On the Best Constants in the Khintchine Inequality

Johannes Wissel and Jϕseph Turian Mentor: Arteom Zvavitch Supervisor: Gideon Schechtman Department of Theoretical Mathematics, Weizmann Institute of Science

Abstract We show a self-contained new proof for the best B_{p} constant in the Khintchine

Inequality for p > 3 using only elementary calculus.

Introduction

Consider the Khintchine Inequality:

A_{p} · (∑_{i=1}^{n} α_{i}^{2})^{\frac{1}{2}} ≤ (E| ∑_{i=1}^{n} ε_{i} · α_{i}|^{p})^{\frac{1}{p}} ≤ B_{p} · (∑_{i=1}^{n} α_{i}^{2})^{\frac{1}{2}} ,

where p ∈ IR, p > 0 and, α_{i} ∈ IR, for all i such that 1 ≤ i ≤ n and ε_{1}, ⋯, ε_{n} are independent random variables that each assume the values ±1 with probabilities \frac{1}{2}.

The actual inequality (in which B_{p} and A_{p} are not necessarily the best) was proved by Khintchine. But we are looking for the best constants A_{p} and B_{p} that satisfy the inequality for all natural numbers n. For all p, a solution and proof was given in [1]. Here we present another proof, for p > 3, which uses only elementary calculus.

Proof

Because of the homogeneity of the inequality, we may assume that ∑_{i=1}^{n} α_{i}^{2} = 1. Therefore, we can write the inequality in the following way:

A_{p} ≤ (E| ∑_{i=1}^{n} α_{i} · ε_{i}|^{p})^{\frac{1}{p}} ≤ B_{p}.

Using the definition of mathematical expectation, we arrive to the following formula:

A_{p}^{p} ≤ ∑_{{ε_{i}=±1, \atop 1≤i≤n}} \frac{1}{2^{n}} | ∑_{i=1}^{n} α_{i} · ε_{i}|^{p} ≤ B_{p}^{p}.

And finally:

A_{p}^{p} · 2^{n} ≤ ∑_{{ε_{i}=±1, \atop 1≤i≤n}} | ∑_{i=1}^{n} α_{i} · ε_{i}|^{p} ≤ B_{p}^{p} · 2^{n}.

As changing the signs of the α_{i} does not change the sum, we may assume that every α is non-negative. For n = 1, we obtain α_{1} = 1. If n ≥ 2 and a_{k} = 0 for some k, without loss of generality let k = n. Then note that:

∑_{{ε_{i}=±1, \atop 1≤i≤n}} | ∑_{i=1}^{n} α_{i} · ε_{i}|^{p} = 2 · ∑_{{ε_{i}=±1, \atop 1≤i≤n−1}} | ∑_{i=1}^{n−1} α_{i} · ε_{i}|^{p},

and that the inequality is therefore

A_{p}^{p} · 2^{n−1} ≤ ∑_{{ε_{i}=±1, \atop 1≤i≤n−1}} | ∑_{i=1}^{n−1} α_{i} · ε_{i}|^{p} ≤ B_{p}^{p} · 2^{n−1}.

Thus, the equality for n reduces to the inequality for n − 1.

We would like to show that to find the best value of B_{p}, it suffices to find the upper bound of

\frac{1}{2^{n}} · ∑_{ε_{i}=±1} | ∑_{i=1}^{n} \frac{ε_{i}}{\sqrt{n}}|^{p}

for all valid ns.

We now define the following function f:

f(α_{1}, …, α_{n}) = ∑_{{ε_{i}=±1, \atop 1≤i≤n}} | ∑_{i=1}^{n} α_{i} · ε_{i}|^{p},

with ∑_{i=1}^{n} α_{i}^{2} = 1.

Theorem 1

f(α_{1}, …, α_{n}) = ∑_{ε_{i}=±1} | ∑_{i=1}^{n} ε_{i}α_{i}|^{p},

under the constraints p > 3 and ∑_{i=1}^{n} α_{i}^{2} = 1, is not maximal for the given α sequence if there exists an α_{j} and α_{k} such that α_{k} > α_{j} > 0.

