# Lévy’s inequality, Rademacher sums, and Kahane’s inequality

Jordan Bell
May 21, 2015

## 1 Lévy’s inequality

Let $(\Omega,\mathscr{A},P)$ be a probability space. A random variable is a Borel measurable function $\Omega\to\mathbb{R}$. For a random variable $X$, we denote by $X_{*}P$ the pushforward measure of $P$ by $X$. $X_{*}P$ is a Borel probability measure on $\mathbb{R}$, called the distribution of $X$. A random variable $X$ is called symmetric when the distribution of $X$ is equal to the distribution of $-X$. Because the collection $\{(-\infty,a]:a\in\mathbb{R}\}$ generates the Borel $\sigma$-algebra of $\mathbb{R}$, the statement that $X_{*}P=(-X)_{*}P$ is equivalent to the statement that for all $a\in\mathbb{R}$,

 $P(\{\omega\in\Omega:X(\omega)\leq a\})=P(\omega\in\Omega:-X(\omega)\leq a\}).$

The following is Lévy’s inequality.11 1 Joe Diestel, Hans Jarchow, and Andrew Tonge, Absolutely Summing Operators, p. 213, Theorem 11.3.

###### Theorem 1 (Lévy’s inequality).

Suppose that $\chi_{k}$, $k\geq 1$, are independent symmetric random variables, that $U$ is a real or complex Banach space, and that $u_{k}\in U$, $k\geq 1$. Then for each $a>0$ and for each $n\geq 1$,

 $P\left(\max_{1\leq k\leq n}\left\|\sum_{1\leq j\leq k}\chi_{j}u_{j}\right\|% \geq a\right)\leq 2\cdot P\left(\left\|\sum_{1\leq j\leq n}\chi_{j}u_{j}\right% \|\geq a\right).$
###### Proof.

Let $S_{0}=0$ and for $1\leq k\leq n$,

 $S_{k}(\omega)=\sum_{j=1}^{k}\chi_{j}(\omega)u_{j},\qquad\omega\in\Omega.$

For $1\leq k\leq n$, the function $\omega\mapsto(\chi_{1}(\omega),\ldots,\chi_{k}(\omega))$ is Borel measurable $\Omega\to\mathbb{R}^{k}$.22 2 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 152, Lemma 4.49. The function $(t_{1},\ldots,t_{k})\mapsto\sum_{j=1}^{k}t_{j}u_{j}$ is continuous $\mathbb{R}^{k}\to U$. And the function $u\mapsto\left\|u\right\|$ is continuous $U\to\mathbb{R}$. Therefore $\omega\mapsto\left\|S_{k}(\omega)\right\|$, the composition of these functions, is Borel measurable $\Omega\to\mathbb{R}$. This then implies that $\omega\mapsto\max_{1\leq k\leq n}\left\|S_{k}(\omega)\right\|$ is Borel measurable $\Omega\to\mathbb{R}$. Let

 $A=\{\omega\in\Omega:\max_{1\leq k\leq n}\left\|S_{k}(\omega)\right\|\geq a\},% \quad B=\{\omega\in\Omega:\left\|S_{n}(\omega)\right\|\geq a\},$

for which $B\subset A$. For $1\leq k\leq n$, let

 $A_{k}=\bigcap_{0\leq j

It is apparent that that $A_{1},\ldots,A_{n}$ are pairwise disjoint and that $A=\bigcup_{k=1}^{n}A_{k}$.

For $1\leq k\leq n$, let

 $T_{n,k}(\omega)=S_{k}(\omega)-\sum_{j=k+1}^{n}\chi_{j}(\omega)u_{j}=\sum_{j=1}% ^{k}\chi_{j}(\omega)u_{j}-\sum_{j=k+1}^{n}\chi_{j}(\omega)u_{j},\qquad\omega% \in\Omega,$

in other words, $S_{n}+T_{n,k}=2S_{k}$. Let

 $U_{k}=A_{k}\cap B,\qquad V_{k}=A_{k}\cap\{\omega\in\Omega:\left\|T_{n,k}(% \omega)\right\|\geq a\}.$

If $\omega\in A_{k}$, then

 $\left\|S_{n}(\omega)+T_{n,k}(\omega)\right\|=2\left\|S_{k}(\omega)\right\|\geq 2a,$

which implies that at least one of the inequalities $\left\|S_{n}(\omega)\right\|\geq a$ or $\left\|T_{n,k}(\omega)\right\|\geq a$ is true. Therefore

