Ball, Haagerup, and distribution functions

Fedor L. Nazarov, Anatoliy N. Podkorytov

Abstract. We discuss an elementary trick that sometimes yields simple proofs of integral inequalities of the kind ∫_{X} (f^{s} − g^{s}) dμ ⩾ 0. We use this trick to obtain “computation-free” proofs of two famous theorems: Ball’s theorem on the sections of a cube and Haagerup’s theorem on the sharp constants in the Khinchin inequality for Rademacher functions.

Introduction

This paper does not contain any new results. Its only aim is to acquaint the reader with a completely elementary trick that often allows one to prove the integral inequality Δ(s) = ∫_{X} (f^{s} − g^{s}) dμ ⩾ 0 for s > s_{0}, if one knows that it holds for s = s_{0} (in particular, if f, g ∈ L^{s}(X , μ), we are interested in the preservation of the inequality between the L^{s}-norms of the functions f and g as s increases). Such estimates have various applications in analysis, especially in Banach space geometry.

In this paper we will use our trick to get rid of ugly numerical computations in the proofs of two famous results. The first chapter is devoted to the beautiful theorem of Keith Ball [1] that claims that the maximal (n−1)-dimensional volume of a section of the n-dimensional unit cube by a hyperplane is \sqrt{2}, i.e., that the maximal section is just the product of the diagonal of the two-dimensional unit square and the (n − 2)-dimensional unit cube. The second chapter deals with the problem of finding the sharp constants in the Khinchin inequality for Rademacher functions, which was first solved by Uffe Haagerup [5]. We believe that in both cases ^{1} our proofs are much simpler (both technically and ideologically), shorter, and more elegant than the original ones.

Our main idea will be to use distribution functions, the tool that proved extremely helpful in the probability theory. In this way we shall present a simple sufficient condition for “almost monotonicity” of the integral Δ(s). More precisely, our condition will allow to conclude that the product s^{−1}Q^{s}Δ(s) is increasing for some Q > 0.

1To be honest, we will prove only one half of Haagerup’s theorem in this paper. The second half can be obtained by our methods as well, but the proof becomes more complicated and less transparent. So, after hesitating for a while, we finally decided not to include it.

Chapter I. Ball’s theorem on the sections of the unit cube.

Let t ∈ ℝ. Let a = (a_{1}, …, a_{n}) be a unit vector in ℝ^{n} (n > 1). By σ(a, t) we will denote the (n − 1)-dimensional volume of the section of the cube [−\frac{1}{2}, \frac{1}{2}]^{n} by the hyperplane {x ∈ ℝ^{n} : (x, a) = t}.

Theorem (Ball). For any t ∈ ℝ and for any unit vector a ∈ ℝ^{n}, one has

σ(a, t) ⩽ \sqrt{2}.

Equality is attained only if t = 0 and the vector a has only two non-zero coordinates, each having absolute value \frac{1}{\sqrt{2}}, i.e., when the section is a product of the

cube [−\frac{1}{2}, \frac{1}{2}]^{n−2} and the diagonal of the square [−\frac{1}{2}, \frac{1}{2}]^{2}.

Among other things, this theorem implies that in Euclidean space of sufficiently large dimension all central sections of the ball of n-dimensional volume 1 have greater (n − 1)-dimensional volumes than the corresponding sections of the unit cube. To prove this (rather counter-intuitive) fact, it is enough to observe that the (n−1)-dimensional volumes of the central sections of the ball of n-dimensional volume 1 tend to \sqrt{e} as n → +∞. Actually, this phenomenon occurs starting from n = 10 (see [2] for details).

Thus, Ball’s theorem provides the simplest counterexample to the famous conjecture of Busemann and Petty [3]. (They conjectured that of two symmetric with respect to the origin convex bodies in ℝ^{n} , the one with greater (n − 1)- dimensional volumes of central sections always has greater n-dimensional volume). The statement of the conjecture is obvious in ℝ^{2}. Recently it has been proved for ℝ^{3} (see [4]). For 5 ⩽ n ⩽ 9 there are counterexamples (see [6]), though not quite as simple as a ball and a cube. The case n = 4 remains controversial to the best of our knowledge. The latest rumor we heard was that the previously claimed counterexample turned out to be faulty and that a proof of the conjecture was found instead. We leave it to the reader to ferret out the truth.

Proof. Fix the unit vector a ∈ ℝ^{n}. If at least one coordinate a_{j} of the vector a vanishes, we can reduce the dimension of the problem. So it is enough to consider the case when all a_{j} ≠ 0. Since the cube [−\frac{1}{2}, \frac{1}{2}]^{n} is symmetric with respect to each coordinate hyperplane, we may also assume below that a_{j} > 0 for each j.

We shall split the proof of Ball’s theorem into several steps.

Step 1. One of the coordinates a_{j} is large. Observe that σ(a, t) ⩽ \frac{1}{a_{j}} for each j = 1, …, n. Indeed, the volume of a section of the unit cube by a hyperplane orthogonal to a equals the volume of its projection to the coordinate hyperplane orthogonal to the j-th ort divided by the cosine of the angle between the vector a and that ort, i.e., by a_{j}. It remains to note that the projection of the whole unit cube to any of the coordinate hyperplanes has volume 1. Thus σ(a, t) < \sqrt{2} if one of the coordinates a_{j} is greater than \frac{1}{\sqrt{2}}.

From now on, we shall assume that a_{j} ⩽ \frac{1}{\sqrt{2}} for all j = 1, …, n.

Step 2. Integral representation for σ(a, t). Consider the function S(t) := σ(a, t). For every u ∈ ℝ, we have

Ŝ(u) = ∫_{ℝ} σ(a, t)e^{−2πiut}dt = ∫_{[− \frac{1}{2} , \frac{1}{2} ]^{n}} e^{−2πiu(x,a)} dx

= ∏_{j=1}^{n} ∫− \frac{1}{2}^{\frac{1}{2}} e^{−2πiua_{j}x_{j}}dx_{j} = ∏_{j=1}^{n} \frac{sin πa_{j}u}{πa_{j}u}.

Taking the inverse Fourier transform, we get

σ(a, t) = S(t) = ∫_{−∞}^{∞} Ŝ(u)e^{2πiut}du = ∫_{−∞}^{∞} e^{2πiut} ∏_{j=1}^{n} \frac{sin πa_{j}u}{πa_{j}u}du.

From this representation one can derive that the naive conjecture that the section of the largest volume is the central section orthogonal to the main diagonal is false. Indeed, the volume of this section (corresponding to t = 0, a_{j} = \frac{1}{\sqrt{n}}) equals

∫_{ℝ} (\frac{sin \frac{πu}{\sqrt{n}}}{\frac{πu}{\sqrt{n}}} )^{n} du = \sqrt{n} ∫_{ℝ} (\frac{sin πv}{πv} )^{n} dv

Using the Laplas asymptotic formula, we find that this volume tends to \sqrt{\frac{6}{π}} < \sqrt{2} as n → ∞.

Step 3. Reduction to Ball’s integral inequality. Since a_{1}^{2} + ⋯ + a_{n}^{2} = 1, the integral representation of σ(a, t) together with the Hölder inequality imply

σ(a, t) ⩽ ∫_{ℝ} ∏_{j=1}^{n} |\frac{sin πa_{j}u}{πa_{j}u}| du ⩽ ∏_{j=1}^{n}(∫_{ℝ} |\frac{sin πa_{j}u}{πa_{j}u}|^{1/a_{j}^{2}} du)^{a_{j}^{2}}.