It follows from Theorem 1 that f is only maximal when the non-zero αs are equal.

Proof of Theorem 1:

We prove Theorem 1 by contradiction. Let us suppose that f is maximal for some α_{1}, ⋯, α_{n}, such that, without loss of generality, α_{2} > α_{1} > 0.

We define following function:

h(θ) = f(k(θ), l(θ), α_{3}, …, α_{n}) = ∑_{ε_{i}=±1} |ε_{1}k(θ) + ε_{2}l(θ) + A_{ε}|^{p},

with A_{ε} as a constant dependent upon the ε sequence, {ε_{i}}_{i=3}^{n}:

A_{ε} = ∑_{i=3}^{n} ε_{i}α_{,}i

and with

k(θ) = \sqrt{(1 − θ)α_{1}^{2} + θα_{2}^{2}} = \sqrt{θ(α_{2}^{2} − α_{1}^{2}) + α_{1}^{2}},
l(θ) = \sqrt{(1 − θ)α_{2}^{2} + θα_{1}^{2}} = \sqrt{θ(α_{1}^{2} − α_{2}^{2}) + α_{2}^{2}},

and θ ∈ [0, 1]. Note that k(0) = α_{1}, l(0) = α_{2}, and k(θ)^{2} + l(θ)^{2} = α_{1}^{2} + α_{2}^{2}, for all θ ∈ [0, 1]. Because α_{2} ≠ 0, k(θ) and l(θ) are not zero-functions.

If we can show that h(θ) increases at θ = 0, then we have a contradiction because f increases with h(θ) and is therefore not maximal for α_{1}, ⋯, α_{n} as assumed.

Lemma 1 h(θ) is increasing at θ = 0.

Proof of Lemma 1:

k′(θ) = \frac{α_{2}^{2} − α_{1}^{2}}{2\sqrt{θ(α_{2}^{2} − α_{1}^{2}) + α_{1}^{2}}}

l′(θ) = \frac{α_{1}^{2} − α_{2}^{2}}{2\sqrt{θ(α_{1}^{2} − α_{2}^{2}) + α_{2}^{2}}}

and therefore k′(0) = \frac{a_{2}^{2}−a_{1}^{2}}{2·a_{1}} and l′(0) = \frac{a_{1}^{2}−a_{2}^{2}}{2·a_{2}} .

Note that k(θ)k′(θ) + l(θ)l′(θ) = 0 for any θ.

Define the following constant: c = \frac{α_{2}^{2}−α_{1}^{2}}{2·α_{1}·α_{2}} . Thus, c > 0.

Note that

k′(0) − l′(0) = \frac{α_{2}^{2}−α_{1}^{2}}{2·α_{1}·α_{2}} · (l(0) + k(0)) = c · (l(0) + k(0))

and

k′(0) + l′(0) = \frac{α_{2}^{2}−α_{1}^{2}}{2·α_{1}·α_{2}} · (l(0) − k(0)) = c · (l(0) − k(0))

Also note that

k(0)l′(0) + k′(0)l(0) = \frac{(α_{2}^{2} − α_{1}^{2})^{2}}{2 · α_{1} · α_{2}} = c · (α_{2}^{2} − α_{1}^{2}).

h′(θ) = p · ∑_{ε_{i}=±1} |ε_{1}k(θ) + ε_{2}l(θ) + A_{ε}|^{p−2}(ε_{1}k(θ) + ε_{2}l(θ) + A_{ε})(ε_{1}k′(θ) + ε_{2}l′(θ))

From this point forward, k (and k′) shall be understood as k(0) (and k′(0)) and l (and l′) as l(0) (and l′(0)).