 $A_{k}=U_{k}\cup V_{k}.$

Because $\chi_{1},\ldots,\chi_{n}$ are independent, the random vector $X=(\chi_{1},\ldots,\chi_{n}):\Omega\to\mathbb{R}^{n}$ has the pushforward measure

 $X_{*}P={\chi_{1}}_{*}P\times\cdots\times{\chi_{n}}_{*}P,$

and for each $1\leq k\leq n$, the random vector $X_{k}=(\chi_{1},\ldots,\chi_{k},-\chi_{k+1},\ldots,-\chi_{n}):\Omega\to\mathbb% {R}^{n}$ has the pushforward measure

 ${X_{k}}_{*}P={\chi_{1}}_{*}P\times\cdots{\chi_{k}}_{*}P\times{(-\chi_{k+1})}_{% *}P\times\cdots{(-\chi_{n})}_{*}P,$

and because each $\chi_{j}$ is symmetric, these pushforward measures are equal. Define $\sigma_{k}:\mathbb{R}^{k}\to\mathbb{R}$ by

 $\sigma_{k}(t_{1},\ldots,t_{k})=\left\|\sum_{j=1}^{k}t_{j}u_{j}\right\|,\qquad(% t_{1},\ldots,t_{k})\in\mathbb{R}^{k},$

define $\sigma_{0}=0$, and set

 $\displaystyle H_{k}$ $\displaystyle=\left(\bigcap_{0\leq j $\displaystyle\cap\{(t_{1},\ldots,t_{n})\in\mathbb{R}^{n}:\sigma_{k}(t_{1},% \ldots,t_{k})\geq a,\sigma_{n}(t_{1},\ldots,t_{n})\geq a\}.$

Because each $\sigma_{j}$ is continuous, $H_{k}$ is a Borel set in $\mathbb{R}^{n}$. Then we have

 $\displaystyle P(U_{k})$ $\displaystyle=P(A_{k}\cap B)$ $\displaystyle=P(X^{-1}(H_{k}))$ $\displaystyle=(X_{*}P)(H_{k})$ $\displaystyle=({X_{k}}_{*}P)(H_{k})$ $\displaystyle=P(X_{k}^{-1}(H_{k}))$ $\displaystyle=P(A_{k}\cap\{\omega\in\Omega:\left\|T_{n,k}(\omega)\right\|\geq a\})$ $\displaystyle=P(V_{k});$

among the above equalities, the two equalities that deserve chewing on are

 $P(A_{k}\cap B)=P(X^{-1}(H_{k}))\quad\textrm{and}\quad P(X_{k}^{-1}(H_{k}))=P(A% _{k}\cap\{\omega\in\Omega:\left\|T_{n,k}(\omega)\right\|\geq a\}).$

Thus we have

 $P(A_{k})=P(U_{k}\cup V_{k})\leq P(U_{k})+P(V_{k})=2P(U_{k})=2P(A_{k}\cap B).$

Therefore

 $\displaystyle P(A)$ $\displaystyle=\sum_{k=1}^{n}P(A_{k})$ $\displaystyle\leq\sum_{k=1}^{n}2P(A_{k}\cap B)$ $\displaystyle=2P(A\cap B)$ $\displaystyle=2P(B),$

proving the claim. ∎

Suppose that $\epsilon_{n}:(\Omega,\mathscr{A},P)\to(\mathbb{R},\mathscr{B}_{\mathbb{R}},\lambda)$, $n\geq 1$, are independent random variables each with the Rademacher distribution: for each $n$,

 ${\epsilon_{n}}_{*}P=\frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1},$

in other words, $P(\epsilon_{n}=1)=\frac{1}{2}$ and $P(\epsilon_{n}=-1)=\frac{1}{2}$.

We now use Lévy’s inequality to prove the following for independent random variables with the Rademacher distribution.33 3 Joe Diestel, Hans Jarchow, and Andrew Tonge, Absolutely Summing Operators, p. 214, Lemma 11.4.

###### Theorem 2.

Suppose that $X$ is a real or complex Banach space, and that $x_{k}\in X$, $k\geq 1$. Then for each $a>0$ and for each $n\geq 1$,

 $P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|\geq 2a\right)\leq 4\left% (P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|\geq a\right)\right)^{2}.$
###### Proof.