Thus, we need only show that every integral ∫_{ℝ} |\frac{sin πa_{j}u}{πa_{j}u}|^{1/a_{j}^{2}} du does not exceed

\sqrt{2} and that equality is possible only when a_{j} = \frac{1}{\sqrt{2}}. Taking into account that \frac{1}{a_{j}^{2}} ⩾ 2 for every j and making the change of variable u → a_{j}u, we see that it would suffice to prove the following Ball’s integral inequality:

∫_{ℝ} |\frac{sin πx}{πx}|^{s} dx < \sqrt{\frac{2}{s}} for every s > 2

(for s = 2, it turns into identity).

Ball’s original proof of this inequality, though quite straightforward, involved a lot of unpalatable numerical computations some of which were almost impossible to verify without a calculator. Below we shall use a completely different approach.

Step 4. Lemma on distribution functions. Let (X , μ) be a measure space. For a non-negative function f : X → [0, +∞), its distribution function F (y), y > 0 is defined by

F (y) := μ{x ∈ X : f(x) > y}.

Clearly, F is a non-increasing function on (0, +∞).

Lemma on distribution functions. Let f and g be any two nonnegative measurable functions on a measure space (X , μ). Let F and G be their distribution functions. Assume that both F (y) and G(y) are finite for every y > 0. Assume also that at some point y_{0} the difference F − G changes sign from − to +, i.e., F (y) ⩽ G(y) for all y ∈ (0, y_{0}) and F (y) ⩾ G(y) for all y > y_{0}.

y_{0} G F

Let S = {s > 0 : f^{s} − g^{s} ∈ L^{1}(X , μ)}. Then the function

φ(s) = \frac{1}{sys_{0}} ∫_{X} (f^{s} − g^{s}) dμ

is increasing on S. In particular, if ∫_{X} (f^{s_{0}} − g^{s_{0}})dμ = 0, then ∫_{X} (f^{s} − g^{s})dμ ⩾ 0 for each s > s_{0}. The equality may hold only if the functions F and G coincide.

Remark. Note that we do not demand finiteness of the integrals ∫_{X} f^{s} dμ and ∫_{X} g^{s} dμ in the definition of the set S. Only the difference f^{s} − g^{s} is required to be Lebesgue integrable with respect to the measure μ. Though this observation is of little importance now, it will play crucial role in Chapter II.

Proof. Observe first of all, that if the difference f −g of two nonnegative functions f and g is Lebesgue integrable and both their distribution functions F (y) and G(y) are finite for all y > 0, then

∫_{X} (f − g) dμ = ∫_{0}^{∞} (F (y) − G(y)) dy.

Indeed, let h(x) = min(f(x), g(x)). The distribution function H(y) of the function h does not exceed F (y) and therefore is finite for every y > 0. Also, the (non-negative) difference f − h does not exceed |f − g|, and thereby is Lebesgue integrable. Now, applying the Fubini theorem to the characteristic function of the set {(x, y) ∈ X × (0, +∞) : h(x) ⩽ y < f(x)}, we obtain

0 ⩽ ∫_{X} (f − h) dμ = ∫_{0}^{∞}(F (y) − H(y)) dy < +∞.

Analogously,

0 ⩽ ∫_{X} (g − h) dμ = ∫_{0}^{∞}(G(y) − H(y)) dy < +∞.

Subtracting the second identity from the first, we obtain the desired statement. Since for s > 0 the distribution function of, say, f^{s} equals F (y^{1/s}), we get

∫_{X} (f^{s} − g^{s}) dμ = ∫_{0}^{∞} (F (y^{1/s}) − G(y^{1/s})) dy = s ∫_{0}^{∞} y^{s−1} (F (y) − G(y)) dy.

Therefore for s > s_{0} (s, s_{0} ∈ S), we have

φ(s) − φ(s_{0}) = \frac{1}{y_{0}} ∫_{0}^{∞} ((\frac{y}{y_{0}} )^{s−1} − (\frac{y}{y_{0}} )^{s_{0}−1}) (F (y) − G(y)) dy ⩾ 0,

because both factors in the integrand simultaneously change their signs at y_{0}. □

Step 5. Proof of Ball’s integral inequality.

Using the Euler-Poisson integral ∫_{ℝ} e^{−πx^{2}}dx = 1, we can rewrite Ball’s inequality as ∫_{0}^{∞} g^{s} < ∫_{0}^{∞} f^{s}, where f(x) = e^{−πx^{2}/2}, g(x) = | \frac{sin πx}{πx}|. We need to prove this inequality for all powers s > 2. Since for s = 2 we have an identity, it is enough to show that the corresponding distribution functions F and G satisfy the condition of the lemma from Step 4 (with X = (0, ∞), dμ = dx, s_{0} = 2). Observe that both the functions f and g are strictly less than 1 on the whole ray (0, +∞) and thereby F (y) = G(y) = 0 for each y ⩾ 1. So, we may restrict ourselves to the values y ∈ (0, 1).

Since f is decreasing on (0, ∞), we have

F (y) = f^{−1}(y) = \sqrt{\frac{2}{π} ln \frac{1}{y}} .

To estimate G(y), consider the quantities y_{m} = max_{[m,m+1]} g, m ∈ ℕ (the graph of g(x) for x ⩾ 1 looks as a sequence of decreasing humps and y_{m} is just the height of the m-th hump in this sequence – see the picture below). Obviously,

\frac{1}{π(m + \frac{1}{2})} < y_{m} < \frac{1}{πm}.

For x ∈ (0, 1), we have

g(x) = ∏_{k=1}^{∞} (1 − \frac{x^{2}}{k^{2}} ) ⩽ ∏_{k=1}^{∞} e^{−x^{2}/k^{2}} = e^{−(πx)^{2}/6} < e^{−πx^{2}/2} = f(x).

Therefore, for all y ∈ (y_{1}, 1),

G(y) = μ{x ∈ (0, 1) : g(x) > y} < μ{x ∈ (0, 1) : f(x) > y} ⩽ F (y),

i.e., the difference F − G is positive on the interval (y_{1}, 1).

The difference F − G must change sign at least once on (0, +∞) because ∫_{0}^{∞} 2y(F (y) − G(y)) dy = ∫_{0}^{∞}(f(x)^{2} − g(x)^{2}) dx = 0. To prove that the change of sign occurs only once, it now suffices to show that F − G is increasing on (0, y_{1}). Since both F and G are continuous decreasing functions, it is enough to prove that |G′(y)| > |F′(y)| on each interval (y_{m+1}, y_{m}), m ∈ ℕ.

For y ∈ (0, y_{1}), the value G(y) can be visualized as the sum of lengths of the thick black intervals on the picture below.

1  ^{2}  ^{3}  ^{4} y_{1} y_{2} y_{3} y  ^{g(x) = | \frac{sin πx}{πx} |}^{G(y) equals the total length of the thick black intervals.}

It is clear from this picture that for each y ∈ (0, y_{1}), y ≠ y_{j}, the function G is differentiable and

|G′(y)| = ∑_{x>0 : g(x)=y} \frac{1}{|g′(x)|}.