h′(0) =  p · ∑_{ε_{i}=±1} |ε_{1}k + ε_{2}l + A_{ε}|^{p−2}(ε_{1}k + ε_{2}l + A_{ε})(ε_{1}k′ + ε_{2}l′)
=  p · ∑_{ε_{i}=±1} |ε_{1}k + ε_{2}l + A_{ε}|^{p−2}(k · k′ + l · l′ + ε_{1} · ε_{2}(k · l′ + k′ · l) +
+ A_{ε}(ε_{1}k′ + ε_{2}l′))
=  p · ∑_{ε_{i}=±1} |ε_{1}k + ε_{2}l + A_{ε}|^{p−2}(ε_{1} · ε_{2} · c · (α_{2}^{2} − α_{1}^{2}) + A_{ε}(ε_{1}k′ + ε_{2}l′))

Replacing ε_{1} and ε_{2} with explicit values, we obtain the following equation:

h′(0) = p · ∑_{{ε_{i}=±1, \atop 3≤i≤n}}  |k + l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) + A_{ε}(k′ + l′)) −
− |k − l − A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) + A_{ε}(k′ − l′)) −
− |k − l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) − A_{ε}(k′ − l′)) +
+ |k + l − A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) − A_{ε}(k′ + l′))
= p · ∑_{{ε_{i}=±1, \atop 3≤i≤n}}  |k + l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) + A_{ε} · c · (l − k) −
− |k − l − A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) + A_{ε} · c · (l + k) −
− |k − l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) − A_{ε} · c · (l + k) +
+ |k + l − A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) − A_{ε} · c · (l − k)

Note that ∀A_{ε} ≥ 0, there is an A_{−ε} ≤ 0 such that |A_{ε}| = |A_{−ε}|. The inverse is also true. Therefore, the equation can be condensed:

h′(0) = 2 · p · ∑_{{ε_{i}=±1, \atop 3≤i≤n}}  |k + l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) + A_{ε} · c · (l − k) −
− |k − l + A_{ε}|^{p−2}(c · (α_{2}^{2} − α_{1}^{2}) − A_{ε} · c · (l + k)
= 2 · p · c · ∑_{{ε_{i}=±1, \atop 3≤i≤n}}  |k + l + A_{ε}|^{p−2}(α_{2}^{2} − α_{1}^{2} + A_{ε} · (l − k) −
− |k − l + A_{ε}|^{p−2}(α_{2}^{2} − α_{1}^{2} − A_{ε} · (l + k)

Now we compare every non-negative A_{ε} with its corresponding A_{−ε}.

h′(0) = 2 · p · c · ∑_{{ε_{i}=±1, \atop Aε≥0}}  |k + l + A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) + A_{ε} · (l − k) +
+ |k + l − A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) − A_{ε} · (l − k) +
− |k − l + A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) − A_{ε} · (l + k) −
− |k − l − A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) + A_{ε} · (l + k).

Note that when the summation constraint is A_{ε} ≥ 0, the sum is greater than or equal to the sum when the summation constraint is A_{ε} > 0, that is

h′(0) ≥ 2 · p · c · ∑_{{ε_{i}=±1, \atop Aε>0}}  |k + l + A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) + A_{ε} · (l − k) +
+ |k + l − A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) − A_{ε} · (l − k) +
− |k − l + A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) − A_{ε} · (l + k) −
− |k − l − A_{ε}|^{p−2}((α_{2}^{2} − α_{1}^{2}) + A_{ε} · (l + k).