Let $S_{0}=0$ and for $1\leq k\leq n$, define

 $S_{k}(\omega)=\sum_{1\leq j\leq k}\epsilon_{j}(\omega)x_{j},\qquad\omega\in\Omega.$

Let

 $A=\left\{\max_{1\leq k\leq n}\left\|S_{k}\right\|\geq a\right\},\quad B=\{% \left\|S_{n}\right\|\geq a\},\quad C=\{\left\|S_{n}\right\|\geq 2a\}.$

Lévy’s inequality tells us that $P(A)\leq 2P(B)$.

For $1\leq k\leq n$, let

 $A_{k}=\bigcap_{0\leq j

and

 $C_{k}=\{\left\|S_{n}-S_{k-1}\right\|\geq a\}.$

If $\omega\in A_{k}\cap C$, then

 $\left\|S_{n}(\omega)-S_{k-1}(\omega)\right\|\geq\left\|S_{n}(\omega)\right\|-% \left\|S_{k-1}(\omega)\right\|\geq 2a-a=a,$

hence $A_{k}\cap C\subset C_{k}$. Then because $C\subset A$ and because $A$ is the disjoint union of $A_{1},\ldots,A_{n}$,

 $P(C)=P(A\cap C)=P\left(\bigcup_{k=1}^{n}(A_{k}\cap C)\right)=\sum_{k=1}^{n}P(A% _{k}\cap C)\leq\sum_{k=1}^{n}P(A_{k}\cap C_{k}).$

Let $1\leq k\leq n$. $P(\epsilon_{k}^{2}=1)=1$, so for almost all $\omega\in\Omega$,

 $\left\|\sum_{j=k}^{n}\epsilon_{j}(\omega)x_{j}\right\|=\left\|\epsilon_{k}(% \omega)\sum_{j=k}^{n}\epsilon_{j}(\omega)x_{j}\right\|=\left\|x_{k}+\sum_{j=k+% 1}^{n}\epsilon_{k}(\omega)\epsilon_{j}(\omega)x_{j}\right\|.$

Thus, for

 $D_{k}=\left\{\left\|x_{k}+\sum_{j=k+1}^{n}\epsilon_{k}\epsilon_{j}x_{j}\right% \|\geq a\right\},$

we have

 $P(C_{k}\triangle D_{k})=0.$

Let $(\delta_{1},\ldots,\delta_{n})\in\{+1,-1\}^{n}$. On the one hand, because $\delta_{j}^{2}=1$ and using that $\epsilon_{1},\ldots,\epsilon_{n}$ are independent,

 $\begin{split}&\displaystyle P(\epsilon_{1}=\delta_{1},\ldots,\epsilon_{k}=% \delta_{k},\epsilon_{k}\epsilon_{k+1}=\delta_{k+1},\ldots,\epsilon_{k}\epsilon% _{n}=\delta_{n})\\ \displaystyle=&\displaystyle P(\epsilon_{1}=\delta_{1},\ldots,\epsilon_{k}=% \delta_{k},\epsilon_{k+1}=\delta_{k}\delta_{k+1},\ldots,\epsilon_{n}=\delta_{k% }\delta_{n})\\ \displaystyle=&\displaystyle P(\epsilon_{1}=\delta_{1})\cdots P(\epsilon_{k}=% \delta_{k})P(\epsilon_{k+1}=\delta_{k}\delta_{k+1})\cdots P(\epsilon_{n}=% \delta_{k}\delta_{n})\\ \displaystyle=&\displaystyle 2^{-n}.\end{split}$

On the other hand, for $k+1\leq j\leq n$ we have

 $\begin{split}&\displaystyle P(\epsilon_{k}\epsilon_{j}=\delta_{j})\\ \displaystyle=&\displaystyle P(\epsilon_{k}\epsilon_{j}=\delta_{j}|\epsilon_{k% }=1)P(\epsilon_{k}=1)+P(\epsilon_{k}\epsilon_{j}=\delta_{j}|\epsilon_{k}=-1)P(% \epsilon_{k}=-1)\\ \displaystyle=&\displaystyle\frac{1}{2}P(\epsilon_{j}=\delta_{j})+\frac{1}{2}P% (\epsilon_{j}=-\delta_{j})\\ \displaystyle=&\displaystyle\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{% 1}{2}\\ \displaystyle=&\displaystyle\frac{1}{2},\end{split}$

and hence

 $P(\epsilon_{1}=\delta_{1})\cdots P(\epsilon_{k}=\delta_{k})P(\epsilon_{k}% \epsilon_{k+1}=\delta_{k+1})\cdots P(\epsilon_{k}\epsilon_{n}=\delta_{n})=2^{-% n}.$