When y ∈ (y_{m+1}, y_{m}), the equation g(x) = y has one root on the interval (0, 1) and two roots on each interval (k, k + 1), k = 1, …, m. Let us estimate |g′(x)| from above at these roots.

• If the root x ∈ (k, k + 1), k ⩾ 1, then

|g′(x)|= |\frac{cos πx}{x} − \frac{sin πx}{πx^{2}} | ⩽\frac{1}{x} (1 + \frac{| sin π(x − k)|}{πx}) ⩽\frac{1}{x} (1 + \frac{π(x − k)}{πk}) =\frac{1}{k};.

• If the root x ∈ (0, 1), then

|g′(x)| = \frac{sin πx − πx cos πx}{πx^{2}} = \frac{1}{πx^{2}} ∫_{0}^{πx} t sin t dt ⩽ \frac{1}{πx^{2}} ∫_{0}^{πx} t dt = \frac{π}{2} .

Thus for every y ∈ (y_{m+1}, y_{m}), we have

|G′(y)| ⩾ \frac{2}{π} + 2 ∑_{k=1}^{m} k = \frac{2}{π} + m + m^{2}.

Recalling that F (y) = \sqrt{\frac{2}{π} ln \frac{1}{y}}, we obtain

|\frac{G′(y)}{F′(y)}| = |G′(y)| y\sqrt{2π ln \frac{1}{y}} ⩾ (\frac{2}{π} + m + m^{2}) y\sqrt{2π ln \frac{1}{y}} .

Since y_{1} < \frac{1}{π} < \frac{1}{\sqrt{e}}, the function y\sqrt{2π ln \frac{1}{y}} is increasing on (0, y_{1}). Recalling that

y > y_{m+1} > \frac{1}{π(m + \frac{3}{2})}, we get

|\frac{G′(y)}{F′(y)}| > \frac{\frac{2}{π} + m + m^{2}}{\frac{3}{2} + m} \sqrt{\frac{2}{π} ln π(m + \frac{3}{2})} > \sqrt{\frac{2}{π} ln \frac{5π}{2}}

(m ⩾ 1 and \frac{2}{π} + m^{2} > \frac{1}{2} + 1 = \frac{3}{2}).

It remains to note that the right hand part is greater than 1 because ln 5x > x for every x ∈ [1, 2], and, in particular, for x = \frac{π}{2} (since the function ln 5x is concave, it is enough to check this estimate at the ends of the interval, where it is obvious).

Ball’s inequality is thus completely proved. □

Chapter II. Haagerup’s theorem on sharp constants in the Khinchin inequality.

Recall that Rademacher functions r_{k}(t) (k = 1, 2, …, t ∈ [0, 1]) are defined by

r_{k}(t) = sign sin(2^{k}πt)

From the probabilistic point of view, they are just independent random variables assuming values +1 and −1 with probability \frac{1}{2} each. One of the most famous results on Rademacher functions is the Khinchin inequality:

A_{p} (∑_{k=1}^{n} a_{k}^{2})^{p/2} ⩽ R_{p}(a) := ∫_{0}^{1} |∑_{k=1}^{n} a_{k}r_{k}(t)|^{p} dt ⩽ B_{p} (∑_{k=1}^{n} a_{k}^{2})^{p/2}

for any real coefficient vector a = (a_{1}, …, a_{n}) and for any 0 < p < +∞. Uffe Haagerup found the sharp values of the constants A_{p} and B_{p} in this inequality for all powers p > 0 (see [5]).

Formulation of Haagerup’s theorem. To find the best constants A_{p} and B_{p} in the Khinchin inequality is the same as to solve the extremal problems

∫_{0}^{1} |∑_{k=1}^{n} a_{k}r_{k}(t)|^{p} dt → min,  ∑_{k} a_{k}^{2} = 1

and

∫_{0}^{1} |∑_{k=1}^{n} a_{k}r_{k}(t)|^{p} dt → max,  ∑_{k} a_{k}^{2} = 1

respectively. There are three “natural” linear combinations ∑_{k} a_{k}r_{k}(t) to look at: • n = 1, a_{1} = 1 (a single Rademacher function). In this case R_{p}(a) = 1.

• n = 2, a_{1} = a_{2} = \frac{1}{\sqrt{2}} (the sum or the difference of two Rademacher func-

tions). In this case R_{p}(a) = 2^{\frac{p−2}{2}} .

• n → +∞, a_{1} = ⋯ = a_{n} = \frac{1}{\sqrt{n}}. In this case it is more convenient to accept the probabilistic point of view and to conclude, using the central limit theorem, that we are actually talking about the normal distribution. Hence, the corresponding (limit) value of R_{p}(a) equals

\frac{1}{\sqrt{2π}} ∫_{ℝ} |x|^{p}e^{− \frac{x^{2}}{2}} dx = \sqrt{\frac{2}{π}} ∫_{0}^{∞} x^{p}e^{− \frac{x^{2}}{2}} dx

= 2^{\frac{p−1}{2}} \sqrt{\frac{2}{π}} ∫_{0}^{∞} y^{\frac{p−1}{2}} e^{−y}dy = \frac{2^{\frac{p}{2}}}{\sqrt{π}} Γ (\frac{p + 1}{2}) = 2^{\frac{p−2}{2}} \frac{Γ (\frac{p+1}{2})}{Γ (\frac{3}{2})}

(the statement of the central limit theorem for bounded symmetric random variables is strong enough to ensure convergence of moments of any order).

Now Haagerup’s theorem can be stated in a very simple way:

Theorem (Haagerup). For each p ∈ (0, +∞), the extremum in either of the extremal problems is one of these three values.

What does this mean in terms of formulae? First of all, note that A_{p} = 1 for p ⩾ 2 and B_{p} = 1 for p ⩽ 2 (this is just the Hölder inequality). So, the only non-trivial tasks are to find A_{p} for p < 2 and to find B_{p} for p > 2. Haagerup’s theorem claims that in these cases the extreme value is delivered by either the sum of two Rademacher functions or by the normal distribution, i.e.,

A_{p} = 2^{\frac{p−2}{2}} min (1, \frac{Γ(\frac{p+1}{2})}{Γ(\frac{3}{2})}); B_{p} = 2^{\frac{p−2}{2}} max (1, \frac{Γ(\frac{p+1}{2})}{Γ(\frac{3}{2})}) .

Using the alternative representation for the Γ-function

Γ(q) = ∫_{0}^{∞} t^{q}e^{−t} \frac{dt}{t} = lim_{n→∞} ∫_{0}^{n} t^{q} (1 − \frac{t}{n})^{n} \frac{dt}{t} = lim_{n→∞} \frac{n^{q} n!}{q(q + 1) … (q + n)},

we conclude that the second derivative \frac{d^{2}}{dq^{2}} ln Γ(q) = ∑_{k=0}^{∞} \frac{1}{(q + k)^{2}} of the function

ln Γ(q) is positive and decreasing on (0, +∞). Since Γ(1) = Γ(2) = 1, we can derive from here that the function ln Γ(q) is convex on (0, +∞) and attains its minimum on the interval [1, 2] somewhere to the left from its middle point \frac{3}{2}. In particular, this implies that the function Γ is increasing on [\frac{3}{2}, +∞). So for p > 2, the maximum of R_{p}(a) is delivered by the normal distribution and

B_{p} = 2^{\frac{p−2}{2}} \frac{Γ(\frac{p+1}{2})}{Γ(\frac{3}{2})}.

