The cross-polytope, the ball, and the cube

1 l^q norms and volume of the unit ball

For $x,y\in {ℝ}^n$,

$$⟨x,y⟩=\sum _{j=1}^n{x}_j{y}_j.$$

Let $e_1,\mathrm{\dots },e_n$ be the standard basis for ${ℝ}^n$.

$$x=\sum _{j=1}^n{x}_j{e}_j=\sum _{j=1}^n⟨x,e_j⟩e_j.$$

For let

$${|x|}_q={\left(\sum _{j=1}^n{|x_j|}^q\right)}^{1/q}$$

and for $q=\mathrm{\infty }$ let

$${|x|}_{\mathrm{\infty }}=\underset{1\le j\le n}{\max }|x_j|.$$

Then for for $1\le q\le \mathrm{\infty }$ let

$$B_q^n=\{x\in {ℝ}^n:{|x|}_q\le 1\}.$$

For $0\le k\le n$ let ${\lambda }_k$ be $k$-dimensional Lebesgue measure on ${ℝ}^n$. We calculate the volume of the unit ball with the ${\mathrm{\ell }}^q$ norm for .

Theorem 1.

For $n\ge 1$ and for ,

$${\lambda }_n(B_q^n)=\frac{{(2\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}.$$

Proof.

For $R\ge 0$ let $V_q^n(R)={\lambda }_n(R\cdot B_q^n)$. For $n=1$,

$$V_q^1(R)={\lambda }_1(R\cdot B_q^1)={\int }_{-R\le x_1\le R}𝑑{\lambda }_1(x_1)=2R.$$

By induction, suppose for some $n$ that

$$V_q^n(R)=\frac{{(2R\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}.$$

Using Fubini’s theorem and the induction hypothesis and doing the change of variable $x_{n+1}=Rt$ we calculate

$V_q^{n+1}(R)$ $={\displaystyle {\int }_{{|x_1|}^q+\mathrm{\cdots }+{|x_n|}^q+{|x_{n+1}|}^q\le R^q}}𝑑{\lambda }_{n+1}(x)$

$={\displaystyle {\int }_{-R\le x_{n+1}\le R}}\left({\displaystyle {\int }_{{|x_1|}^q+\mathrm{\cdots }+{|x_n|}_q\le R^q-{|x_{n+1}|}^q}}𝑑{\lambda }_n(x_1,\mathrm{\dots },x_n)\right)𝑑{\lambda }_1(x_{n+1})$

$={\displaystyle {\int }_{-R\le x_{n+1}\le R}}{V}_q^n({(R^q-{|x_{n+1}|}^q)}^{1/q})𝑑{\lambda }_1(x_{n+1})$

$={\displaystyle {\int }_{-R\le x_{n+1}\le R}}{\displaystyle \frac{{(2{(R^q-{|x_{n+1}|}^q)}^{1/q}\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}}𝑑{\lambda }_1(x_{n+1})$

$={\displaystyle \frac{{(2\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}}{\displaystyle {\int }_{-1\le t\le 1}}{(R^q-{|Rt|}^q)}^{n/q}\cdot R𝑑{\lambda }_1(t)$

$={\displaystyle \frac{{(2\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}}\cdot R^{n+1}\cdot 2{\displaystyle {\int }_{0\le t\le 1}}{(1-t^q)}^{n/q}𝑑{\lambda }_1(t).$

Now, doing the change of variable $u=t^q$, namely $t=u^{1/q}$ with $t^{\prime }=\frac{1}{q}{u}^{\frac{1}{q}-1}$ and using the beta function $B(a,b)={\int }_0^1{u}^{a-1}{(1-u)}^{b-1}𝑑{\lambda }_1(u)$,

${\displaystyle {\int }_{0\le t\le 1}}{(1-t^q)}^{n/q}𝑑{\lambda }_1(t)$ $={\displaystyle {\int }_{0\le u\le 1}}{(1-u)}^{n/q}\cdot {\displaystyle \frac{1}{q}}{u}^{\frac{1}{q}-1}𝑑{\lambda }_1(u)$

