such that max_{j} a_{j} ⩽ a(ξ_{0}) and for all ε > 0 one has

C_{ε}(ξ) ⩽ A(ξ_{0})ε

where the constants a(ξ_{0}) and A(ξ_{0}) depend a little bit on the sample random variable ξ_{0}, but not on ε. So, any singularity in the distribution smoothes out very fast.

At last it is worth mentioning that if the Fourier transform of the density of a random variable ξ_{0} is positive, then always

C_{ε}(ξ_{0}) ⩽ 2ℙ{|ξ_{0}| ⩽ 2ε}

and therefore under this assumption ξ_{0} is the worst guy in the family F. In particular, if 𝔼ξ_{0} = 0, 𝔼ξ_{0}^{2} < ∞, then

∥ξ∥_{0} ≍ ∥ξ∥_{2} ∀ξ ∈ F ⟺ ∥ξ_{0}∥_{0} > 0.

To get a typical example of such a random variable, just subtract independently any distribution from itself. Zero-Khinchin inequality: Let ξ_{0} be any random variable with values in ℝ (everything works in ℝ^{n} too without any essential change in the proof) satisfying 𝔼ξ_{0} = 0 and 𝔼ξ_{0}^{2} = 1. Let again

F = {∑_{j=1}^{N} a_{j}ξ_{j} : N ⩾ 1; ∑_{j} a_{j}^{2} = 1; ξ_{j} are independent and equidistributed with ξ_{0}}

Then

inf{∥ξ∥_{0} : ξ ∈ F} > 0 ⟺ ∥ξ_{0}∥_{0} > 0 & ∥ξ_{0} − ξ′_{0}∥_{0} > 0

(we will always denote by ξ′ a random variable equidistributed with ξ and independent of it). Proof of the Zero-Khinchin inequality. ⟹ is trivial. Let us prove ⟸.

Let

L(x) = \{ \begin{matrix}log \frac{1}{|x|}, & if |x| < 1; \\ 0 & otherwise \end{matrix} (x ∈ ℝ).

We need to show that

sup_{ξ∈F} 𝔼L(ξ) < +∞.

Key observation:

The Fourier transform L̂(t) ⩾ 0 everywhere and therefore for any two independent random variables ξ and η

𝔼L(ξ − η) ⩽ \{𝔼L(ξ − ξ′)\}^{\frac{1}{2}} \{𝔼L(η − η′)\}^{\frac{1}{2}} .

Step 1: For every ξ ∈ F

𝔼L(ξ − ξ′) ⩽ A𝔼L(ξ_{0} − ξ′_{0}).

Indeed, let ξ = ∑_{j} a_{j}ξ_{j}. Let η_{0} = ξ_{0} −ξ′_{0}, then η = ξ −ξ′ = ∑_{j} a_{j}η_{j} and, according to Little Level Theorem,

𝔼L(η) = ∫_{0}^{1} \frac{ℙ\{|η| < ε\}}{ε}dε ⩽ ∫_{0}^{1} \frac{C_{ε}(η)}{ε}dε

⩽ A ∫_{0}^{1} \frac{C_{ε}(η_{0})}{ε}dε ⩽ 4A ∫_{0}^{1} \frac{ℙ\{|η_{0}| < ε\}}{ε}dε = 4A𝔼L(η_{0}).

Step 2: Suppose that ξ = ∑_{j} a_{j}ξ_{j} and that all |a_{j}| ⩽ \frac{1}{\sqrt{2}}. Then ξ = αη − βζ where η, ζ ∈ F; α, β ∈ [\frac{1}{2}, 1], α^{2} + β^{2} = 1 and we immediately get

𝔼L(ξ) = 𝔼L(αη − βζ) ⩽ \{𝔼L(α(η − η′))\}^{\frac{1}{2}} \{𝔼L(β(ζ − ζ′))\}^{\frac{1}{2}} .

It remains to note that

𝔼L(α(η − η′)) ⩽ log \frac{1}{α} + 𝔼L(η − η′) ⩽ log 2 + A𝔼L(ξ_{0} − ξ′_{0}).

So, for vast majority of random variables in F only the condition ∥ξ_{0} − ξ′_{0}∥_{0} > 0 matters. Step 3: The rest are “small perturbations” of ξ_{0}. Forgetting about a non-essential constant factor close to 1, we see that our task is to get an upper bound for 𝔼L(ξ_{0} − aη) where η ∈ F, a ∈ (0, 1).

Split the probability space Ω into 3 parts:

Ω_{1} = \{ω : ξ_{0}(ω) ⩽ \sqrt{a}\};  Ω_{2} = \{ω : η(ω) ⩾ \frac{1}{2\sqrt{a}}\};  Ω_{3} = Ω \ (Ω_{1} ∪ Ω_{2}).

Note that for every ω ∈ Ω_{3}

|ξ_{0}(ω) − aη(ω)| ⩾ \frac{ξ_{0}}{2}

and thereby

∫_{Ω_{3}} L(ξ_{0}(ω) − aη(ω)) ⩽ 𝔼L(\frac{ξ_{0}}{2} ) ⩽ log 2 + 𝔼L(ξ_{0}).

Note also that everywhere

L(ξ_{0} − aη) ⩽ log \frac{1}{a} + L(a^{−1}ξ_{0} − η).

Since 𝔼L(a^{−1}ξ_{0}−η) ⩽ \{𝔼L(a^{−1}(ξ_{0}−ξ′_{0}))\}^{\frac{1}{2}} \{𝔼L(η−η′)\}^{\frac{1}{2}} ⩽ \{𝔼L(ξ_{0}−ξ′_{0})\}^{\frac{1}{2}} \{𝔼L(η−η′)\}^{\frac{1}{2}} , it remains to check that (log \frac{1}{a})ℙ\{Ω_{1} ∪ Ω_{2}\} is bounded by some constant not depending on a.

But

ℙ\{Ω_{1}\} ⩽ 2\frac{𝔼L(ξ_{0})}{log \frac{1}{a}}

while

ℙ{Ω_{2}} ⩽ 4a𝔼η^{2} = 4a

and we are done.