The case p < 2 is slightly more complicated because there exists q_{0} ∈ (1, \frac{3}{2}) such that Γ(q) > Γ(\frac{3}{2}) for 0 < q < q_{0}, Γ(q_{0}) = Γ(\frac{3}{2}), and Γ(q) < Γ(\frac{3}{2}) for q_{0} < q < \frac{3}{2}. Therefore

A_{p} = \begin{cases}2^{\frac{p−2}{2}} , & if 0 < p ⩽ p_{0}; \\ 2^{\frac{p−2}{2}} \frac{Γ( \frac{p+1}{2} )}{Γ( \frac{3}{2} )}, & if p_{0} ⩽ p ⩽ 2 \end{cases}

where p_{0} = 2q_{0} − 1 ∈ (1, 2) is the only nontrivial positive root of the equation Γ(\frac{p+1}{2}) = Γ(\frac{3}{2}).

Our aim in this chapter will be to give a (relatively) simple proof of one half of Haagerup’s theorem. Namely, we shall consider the case p < 2 (or, which is quite the same, shall find the sharp value of A_{p} for all p > 0). Haagerup also considers the cases p < 2 and p > 2 separately and it is almost impossible to decide for which of them his original proof is more painstaking.

Proof. Again, we will split the proof into several steps.

Step 1. Reduction of the case 0 < p < p_{0} to the case p = p_{0}. Assume that we already know that A_{p_{0}} = 2^{\frac{p_{0} −2}{2}} , i.e., that R_{p}(a) ⩾ 2^{\frac{p_{0} −2}{2}} for any a = (a_{1}, …, a_{n}) ∈ ℝ^{n} satisfying ∑_{k} a_{k}^{2} = 1. Let now 0 < p < p_{0}. Then, using the Hölder inequality

R_{p_{0}}(a) ⩽ R_{p}(a)^{\frac{2−p_{0}}{2−p}} R_{2}(a)^{\frac{p_{0} −p}{2−p}}

and taking into account that R_{2}(a) = ∑_{k} a_{k}^{2} = 1, we conclude that

R_{p}(a) ⩾ (2^{\frac{p_{0} −2}{2}} )^{\frac{2−p}{2−p_{0}}} = 2^{\frac{p−2}{2}} ,

so A_{p} = 2^{\frac{p−2}{2}} for all 0 < p < p_{0}. Thus we need only consider the powers p_{0} ⩽ p ⩽ 2. Step 2. Integral representation for R_{p}(a). Let 0 < p < 2. Then for every s ∈ ℝ,

|s|^{p} = C_{p} ∫_{0}^{∞} \frac{1 − cos us}{u^{p+1}} du

where the constant C_{p} > 0 depends on p only. Plugging s = ∑_{1}^{n} a_{k}r_{k}(t) into this identity and integrating over t ∈ [0, 1], we find

∫_{0}^{1} |∑_{1}^{n} a_{k}r_{k}(t)|^{p} dt = C_{p} ∫_{0}^{∞} \frac{1 − ∫_{0}^{1} cos (u ∑_{1}^{n} a_{k}r_{k}(t)) dt}{u^{p+1}} du.

Observing that

∫_{0}^{1} cos (u∑_{1}^{n} a_{k}r_{k}(t)) = Re ∫_{0}^{1} e^{iu ∑_{1}^{n} a_{k}r_{k}(t)} = ∏_{1}^{n} cos a_{k}u,

we finally arrive at Haagerup’s integral representation:

R_{p}(a) = ∫_{0}^{1} |∑_{1}^{n} a_{k}r_{k}(t)|^{p} dt = C_{p} ∫_{0}^{∞} \frac{1 − ∏_{1}^{n} cos a_{k}u}{u^{p+1}} du.

Step 3. The case of small coefficients: reduction to Haagerup’s integral inequality. Let us again assume that ∑_{j} a_{k}^{2} = 1. Young’s inequality yields

∏_{1}^{n} cos a_{k}u ⩽ ∏_{1}^{n} | cos a_{k}u| ⩽ ∑_{1}^{n} a_{k}^{2} | cos a_{k}u|^{1/a_{k}^{2}} = 1 − ∑_{1}^{n} a_{k}^{2}(1 − | cos a_{k}u|^{1/a_{k}^{2}})

and therefore

∫_{0}^{1} |∑_{1}^{n} a_{k}r_{k}(t)|^{p} dt ⩾ ∑_{1}^{n} a_{k}^{2} I_{p}(\frac{1}{a_{k}^{2}} )  (∗)

where

I_{p}(s) = C_{p} ∫_{0}^{∞} (1 − |cos \frac{u}{\sqrt{s}}|^{s}) \frac{du}{u^{p+1}} (s > 0).

The value of the integral I_{p}(s) can be computed in two cases:

• Case 1: s = 2. It follows immediately from Haagerup’s integral representation for n = 2, a_{1} = a_{2} = \frac{1}{\sqrt{2}} that

I_{p}(2) = 2^{\frac{p−2}{2}} .

• Case 2: s = ∞. The trivial estimate ln cos θ = O(θ^{2}) (θ → 0) implies that 1 − |cos \frac{u}{\sqrt{s}} |^{s} ⩽ 1 − e^{− Const u^{2}} ⩽ Const u^{2} for 0 < u < 1 and sufficiently large s. Since this difference also never exceeds 1, we may apply Lebesgue’s dominated convergence theorem and conclude that the limit lim_{s→∞} I_{p}(s) does exist and equals

I_{p}(∞) = C_{p} ∫_{0}^{∞} (1 − e^{− \frac{u^{2}}{2}} ) \frac{du}{u^{p+1}} .

On the other hand, for every even n ∈ ℕ, Haagerup’s integral representation with a_{1} = ⋯ = a_{n} = \frac{1}{\sqrt{n}} yields

R_{p}(\frac{1}{\sqrt{n}}, …, \frac{1}{\sqrt{n}}) = I_{p}(n).

As we have seen above, the left hand part tends to 2^{\frac{p−2}{2}} \frac{Γ( \frac{p+1}{2} )}{Γ( \frac{3}{2} )} (which is

our hypothetical value for A_{p}) as n → ∞. Thus we also have the identity

I_{p}(∞) = 2^{\frac{p−2}{2}} \frac{Γ(\frac{p+1}{2})}{Γ(\frac{3}{2})}.

Recall that I_{p}(2) ⩾ I_{p}(∞) for all p ∈ [p_{0}, 2). If we could only show that I_{p}(s) ⩾ I_{p}(∞) for all other s ∈ [1, +∞) as well, then the inequality (∗) would yield

∫_{0}^{1} |∑_{1}^{n} a_{k}r_{k}(t)|^{p} dt ⩾ (∑_{1}^{n} a_{k}^{2}) I_{p}(∞) = I_{p}(∞) = 2^{\frac{p−2}{2}} \frac{Γ (\frac{p+1}{2})}{Γ (\frac{3}{2})},

proving Haagerup’s theorem. Note now that

I_{p}(s) − I_{p}(∞) = C_{p} ∫_{0}^{∞} (1 − |cos \frac{u}{\sqrt{s}}|^{s}) \frac{du}{u^{p+1}} − C_{p} ∫_{0}^{∞} (1 − e^{− \frac{u^{2}}{2}} ) \frac{du}{u^{p+1}}

=^{(u=x\sqrt{s})} C_{p}s^{\frac{p}{2}} ∫_{0}^{∞} (e^{−s \frac{x^{2}}{2}} − |cos x|^{s}) \frac{dx}{x^{p+1}} .