$={\displaystyle \frac{1}{q}}B({\displaystyle \frac{1}{q}},{\displaystyle \frac{n}{q}}+1).$

But $B(a,b)=\frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)}$, and using $\mathrm{\Gamma }(a+1)=a\mathrm{\Gamma }(a)$,

$$\frac{1}{q}B(\frac{1}{q},\frac{n}{q}+1)=\frac{1}{q}\frac{\mathrm{\Gamma }\left(\frac{1}{q}\right)\mathrm{\Gamma }\left(\frac{n}{q}+1\right)}{\mathrm{\Gamma }\left(\frac{1}{q}+\frac{n}{q}+1\right)}=\frac{\mathrm{\Gamma }\left(\frac{1}{q}+1\right)\mathrm{\Gamma }\left(\frac{n}{q}+1\right)}{\mathrm{\Gamma }\left(\frac{n+1}{q}+1\right)}.$$

Therefore

$V_q^{n+1}(R)$ $={\displaystyle \frac{{(2\mathrm{\Gamma }(\frac{1}{q}+1))}^n}{\mathrm{\Gamma }(\frac{n}{q}+1)}}\cdot R^{n+1}\cdot 2\cdot {\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{q}+1\right)\mathrm{\Gamma }\left(\frac{n}{q}+1\right)}{\mathrm{\Gamma }\left(\frac{n+1}{q}+1\right)}}$

$={\displaystyle \frac{{(2R\mathrm{\Gamma }(\frac{1}{q}+1))}^{n+1}}{\mathrm{\Gamma }\left(\frac{n+1}{q}+1\right)}},$

which proves the claim. ∎

$B_1^n$ is an $n$-dimensional cross-polytope, $B_2^n$ is an $n$-dimensional Euclidean ball, and $B_{\mathrm{\infty }}^n$ is an $n$-dimensional cube.

$${\lambda }_n(B_1^n)=\frac{2^n}{n!},{\lambda }_n(B_2^n)=\frac{{\pi }^{n/2}}{\mathrm{\Gamma }(\frac{n}{2}+1)},{\lambda }_n(B_{\mathrm{\infty }}^n)=2^n,$$

using $\mathrm{\Gamma }(n+1)=n!$ and $\mathrm{\Gamma }(\frac{3}{2})=\frac{\sqrt{\pi }}{2}$.

2 Intersection of a hyperplane and the cube

Let $\xi \in S^{n-1}$ and $t\in ℝ$, and define

$$P_{\xi ,t}=\{x\in {ℝ}^n:⟨x,\xi ⟩=t\}.$$

In particular,

$${\xi }^{⟂}=P_{\xi ,0}.$$

Let

$$A_{\xi }(t)={\lambda }_{n-1}(P_{\xi ,t}\cap B_{\mathrm{\infty }}^n)={\int }_{P_{\xi ,t}}{1}_{B_{\mathrm{\infty }}^n}(x)𝑑{\lambda }_{n-1}(x).$$

Theorem 2.

For $\xi \in S^{n-1}$ and $t\in ℝ$,

$$A_{\xi }(t)=\frac{2^n}{\pi }{\int }_0^{\mathrm{\infty }}\cos tr\cdot \prod _{i=1}^n\frac{2}{{\xi }_{i}r}\sin {\xi }_{i}rdr.$$

Proof.