Therefore, for each $1\leq k\leq n$ and for each $(\delta_{1},\ldots,\delta_{n})\in\{+1,-1\}^{n}$,

 $\begin{split}&\displaystyle P(\epsilon_{1}=\delta_{1},\ldots,\epsilon_{k}=% \delta_{k},\epsilon_{k}\epsilon_{k+1}=\delta_{k+1},\ldots,\epsilon_{k}\epsilon% _{n}=\delta_{n})\\ \displaystyle=&\displaystyle P(\epsilon_{1}=\delta_{1})\cdots P(\epsilon_{k}=% \delta_{k})P(\epsilon_{k}\epsilon_{k+1}=\delta_{k+1})\cdots P(\epsilon_{k}% \epsilon_{n}=\delta_{n}).\end{split}$

But for almost all $\omega\in\Omega$,

 $(\epsilon_{1}(\omega),\ldots,\epsilon_{k}(\omega),\epsilon_{k}(\omega)\epsilon% _{k+1}(\omega),\ldots,\epsilon_{k}(\omega)\epsilon_{n}(\omega))\in\{+1,-1\}^{n},$

so it follows that

 $\epsilon_{1},\ldots,\epsilon_{k},\epsilon_{k}\epsilon_{k+1},\ldots,\epsilon_{k% }\epsilon_{n}$

are independent random variables. We check that $A_{k}\in\sigma(\epsilon_{1},\ldots,\epsilon_{k})$ and $D_{k}\in\sigma(\sigma_{k}\sigma_{k+1},\ldots,\sigma_{k}\sigma_{n})$, and what we have just established means that these $\sigma$-algebras are independent, so

 $P(A_{k}\cap D_{k})=P(A_{k})P(D_{k}).$

But

 $A_{k}\cap(C_{k}\triangle D_{k})=(A_{k}\cap C_{k})\triangle(A_{k}\cap D_{k}),$

so, because $P(C_{k}\triangle D_{k})=0$,

 $P(A_{k}\cap C_{k})=P(A_{k}\cap D_{k})=P(A_{k})P(D_{k})=P(A_{k})P(C_{k}).$

We had already established that $P(C)\leq\sum_{k=1}^{n}P(A_{k}\cap C_{k})$. Using this with the above, and the fact that $A$ is the disjoint union of $A_{1},\ldots,A_{n}$, we obtain

 $\displaystyle P(C)$ $\displaystyle\leq\sum_{k=1}^{n}P(A_{k}\cap C_{k})$ $\displaystyle=\sum_{k=1}^{n}P(A_{k})P(C_{k})$ $\displaystyle\leq\left(\sum_{k=1}^{n}P(A_{k})\right)\max_{1\leq k\leq n}P(C_{k})$ $\displaystyle=P\left(\bigcup_{k=1}^{n}A_{k}\right)\max_{1\leq k\leq n}P(C_{k})$ $\displaystyle=P(A)\max_{1\leq k\leq n}P(C_{k}).$

As we stated before, we have from Lévy’s inequality that $P(A)\leq 2P(B)$, with which

 $P(C)\leq 2P(B)\max_{1\leq k\leq n}P(C_{k}).$

To prove the claim it thus suffices to show that

 $\max_{1\leq k\leq n}P(C_{k})\leq 2P(B).$

Let $1\leq k\leq n$. For $\delta=(\delta_{1},\ldots,\delta_{k-1})\in\{+1,-1\}^{k-1}$, let let $H_{k,\delta,+}$ be those $(t_{1},\ldots,t_{n})\in\mathbb{R}^{n}$ such that (i) for each $1\leq j\leq k-1$, $t_{j}=\delta_{j}$, (ii) $\left\|\sum_{j=k}^{n}t_{j}x_{j}\right\|\geq a$, and (iii)

 $\left\|\sum_{j=1}^{n}t_{j}x_{j}\right\|\geq a,$

and let $H_{k,\delta,-}$ be those $(t_{1},\ldots,t_{n})\in\mathbb{R}^{n}$ satisfying (i) and (ii) and