So, the desired estimate I_{p}(s) ⩾ I_{p}(∞) is equivalent to Haagerup’s integral inequality

H(p, s) := ∫_{0}^{∞} (e^{−s \frac{x^{2}}{2}} − |cos x|^{s}) \frac{dx}{x^{p+1}} ⩾ 0,

which again is an inequality of the kind ∫_{X} (f^{s} − g^{s}) dμ ⩾ 0, only now with X =

(0, +∞), f(x) = e^{−x^{2}/2}, g(x) = | cos x|, and dμ = \frac{dx}{x^{p+1}} .

Unfortunately, there is no hope to prove the inequality H(p, s) ⩾ 0 for every p ∈ [p_{0}, 2), s ⩾ 1 by our distribution function technique. Indeed, since H(p_{0}, 2) = 0, such a proof of the estimate H(p_{0}, s) > 0 for s > 2 would automatically be a proof of the estimate H(p_{0}, s) < 0 for s < 2 and vice versa. Fortunately, it is still possible to show that H(p, s) ⩾ 0 for s ⩾ 2, which is enough to establish Haagerup’s theorem for the case when all the coefficients a_{k} do not exceed \frac{1}{\sqrt{2}}.

Step 4. Modified lemma on distribution functions and the proof of Haagerup’s integral inequality. Our lemma on distribution functions from Chapter I cannot be applied directly now because the corresponding distribution functions F (y) and G(y) are infinite for small values of y. Nevertheless, the modified distribution functions

F_{∗}(y) = μ{x ∈ X : f(x) < y} and G_{∗}(y) = μ{x ∈ X : g(x) < y}

are finite for all 0 < y < 1, so we may use the following

Modified lemma on distribution functions. Let Y > 0. Let f, g : X → [0, Y] be any two measurable functions on a measure space (X , μ). Let F_{∗} and G_{∗} be their modified distribution functions. Assume that both F_{∗}(y) and G_{∗}(y) are finite for every y ∈ (0, Y). Assume also that at some point y_{0} the difference F_{∗} −G_{∗} changes sign from + to −, i.e., F_{∗}(y) ⩾ G_{∗}(y) for all y ∈ (0, y_{0}) and F_{∗}(y) ⩽ G_{∗}(y) for all y ∈ (y_{0}, Y).

Y y_{0} G_{∗}F_{∗}

Let again S = {s > 0 : f^{s} − g^{s} ∈ L^{1}(X , μ)}. Then the function

φ(s) = \frac{1}{sy_{0}^{s}} ∫_{X} (f^{s} − g^{s}) dμ

is increasing on S. In particular, if ∫_{X} (f^{s_{0}} − g^{s_{0}})dμ ⩾ 0, then ∫_{X} (f^{s} − g^{s})dμ ⩾ 0 for each s > s_{0}.

The proof is essentially the same as in Chapter I.

Taking X , μ, f, g as above and putting Y = 1, s_{0} = 2, we see that to prove Haagerup’s integral inequality in the desired range p_{0} ⩽ p ⩽ 2, s ⩾ 2, we need only show that the corresponding modified distribution functions F_{∗} and G_{∗} satisfy the conditions of the modified lemma. These functions are not hard to compute: for y ∈ (0, 1), we have

F_{∗}(y)=μ\{x > 0 : e^{− \frac{x^{2}}{2}} < y\}=μ (\sqrt{2 ln \frac{1}{y}} , +∞)=∫_{√\overline{2 ln(1/y)}}^{+∞} \frac{dx}{x^{p+1}} = \frac{1}{p}(2 ln \frac{1}{y})^{−p/2},

G_{∗}(y)=μ{x > 0 : | cos x| < y} = μ ⋃_{k=0}^{∞}(πk + arccos y, πk + π − arccos y)

=∑_{k=0}^{∞} ∫_{πk+arccos y}^{πk+π−arccos y} \frac{dx}{x^{p+1}} = \frac{1}{p} ∑_{k=0}^{∞} (\frac{1}{(πk + arccos y)^{p}} − \frac{1}{(πk + π − arccos y)^{p}} ) .

Make (decreasing) change of variable y = cos t, t ∈ (0, \frac{π}{2} ). Then our aim is to show that the difference L(t)^{−p/2} − S(t), where

L(t) = (pF_{∗}(cos t))^{−2/p} = 2 ln \frac{1}{cos t},
S(t) = pG_{∗}(cos t) = \frac{1}{t^{p}} − \frac{1}{(π − t)^{p}} + \frac{1}{(π + t)^{p}} − \frac{1}{(2π − t)^{p}} + \frac{1}{(2π + t)^{p}} − …,

changes its sign only once on the interval (0, \frac{π}{2} ), this change being from − to +, or, which is the same, that the difference

Φ(t) = L(t) − S(t)^{−p/2}

changes its sign only once on the interval (0, \frac{π}{2} ), this change being from + to −.

The behavior of Φ(t) on the interval (0, \frac{π}{2} ) is quite complicated: there is no obvious monotonicity or concavity that would allow us to give a simple proof of the desired property by real variable methods. Nevertheless our functions are simpler than the distribution functions from the proof of Ball’s integral inequality in another respect: they both can be extended analytically to the complex plane. So, it seems reasonable to try complex variable methods. We will prove the following

Lemma on zeroes. Let 1 < p < 2. Then the analytic extension Φ(z) of the difference Φ(t) to the vertical strip \{0 < Re z < \frac{π}{2} \} has exactly one zero in that strip. This zero z_{0} is real and the sign of Φ(t) changes from + to − at z_{0}.

Proof. Since the minuend L(t) and the subtrahend S(t)^{−2/p} involve logarithms and fractional powers, some comment is due on how the corresponding branches in their extensions L(z) and S(z)^{−2/p} must be chosen. Since our complex-valued functions L(z) and S(z) of complex variable z are going to be extensions of the real-valued functions L(t) and S(t) of real variable t, we should have the symmetry L(\overline{z}) = \overline{L(z)} and S(\overline{z}) = \overline{S(z)}. So, it is enough to trace the branches in the upper half-strip, i.e., when Im z ⩾ 0.

For every such z, we have Im cos z = (e^{− Im z} − e^{Im z}) sin Re z < 0. So, cos z stays in the lower half-plane and, thereby, \frac{1}{cos z} stays in the upper half-plane as z runs over the upper half-strip. Besides, the values of L(z) on the interval (0, \frac{π}{2} ) should be real. Thus, the branch of logarithm we need to choose is the one with the imaginary part between 0 and π.

Let us now turn to S(z). Note first of all that for p > 1, the series in the definition of S(z) converges absolutely for every z in the upper half-strip. Now observe that for every z in the upper half-strip, all the complex numbers z, π + z, 2π + z, … belong to the right upper quadrant {Re ζ > 0, Im ζ ⩾ 0}. The branch of the function ζ → ζ^{−p} that is positive on the positive real semiaxis sends this quadrant to the angle Δ_{p} := \{−\frac{πp}{2} < arg ζ ⩽ 0\} (since p < 2, this angle is a subset of the lower half-plane). On the other hand, the numbers π − z, 2π − z, … belong to the right lower quadrant, which is sent by the mapping ζ → − ζ^{−p} to the same angle Δ_{p}. So, the whole sum S(z) stays in this angle when z runs over the upper half-strip and the branch of the power −2/p in the subtrahend S(z)^{−2/p} should be the one that sends the angle Δ_{p} to the upper half-plane (which is the “natural” branch in this angle).