Then by Fubini’s theorem,

${\widehat{A}}_{\xi }(\tau )$ $={\displaystyle {\int }_{ℝ}}{A}_{\xi }(t)e^{-2\pi it\tau }𝑑{\lambda }_1(t)$

$={\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{P_{\xi ,t}}}{1}_{B_{\mathrm{\infty }}^n}(x)e^{-2\pi i⟨x,\xi ⟩\tau }𝑑{\lambda }_{n-1}(x)\right)𝑑{\lambda }_1(t)$

$={\displaystyle {\int }_{{ℝ}^n}}{1}_{B_{\mathrm{\infty }}^n}(x)e^{-2\pi i⟨x,\xi ⟩\tau }𝑑{\lambda }_n(x).$

Now,

$$1_{B_{\mathrm{\infty }}^n}(x)=(1_{B_{\mathrm{\infty }}^1}\otimes \mathrm{\cdots }\otimes 1_{B_{\mathrm{\infty }}^1})(x)=\prod _{i=1}^n{1}_{B_{\mathrm{\infty }}^1}(x_i),$$

whence, by Fubini’s theorem,

${\displaystyle {\int }_{{ℝ}^n}}{1}_{B_{\mathrm{\infty }}^n}(x)e^{-2\pi i⟨x,\xi ⟩\tau }𝑑{\lambda }_n(x)$ $={\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle \prod _{i=1}^n}{1}_{B_{\mathrm{\infty }}^1}(x_i)e^{-2\pi i{x}_i{\xi }_i\tau }\right)𝑑{\lambda }_n(x)$

$={\displaystyle \prod _{i=1}^n}{\displaystyle {\int }_{ℝ}}{1}_{B_{\mathrm{\infty }}^1}(x_i)e^{-2\pi i{x}_i{\xi }_i\tau }𝑑{\lambda }_1(x_i).$

But, when ${\xi }_i\tau \ne 0$,

$${\int }_{ℝ}{1}_{B_{\mathrm{\infty }}^1}(x_i)e^{-2\pi i{x}_i{\xi }_i\tau }𝑑{\lambda }_1(x_i)={\int }_{-1}^1{e}^{-2\pi i{x}_i{\xi }_i\tau }𝑑{\lambda }_1(x_i)=\frac{1}{\pi {\xi }_i\tau }\sin 2\pi {\xi }_i\tau ,$$

thus

$${\widehat{A}}_{\xi }(\tau )=\prod _{i=1}^n\frac{1}{\pi {\xi }_i\tau }\sin 2\pi {\xi }_i\tau .$$

By the Fourier inversion theorem, using that ${\widehat{A}}_{\xi }$ is an even function,

$A_{\xi }(t)$ $={\displaystyle {\int }_{ℝ}}{\widehat{A}}_{\xi }(\tau )e^{2\pi it\tau }𝑑{\lambda }_1(\tau )$

$={\displaystyle {\int }_{ℝ}}{\widehat{A}}_{\xi }(\tau )\cos 2\pi t\tau d{\lambda }_1(\tau )$

$=2{\displaystyle {\int }_0^{\mathrm{\infty }}}{\widehat{A}}_{\xi }(\tau )\cos 2\pi t\tau d\tau$

$=2{\displaystyle {\int }_0^{\mathrm{\infty }}}\cos 2\pi t\tau \cdot {\displaystyle \prod _{i=1}^n}{\displaystyle \frac{1}{\pi {\xi }_i\tau }}\sin 2\pi {\xi }_i\tau d\tau$

$={\displaystyle \frac{2^n}{\pi }}{\displaystyle {\int }_0^{\mathrm{\infty }}}\cos tr\cdot {\displaystyle \prod _{i=1}^n}{\displaystyle \frac{2}{{\xi }_{i}r}}\sin {\xi }_{i}rdr.$

∎

3 Schwartz functions

Let $𝒮$ be the Fréchet space of Schwartz function ${ℝ}^n\to ℂ$ and let ${𝒮}^{\prime }$ be the locally convex space of tempered distributions $𝒮\to ℂ$. If $f:{ℝ}^n\to ℂ$ is locally integrable and there is some $N$ such that

$${\int }_{{|x|}_2\le R}|f(x)|𝑑{\lambda }_n(x)=O(R^N),R\to \mathrm{\infty },$$

it is a fact that

$$\varphi \mapsto ⟨f,\varphi ⟩={\int }_{{ℝ}^n}f(x)\varphi (x)𝑑{\lambda }_n(x),\varphi \in 𝒮,$$

is a tempered distribution.