 $\left\|\sum_{j=1}^{k-1}t_{j}x_{j}-\sum_{j=k}^{n}t_{j}x_{j}\right\|\geq a.$

Let

 $X=(\epsilon_{1},\ldots,\epsilon_{n}):\Omega\to\mathbb{R}^{n}$

and let

 $X_{k}=(\epsilon_{1},\ldots,\epsilon_{k-1},-\epsilon_{k},\ldots,-\epsilon_{n}):% \Omega\to\mathbb{R}^{n},$

which have the same distribution because $\epsilon_{1},\ldots,\epsilon_{n}$ are independent and symmetric. Then

 $\displaystyle P(X^{-1}(H_{k,\delta,+}))$ $\displaystyle=(X_{*}P)(H_{k,\delta,+})$ $\displaystyle=({X_{k}}_{*}P)(H_{k,\delta,+})$ $\displaystyle=P(X_{k}^{-1}(H_{k,\delta,+}))$ $\displaystyle=P(X^{-1}(H_{k,\delta,-})).$

Set

 $C_{k,\delta,+}=\{X\in H_{k,\delta,+}\},\qquad C_{k,\delta,-}=\{X\in H_{k,% \delta,-}\},$

for which we thus have

 $P(C_{k,\delta,+})=P(C_{k,\delta,-}).$

We can write $C_{k,\delta,+}$ and $C_{k,\delta,-}$ as

 $C_{k,\delta,+}=\left(\bigcap_{0\leq j

and

 $C_{k,\delta,-}=\left(\bigcap_{0\leq j

If $\omega\in C_{k}$ then, because $\left\|S_{n}(\omega)-S_{k-1}(\omega)\right\|\geq a$,

 $\displaystyle 2a$ $\displaystyle\leq 2\left\|S_{n}(\omega)-S_{k-1}(\omega)\right\|$ $\displaystyle=\left\|S_{n}(\omega)+(S_{n}(\omega)-2S_{k-1}(\omega))\right\|$ $\displaystyle\leq\left\|S_{n}(\omega)\right\|+\left\|S_{n}(\omega)-2S_{k-1}(% \omega)\right\|,$

so at least one of the inequalities $\left\|S_{n}(\omega)\right\|\geq a$ and $\left\|S_{n}(\omega)-2S_{k-1}(\omega)\right\|\geq a$ is true, and hence

 $C_{k}\subset\{\left\|S_{n}\right\|\geq a\}\cup\{\left\|S_{n}-2S_{k-1}\right\|% \geq a\}.$

It follows that

 $C_{k}\cap\left(\bigcap_{0\leq j

Therefore, using the fact that for almost all $\omega\in\Omega$,

 $(\epsilon_{1}(\omega),\ldots,\epsilon_{k-1}(\omega))\in\{+1,-1\}^{k-1},$

and

 $C_{k,\delta,+}=\left(\bigcap_{0\leq j

we get

 $\displaystyle P(C_{k})$ $\displaystyle=\sum_{\delta}P\left(C_{k}\cap\bigcap_{0\leq j $\displaystyle=\sum_{\delta}P(C_{k,\delta,+}\cup C_{k,\delta,-})$ $\displaystyle=2\sum_{\delta}P(C_{k,\delta,+})$ $\displaystyle\leq 2\sum_{\delta}P\left(B\cap\bigcap_{0\leq j $\displaystyle=2P(B),$

and thus

 $\max_{1\leq k\leq n}P(C_{k})\leq 2P(B),$

which proves the claim. ∎

## 3 Kahane’s inequality

By $E(X)^{r}$ we mean $(E(X))^{r}$. The following is Kahane’s inequality.44 4 Joe Diestel, Hans Jarchow, and Andrew Tonge, Absolutely Summing Operators, p. 211, Theorem 11.1.

###### Theorem 3 (Kahane’s inequality).

For $0, there is some $K_{p,q}>0$ such that if $X$ is a real or complex Banach space and $x_{k}\in X$, $k\geq 1$, then for each $n$,

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|^{q}\right)^{1/q}\leq K_{% p,q}\cdot E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|^{p}\right)^{1/% p}.$
###### Proof.