Note now that the functions L(z) and S(z) are analytic everywhere in the closed strip \{0 ⩽ Re z ⩽ \frac{π}{2} \} except two singular points 0 and \frac{π}{2} . Let us look at what happens when z approaches these singularities.

When z → 0, we have L(z) = 2 ln \frac{1}{cos z} = z^{2} + \frac{1}{6}z^{4} + o(z^{4}) while S(z) = z^{−p} + O(z) = z^{−p}(1 + o(z^{2})) because p > 1. Thus

Φ(z) = z^{2} + \frac{z^{4}}{6} + o(z^{4}) − z^{2}(1 + o(z^{2})) = \frac{z^{4}}{6} + o(z^{4}) as z → 0.

Note now that \frac{1}{cos z} = \frac{1}{sin(\frac{π}{2} − z)} = O(\frac{1}{|\frac{π}{2} − z|}) and L(z) = O(ln \frac{1}{|\frac{π}{2} − z|}) as

z → \frac{π}{2} . On the other hand, \frac{π}{2} is a simple zero of the function S(z) and therefore

S(z) = D (\frac{π}{2} −z)(1+o(1)) where D = −S′(\frac{π}{2} ) = 2p [(\frac{2}{π} )^{p+1} + (\frac{2}{3π} )^{p+1} + …] > 0. Thus

Φ(z) ≈ −D^{−2/p} (\frac{π}{2} − z)^{−2/p} as z → \frac{π}{2} .

From these asymptotic relations, we can conclude that the function Φ(t) is positive near 0 and negative near \frac{π}{2} . So it has at least one root on the interval (0, \frac{π}{2} ) and, if this root is unique, changes its sign from + to − there.

Now let us estimate the increment of the argument of the function Φ(z) over the boundary of the region Ω, the upper half Ω_{+} of which is shown on the picture below.

[figure: iy; Ω_{+}; Γ_{3}; Γ_{2}; Γ_{4}; Γ_{1}; Γ_{5}; 0; \frac{π}{2}]

Note that Δ_{∂Ω} arg Φ = 2Δ_{∂Ω_{+}} arg Φ due to the symmetry Φ(\overline{z}) = \overline{Φ(z)}. The boundary ∂Ω_{+} consists of 5 parts Γ_{1}, …, Γ_{5}. We assume that the radii of the circle arcs Γ_{1} and Γ_{5} are very small and the height y of Γ_{3} is very large.

Estimates for Γ_{1} and Γ_{5}. From the asymptotic relations above, we immediately obtain that the function Φ(z) does not vanish on Γ_{1} and Γ_{5} and

Δ_{Γ_{1}} arg Φ ≈ −2π; Δ_{Γ_{5}} arg Φ ≈ \frac{π}{p} .

Estimate for Γ_{3}. Clearly,

L(z) = 2 ln \frac{1}{cos z} = 2 ln \frac{2}{e^{iz} + e^{−iz}} = 2iz + O(1) as Im z → +∞, z ∈ Ω_{+}.

As to S(z), we have

Im \frac{1}{z^{p}} , Im \frac{1}{(π + z)^{p}} , ⋯ < 0 and Im \frac{1}{(π − z)^{p}} , Im \frac{1}{(2π − z)^{p}} , ⋯ > 0

for every z ∈ Ω_{+}. What is more, the absolute values of the imaginary parts of the first Im z ≈ |z| terms in the series for S(z) are not less than C|z|^{−p} with some absolute C > 0. Hence, |S(z)| ⩾ | Im S(z)| ⩾ Const |z|^{1−p} and, thereby,

S(z)^{−2/p} = O(|z|^{2(p−1)/p}) = o(|z|)

on Γ_{3} (recall that p < 2). Thus Φ(z) ≈ 2iz on Γ_{3}, so Φ does not vanish on Γ_{3} and

Δ_{Γ_{3}} arg Φ ≈ 0.

Estimate for Γ_{2}. For every z = it, t > 0 on the vertical side Γ_{2}, we have L(z) = 2 ln \frac{2}{e^{−t}+e^{t}} ∈ ℝ. Representing the function S(z) as

S(z) = S(it) = \frac{1}{(it)^{p}} − [\frac{1}{(π − it)^{p}} − \frac{1}{(π + it)^{p}} ] − [\frac{1}{(2π − it)^{p}} − \frac{1}{(2π + it)^{p}} ] − …

and observing that every difference in brackets is a “positive imaginary” number, we arrive at the following picture:

[figure: 0; \frac{πp}{2}; \frac{1}{(it)^{p}}; [...]; [...]; [...]; S(it)]

Hence 0 < arg [S(it)^{−2/p}] < π, i.e., Im [S(it)^{−2/p}] > 0. Therefore Im Φ(it) < 0 for all t > 0, or, by other words, Φ(z) stays in the (open) lower half-plane as z runs over Γ_{2}. Therefore Φ does not vanish on Γ_{2} and

Δ_{Γ_{2}} arg Φ ⩽ π.

Estimate for Γ_{4}. This is the most interesting part of the contour ∂Ω_{+}. On Γ_{4} we have z = \frac{π}{2} + it, t > 0. Hence

L(z) = 2 ln \frac{2}{ie^{−t} − ie^{t}} = 2 ln i + 2 ln \frac{2}{e^{t} − e^{−t}} = πi + 2 ln \frac{2}{e^{t} − e^{−t}} .

Meanwhile,

S(z) = [\frac{1}{(\frac{π}{2} + it)^{p}} − \frac{1}{(\frac{π}{2} − it)^{p}} ] + [\frac{1}{(\frac{3π}{2} + it)^{p}} − \frac{1}{(\frac{3π}{2} − it)^{p}} ] + ….

Every difference in brackets is a “negative imaginary” number. Thus S(z) ∈ −iℝ_{+} and thereby S(z)^{−2/p} ∈ e^{iπ/p} ℝ_{+}. So, when t runs up from 0 to +∞, the minuend L(\frac{π}{2} + it) runs over the horizontal line Im z = π from right to left, while the subtrahend S(\frac{π}{2} + it)^{−2/p} wanders over the ray arg z = \frac{π}{p} (see the picture below).

[figure: 0; a; πi; e^{i \frac{π}{p}} ℝ_{+}]

As long as L(\frac{π}{2} + it) remains to the right from this ray, the difference Φ = L − S^{−2/p} stays in the “upper right” half-plane bounded by the line e^{iπ/p} ℝ. So, the corresponding increment of arg Φ does not exceed π. Once L(\frac{π}{2} + it) has crossed this ray, it never leaves the “left lower” half-plane bounded by the same line and again arg Φ cannot change by more than π. Thus, if Φ does not vanish on Γ_{4}, we have

Δ_{Γ_{4}}Φ ⩽ 2π.

Bringing these estimates together and recalling that Δ_{∂Ω} arg Φ = 2 Δ_{∂Ω_{+}} arg Φ, we see that in this case the total change of arg Φ over the boundary ∂Ω does not exceed (“up to epsilon”) 2(−2π + π + 0 + 2π + \frac{π}{p}) = 2π(1 + \frac{1}{p}). For p > 1, this quantity is strictly less than 4π and thereby there can be no second root of Φ in the strip 0 < Re z < \frac{π}{2} .