Let us prove that each summand is greater than 0, which will prove that h′(0) > 0. For each summand, define the constants: x_{ε} = \frac{l+k}{A_{ε}} and y_{ε} = \frac{l−k}{A_{ε}}. Note that x_{ε} > y_{ε} > 0. Then:

h′(0) ≥ 2 · p · c · ∑_{{ε_{i}=±1, \atop Aε>0}} A_{ε}^{p} · ( |x_{ε} + 1|^{p−2}(x_{ε} · y_{ε} + y_{ε}) +
+ |x_{ε} − 1|^{p−2}(x_{ε} · y_{ε} − y_{ε}) −
− |y_{ε} − 1|^{p−2}(x_{ε} · y_{ε} − x_{ε}) −
− |y_{ε} + 1|^{p−2}(x_{ε} · y_{ε} + x_{ε}))

Therefore, it is enough to prove that for all x > y > 0

γ_{x}(y) =  (|x_{ε} + 1|^{p−2}(x_{ε} + 1) + |x_{ε} − 1|^{p−2}(x_{ε} − 1)) · y −
− (|y_{ε} − 1|^{p−2}(y_{ε} − 1) + |y_{ε} + 1|^{p−2}(y_{ε} + 1)) · x > 0

Note that γ_{x}(0) = 0 and γ_{x}(x) = 0. Thus, if we can prove that γ′′_{x}(y) is less than 0, we will prove that γ_{x}(y) is greater than 0 for valid values of x and y.

γ′_{x}(y) = (|x + 1|^{p−2} · (x + 1) + |x − 1|^{p−2} · (x − 1)) − x · (p − 1) · (|y − 1|^{p−2} + |y + 1|^{p−2}).

Case 1: y − 1 = 0

The question that arises is if γ′_{x}(y) is differentiable, or, more specifically, if δ(y) = |y−1|^{p−2} is differentiable at y = 1. δ′(y) = (p − 2) · (y − 1)^{p−3} for y − 1 > 0 δ′(y) = −(p − 2) · (1 − y)^{p−3} for y − 1 < 0 By the definition of the derivative,

δ′(y_{0}) = lim_{y→y_{0}} \frac{|y − 1|^{p−2} − |y_{0} − 1|^{p−2}}{y − y_{0}} ,

(if the limit exists).

To prove that the limit exists, we take the limit from both sides.

lim_{y→1^{+}} \frac{|y − 1|^{p−2}}{y − 1} = lim_{y→1^{+}} |y − 1|^{p−3} = 0
lim_{y→1^{−}} \frac{|y − 1|^{p−2}}{y − 1} = − lim_{y→1^{−}} \frac{|y − 1|^{p−2}}{|y − 1|} = − lim_{y→y^{−}} |y − 1|^{p−3} = 0

So thus δ′(1) = 0 and γ′′_{x}(y) < 0.

Otherwise, when y − 1 ≠ 0,

γ′′_{x}(y) = −x · (p − 1) · (p − 2) · (|y − 1|^{p−4} · (y − 1) + |y + 1|^{p−4} · (y + 1)).

Case 2: y − 1 > 0

γ′′_{x}(y) = −x · (p − 1) · (p − 2) · (|1 + y|^{p−3} + |1 − y|^{p−3}) < 0.

Case 3: y − 1 < 0

γ′′_{x}(y) = −x · (p − 1) · (p − 2) · (|1 + y|^{p−3} − |1 − y|^{p−3}) < 0.

Therefore, γ_{x}(y) is greater than 0 for x > y > 0. Note that this inequality does not hold for p ≤ 3 Thus, h′(0) > 0 and Lemma 1 has been proven.

Thus, Theorem 1 has been proven.

Q.E.D.

Conclusion

Thus, using elementary calculus, the best B_{p} constant for the Khintchine Inequality has been derived for p > 3.

Suggestions for Future Research

It could be possible to create a proof for 2 < p < 3 using a similar methodology to that of this project. If we can show that h′(0) = 0 implies h′′(0) > 0, then we could also get the contradiction with which we proved Theorem 1.

Acknowledgments

We would like to thank our mentor, Arteom Zvavitch, for all his support and inspiration throughout this project. We would also like to the coordinators and staff of the 29th Annual Bessie F. Lawrence Summer Science Institute and the American Committee for the Weizmann Institute of Science for giving us the opportunity to “transport the sun.”

References

[1] Haagerup U. (1981). The best constants in the Khintchine inequality, Studia Mathematica 70, pp. 232-283.