Suppose that $0; when $p\geq q$ the claim is immediate with $K_{p,q}=1$. Let

 $M=E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|^{p}\right)^{1/p};$

if $M=0$ we check that the claim is $0\leq K_{p,q}\cdot 0$, which is true for, say, $K_{p,q}=1$. Otherwise, $M>0$, and let $u_{k}=\frac{x_{k}}{M}$, $1\leq k\leq n$, for which

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{p}\right)=E\left(\left% \|\sum_{k=1}^{n}\epsilon_{k}\frac{x_{k}}{M}\right\|^{p}\right)=1.$ (1)

Using Chebyshev’s inequality,

 $P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq 8^{1/p}\right)=P% \left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{p}\geq 8\right)\leq\frac% {1}{8}E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{p}\right)=\frac{1% }{8}.$

Assume for induction that for some $l\geq 0$ we have

 $P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq 2^{l}\cdot 8^{1/p}% \right)\leq\frac{1}{4}\cdot 2^{-2^{l}};$ (2)

the above shows that this is true for $l=0$. Applying Theorem 2 and then (2),

 $P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq 2^{l+1}\cdot 8^{1/p}% \right)\leq 4\left(P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq 2% ^{l}\cdot 8^{1/p}\right)\right)^{2}\leq\frac{1}{4}\cdot 2^{-2^{l+1}},$

which shows that (2) is true for all $l\geq 0$.

Generally, for $0, if $X:\Omega\to\mathbb{R}$ is a random variable for which $P(X\geq 0)=1$, then

 $E(X^{q})=\int_{0}^{\infty}qs^{q-1}P(X\geq s)ds;$

the right-hand side is finite if and only if $X\in L^{q}(P)$. Using this,

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{q}\right)=\int_{0}^{% \infty}qs^{q-1}P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq s% \right)ds.$ (3)

Let $\alpha_{0}=$ and for $l\geq 1$ let $\alpha_{l}=2^{l-1}\cdot 8^{1/p}$, and define

 $f(s)=qs^{q-1}P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|\geq s\right% ),\qquad s\geq 0.$

Using (3) and then (2),

 $\displaystyle E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{q}\right)$ $\displaystyle=\int_{0}^{\infty}f(s)ds$ $\displaystyle=\int_{0}^{\alpha_{1}}f(s)ds+\sum_{l=0}^{\infty}\int_{\alpha_{l+1% }}^{\alpha_{l+2}}f(s)ds$ $\displaystyle\leq\int_{0}^{\alpha_{1}}qs^{q-1}ds+\sum_{l=0}^{\infty}\int_{% \alpha_{l+1}}^{\alpha_{l+2}}qs^{q-1}P\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_% {k}\right\|\geq\alpha_{l+1}\right)ds$ $\displaystyle\leq\alpha_{1}^{q}+\sum_{l=0}^{\infty}\int_{\alpha_{l+1}}^{\alpha% _{l+2}}qs^{q-1}\frac{1}{4}\cdot 2^{-2^{l}}ds$ $\displaystyle=8^{q/p}+\frac{1}{4}\sum_{l=0}^{\infty}2^{-2^{l}}(\alpha_{l+2}^{q% }-\alpha_{l+1}^{q}),$

and we define $K_{p,q}$ by taking $K_{p,q}^{q}$ to be equal to the above. Thus

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{q}\right)^{1/q}\leq K_{% p,q},$

and therefore, by (1),

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{q}\right)^{1/q}\leq K_{% p,q}\cdot E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}u_{k}\right\|^{p}\right)^{1/% p}.$

Finally, as $u_{k}=\frac{x_{k}}{M}$,

 $E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|^{q}\right)^{1/q}\leq K_{% p,q}\cdot E\left(\left\|\sum_{k=1}^{n}\epsilon_{k}x_{k}\right\|^{p}\right)^{1/% p},$

which proves the claim. ∎

In the above proof of Kahane’s inequality, for $p=1$ and $q=2$ we have

 $\displaystyle K_{1,2}^{2}$ $\displaystyle=8^{2}+\frac{1}{4}\sum_{l=0}^{\infty}2^{-2^{l}}(\alpha_{l+2}^{2}-% \alpha_{l+1}^{2})$ $\displaystyle=64+16\sum_{l=0}^{\infty}2^{-2^{l}}(2^{2l+2}-2^{2l})$ $\displaystyle=64+48\sum_{l=0}^{\infty}2^{-2^{l}}2^{2l},$

for which

 $K_{1,2}=14.006\ldots.$

In fact, the inequality is true with $K_{1,2}=\sqrt{2}=1.414\ldots$.55 5 R. Latała and K. Oleszkiewicz, On the best constant in the Khinchin-Kahane inequality, Studia Math. 109 (1994), no. 1, 101–104.