Suppose now that Φ vanishes at some point z_{0} ∈ Γ_{4}. Replace a small interval of Γ_{4} centered at z_{0} by the clockwise oriented semicircle bypassing z_{0} from the left. Then we shall get an extra negative change of argument by π (if the root is simple) or more (if it is multiple) when going over this semicircle. And the more roots we have and the higher their multiplicity is, the more we gain on the negative side! ^{2} So if Φ had any zeroes on Γ_{4}, we would end up with even better estimate Δ_{∂Ω} arg Φ < 2π, which would prohibite Φ to have roots in the strip at all! But we know that at least one such root exists, ruling this possibility out.

Haagerup’s integral inequality is thus completely proved. □

Step 5. What to do if one of the coefficients a_{k} is large? From now on, it will be more convenient to replace our usual normalization condition ∑_{k=1}^{n} a_{k}^{2} = 1 by the condition a_{1} = 1. Our aim is to prove the inequality

R_{p}(a) = ∫_{0}^{1} |∑_{k=1}^{n} a_{k}r_{k}(t)|^{p} dt ⩾ A_{p}(1 + a_{2}^{2} + ⋯ + a_{n}^{2})^{p/2} =: A_{p} φ_{p}(a_{2}^{2} + ⋯ + a_{n}^{2})

2Actually this last remark is empty because from what we said above, it is clear that Φ can vanish on Γ_{4} only at the point where L = a and only if S( \frac{π}{2} +it)^{−2/p} reaches the point a simultaneously with L( \frac{π}{2} + it). Also, this root must be simple because the speed at which L( \frac{π}{2} + it) runs over the line ℝ + iπ never vanishes.

with A_{p} = 2^{\frac{p−2}{2}} \frac{Γ( \frac{p+1}{2} )}{Γ( \frac{3}{2} )} for all p ∈ [p_{0}, 2), a = (a_{1}, …, a_{n}) ∈ ℝ^{n}. (Here and below

φ_{p}(x) stands for (1 + x)^{p/2})

The idea of the proof We already know that the desired inequality holds when max_{k} a_{k}^{2} ⩽ \frac{1}{2} ∑_{k} a_{k}^{2} (this is just the result of Step 4, only slightly reformulated). How to get the desired estimate in full generality? We will try to use the almost obvious formula

R_{p}(a) = \frac{R_{p}(a^{+}) + R_{p}(a^{−})}{2}

where a^{±} = (a_{1}, …, a_{n−2}, a_{n−1} ± a_{n}) (n ⩾ 3) to reduce the bad case when one of the coefficients is large to the case when the number of coefficients is n − 1 instead of n. Formally that means that we are going to use induction on n. Since for n ⩾ 3 we have the identity

a_{2}^{2} + ⋯ + a_{n}^{2} = \frac{1}{2} \{[(a_{2}^{−})^{2} + ⋯ + (a_{n−1}^{−})^{2}] + [(a_{2}^{+})^{2} + ⋯ + (a_{n−1}^{+})^{2}]\} ,

the induction step would be fairly easy if the function φ(x) = (1 + x)^{p/2} were convex. Then, denoting for brevity a_{2}^{2} + ⋯ + a_{n}^{2} by x and (a_{2}^{±})^{2} + ⋯ + (a_{n−1}^{±})^{2} by x^{±}, we could make the induction step in just one line:

R_{p}(a) = \frac{R_{p}(a^{+}) + R_{p}(a^{−})}{2} ⩾^{induction assumption} A_{p} \frac{φ_{p}(x^{+}) + φ_{p}(x^{−})}{2}

⩾^{convexity} A_{p} φ_{p} (\frac{x^{+} + x^{−}}{2} ) = A_{p}φ_{p}(x).

Unfortunately there is one small, but annoying obstacle on this otherwise excellent way: for p < 2 the function φ_{p}(x) is concave! When proving inequalities by induction, the standard trick that allows to get rid of such “small but annoying” obstacles is to strengthen the statement. In our case it would be ideal to find a convex function Φ_{p}(x) that is pointwise not less than φ_{p}(x) and to prove the desired estimate with Φ_{p} in place of φ_{p}, i.e., to prove the estimate

R_{p}(a) ⩾ A_{p} Φ_{p}(a_{2}^{2} + ⋯ + a_{n}^{2}).  (∗∗)

Unfortunately, there is no finite function Φ_{p}(x) dominating φ_{p}(x) and convex on the whole ray [0, +∞) either. But we do not actually need that much: we can always reshuffle the coefficients a_{1}, …, a_{n} in the original inequality so that a_{1} = 1 is the largest of them and then just to refer to the result of Step 4 when there is no large coefficient, i.e., when x ⩾ 1. So, it seems that we only need to change our function φ_{p}(x) on the interval [0, 1]. It is really the case and the function that will work is

Φ_{p}(x) = \begin{cases}φ_{p}(x), & if x ⩾ 1; \\ 2φ_{p}(1) − φ_{p}(2 − x), & if 0 ⩽ x ⩽ 1. \end{cases}

In other words, Φ_{p} is the function, whose graph is obtained by replacing the part of the graph of φ_{p} on [0, 1] by the symmetric image of the graph of φ_{p} on [1, 2] with respect to the point (1, φ_{p}(1)) = (1, 2^{p/2}) (see the picture below).

[figure: 1; 2; 1; φ_{p}(x); Φ_{p}(x) = φ_{p}(x); Φ_{p}(x)]

On the interval [0, 1), the graphs of the functions Φ_{p} and φ_{p} are separated by their common tangent line at the point (1, φ_{p}(1)) and therefore Φ_{p}(x) > φ_{p}(x) when 0 ⩽ x < 1, and Φ_{p}(x) = φ_{p}(x) when x ⩾ 1. Now we are ready to make

The induction step. Observe first of all that

Φ_{p} (\frac{x′ + x′′}{2} ) ⩽ \frac{Φ_{p}(x′) + Φ_{p}(x′′)}{2} whenever x′, x′′ ⩾ 0 and \frac{x′ + x′′}{2} ⩽ 1.

(This property is a little bit stronger than just convexity on [0, 1], because we do not assume that max{x′, x′′} ⩽ 1. It was the desire to have this “extended convexity” that prevented us from just replacing the part of the graph of φ_{p} on [0, 1] by the corresponding piece of the tangent line).

Indeed, we may assume without loss of generality that x′ < x′′. If x′′ ⩽ 1, the desired inequality follows immediately from the convexity of Φ_{p} on [0, 1]. Suppose now that x′′ > 1. Notice that

\frac{∂}{∂x′} Φ_{p}(\frac{x′ + x′′}{2} ) = \frac{1}{2} \frac{dΦ_{p}}{dx} |_{x= \frac{x′+x′′}{2}} ⩾ \frac{1}{2} \frac{dΦ_{p}}{dx} |_{x=x′} = \frac{∂}{∂x′} \frac{Φ_{p}(x′) + Φ_{p}(x′′)}{2}

whenever x′ < x′′, \frac{x′+x′′}{2} ⩽ 1, because the derivative Φ′_{p} is increasing on the inter-

val [0, 1]. Therefore the function Φ_{p}(\frac{x′+x′′}{2} ) increases in x′ faster then \frac{Φ_{p}(x′)+Φ_{p}(x′′)}{2}, and it is enough to check our inequality for the maximal possible value of x′, i.e., for x′ = 2 − x′′. But in this case our inequality turns into identity due to the definition of Φ_{p}.

Now suppose that n ⩾ 3 and that the estimate (∗∗) is already established for linear combinations of n − 1 Rademacher functions. Let a = (a_{1}, a_{2}, …, a_{n}) ∈ ℝ^{n}, a_{1} = 1. Denote as before x = a_{2}^{2} + ⋯ + a_{n}^{2}. Three cases are possible:

• a_{1} = 1 is not the largest coefficient. Then x > 1 and (∗∗) coincides with the homogeneous estimate R_{p}(a) ⩾ A_{p}(a_{1}^{2} + ⋯ + a_{n}^{2})^{p/2}. Rearranging the coefficients in the order of decreasing absolute value (this doesn’t change either part of the homogeneous estimate) and renormalizing so that the first coefficient is again 1, we shall arrive at some new version of the inequality (∗∗) in which a_{1} = 1 is the largest coefficient. This new version is not weaker than the homogeneous estimate, i.e., not weaker than the original version we wanted to prove.

• a_{1} = 1 is the largest coefficient, x ⩾ 1. In this case, (∗∗) still coincides with the homogeneous estimate and max_{k} a_{k}^{2} = 1 ⩽ \frac{1}{2}(1 + x) = \frac{1}{2} ∑_{k} a_{k}^{2}. So we may just refer to Step 4.

• a_{1} = 1 is the largest coefficient, x < 1. Then (in the same notation as above)

R_{p}(a) = \frac{R_{p}(a^{+}) + R_{p}(a^{−})}{2} ⩾^{induction assumption} A_{p} \frac{Φ_{p}(x^{+}) + Φ_{p}(x^{−})}{2}

⩾ A_{p} Φ_{p} (\frac{x^{+} + x^{−}}{2} ) = A_{p}Φ_{p}(x)

(because x^{±} ⩾ 0 and \frac{x^{−} + x^{+}}{2} = x < 1).

So it remains only to prove

The induction base n = 2. We need to prove the inequality

R_{p}(1, a_{2}) ⩾ A_{p}Φ_{p}(a_{2}^{2}).

Again we may assume that a_{1} = 1 is the largest of the two coefficients, i.e., a_{2}^{2} = x ⩽ 1 (otherwise, after exchanging a_{1} and a_{2} and renormalizing, we shall obtain a stronger inequality to prove). Computing the left hand part and recalling the definition of Φ_{p}(x) and our hypothetical formula for A_{p}, we can rewrite it as

\frac{(1 + \sqrt{x})^{p} + (1 − \sqrt{x})^{p}}{2} ⩾ 2^{\frac{p−2}{2}} \frac{Γ(\frac{p+1}{2})}{Γ(\frac{3}{2})} [2 · 2^{p/2} − (3 − x)^{p/2}] (0 ⩽ x ⩽ 1).

The ratio \frac{Γ( \frac{p+1}{2} )}{Γ( \frac{3}{2} )} does not exceed 1 for all p ∈ [p_{0}, 2), so it is enough to prove the estimate

\frac{(1 + \sqrt{x})^{p} + (1 − \sqrt{x})^{p}}{2} ⩾ 2^{\frac{p−2}{2}} [2 · 2^{p/2} − (3 − x)^{p/2}] = 2^{p−1} [2 − (\frac{3 − x}{2})^{p/2}]

At first glance, this strengthening of the statement to prove may seem completely unmotivated, but if we recall that for p = p_{0} this ratio equals 1 and that we do not know (or, at least, do not want to know) what p_{0} is, it becomes quite natural.

Dividing both parts by 2^{p−1} and transferring (\frac{3 − x}{2})^{p/2} to the left hand side, we can restate our estimate as

(\frac{1 + \sqrt{x}}{2})^{p} + (\frac{1 − \sqrt{x}}{2})^{p} + (\frac{3 − x}{2})^{p/2} ⩽ 2.

Now denote

a = \frac{1 + \sqrt{x}}{2}, b = \frac{1 − \sqrt{x}}{2}.

Taking into account that 1 − x = 4ab, we can rewrite our inequality as

a^{p} + b^{p} + (1 + 2ab)^{p/2} ⩽ 2 (p ⩽ 2; a, b ⩾ 0; a + b = 1).

Since for p = 2 we have an identity and since the left hand part is convex in p, it is sufficient to show that

\frac{d}{dp}|_{p=2} [a^{p} + b^{p} + (1 + 2ab)^{p/2}] ⩽ 0,

or, which is the same, that

a^{2} ln a + b^{2} ln b + \frac{1}{2}(1 + 2ab) ln(1 + 2ab) ⩽ 0.

Note now that \frac{d^{2}}{dt^{2}} [(1+t) ln(1+t)] = \frac{1}{1 + t} ⩽ 1 for every t ⩾ 0. Thereby, using the

second order Taylor formula at the point t_{0} = 0, we obtain (1+t) ln(1+t) ⩽ t+ \frac{1}{2}t^{2}. In particular,

\frac{1}{2}(1 + 2ab) ln(1 + 2ab) ⩽ \frac{1}{2} (2ab + \frac{1}{2}(2ab)^{2}) = ab + (ab)^{2}.

On the other hand, a^{2} ln a = a^{2} ln(1 − b) = −a^{2} (b + \frac{b^{2}}{2} + …) ⩽ −a^{2} (b + \frac{b^{2}}{2} )

and, similarly, b^{2} ln b ⩽ −b^{2} (a + \frac{a^{2}}{2} ) . Thus

a^{2} ln a+b^{2} ln b ⩽ −a^{2} (b + \frac{b^{2}}{2} )−b^{2} (a + \frac{a^{2}}{2} ) = −ab(a+b)−(ab)^{2} = −ab−(ab)^{2}.

Haagerup’s theorem (case p < 2) is thus completely proved. □

22 F.L. Nazarov, A.N. Podkorytov

References

[1] K. Ball, Cube slicing in ℝ^{n}, Proc. Amer. Math. Soc., 97 (3) (1986), 465–473.

[2] K. Ball, Some remarks on the geometry of convex sets, Geometric aspects of functional analysis (1986/87), 251–260, Lecture Notes in Math., 1317, Springer, Berlin-New York (1988).

[3] H. Busemann, C. M. Petty, Problems on convex bodies, Math. Scand., 4 (1956), 88–94.

[4] R. J. Gardiner, A positive answer to the Busemann-Petty problem in three dimensions, Ann. of Math. (2) 140 (1994), no. 2, 435–447.

[5] U. Haagerup, The best constants in the Khintchine inequality, Studia Math., 70 (3) (1982), 231–283.

[6] G. Y. Zhang, Intersection bodies and the Busemann-Petty inequalities in ℝ^{4}, Ann. of Math. (2) 140 (1994), no. 2, 331–346.

F.L. Nazarov Department of Mathematics, Michigan State University East Lansing, MI 48824, USA

E-mail address: fedja@math.msu.edu

A. N. Podkorytov Department of Mathematics, St. Petersburg State University Bibliotechnaya pl. 2 Peterhof, St. Petersburg 198904, Russia

E-mail address: pan@pan.usr.pu.ru