Lemma 3.

For $1\le q\le \mathrm{\infty }$ and for , ${|x|}_q^{-h}$ is a tempered distribution.

Proof.

For $1\le q\le 2$,

$${|x|}_2\le {|x|}_q\le n^{\frac{1}{q}-\frac{1}{2}}{|x|}_2,$$

and for $2\le q\le \mathrm{\infty }$,

$${|x|}_q\le {|x|}_2\le n^{\frac{1}{2}-\frac{1}{q}}{|x|}_q.$$

Then for $1\le q\le 2$ and for , using polar coordinates and as $\sigma (S^{n-1})=\frac{2{\pi }^{(n+1)/2}}{\mathrm{\Gamma }(\frac{n+1}{2})}$,

${\displaystyle {\int }_{{|x|}_2\le R}}{|x|}_q^{-h}𝑑{\lambda }_n(x)$ $\le {\displaystyle {\int }_{{|x|}_2\le R}}{|x|}_2^{-h}𝑑{\lambda }_n(x)$

$={\displaystyle {\int }_{S^{n-1}}}\left({\displaystyle {\int }_0^{\mathrm{\infty }}}{r}^{-h}\cdot r^{n-1}𝑑r\right)𝑑\sigma$

$={\displaystyle \frac{2{\pi }^{(n+1)/2}}{\mathrm{\Gamma }(\frac{n+1}{2})}}\cdot {\displaystyle {\int }_0^R}{r}^{-h+n-1}𝑑r$

$={{\displaystyle \frac{2{\pi }^{(n+1)/2}}{\mathrm{\Gamma }(\frac{n+1}{2})}}\cdot {\displaystyle \frac{r^{-h+n}}{-h+n}}|}_0^R$

$={\displaystyle \frac{2{\pi }^{(n+1)/2}}{\mathrm{\Gamma }(\frac{n+1}{2})}}\cdot {\displaystyle \frac{R^{-h+n}}{-h+n}}$

$=O(R^{-h+n}).$

For $2\le q\le \mathrm{\infty }$ and for ,

${\displaystyle {\int }_{{|x|}_2\le R}}{|x|}_q^{-h}𝑑{\lambda }_n(x)$ $\le {\displaystyle {\int }_{{|x|}_2\le R}}{(n^{-\frac{1}{2}+\frac{1}{q}}{|x|}_2)}^{-h}𝑑{\lambda }_n(x)$

$=n^{\frac{h}{2}-\frac{h}{q}}{\displaystyle {\int }_{{|x|}_2\le R}}{|x|}_2^{-h}𝑑{\lambda }_n(x)$

$=n^{\frac{h}{2}-\frac{h}{q}}\cdot {\displaystyle \frac{2{\pi }^{(n+1)/2}}{\mathrm{\Gamma }(\frac{n+1}{2})}}\cdot {\displaystyle \frac{R^{-h+n}}{-h+n}}$

$=O(R^{-h+n}).$

∎

For $\varphi \in 𝒮$ let

$$\widehat{\varphi }(\xi )={\int }_{{ℝ}^n}\varphi (x)e^{-2\pi i⟨x,\xi ⟩}𝑑{\lambda }_n(x).$$

For define $c_q:ℝ\to ℝ$ by

$$c_q(z)=e^{-{|z|}^q},z\in ℝ,$$

which belongs to $𝒮(ℝ)$, and let ${\gamma }_q={\widehat{c}}_q$.

For a tempered distribution $T$,

$$⟨\widehat{T},\varphi ⟩=⟨T,\widehat{\varphi }⟩,\varphi \in 𝒮.$$

Define $f_{q,h}(x)={|x|}_q^{-h}$. We calculate the Fourier transform of the tempered distribution $f_{q,h}$.

Theorem 4.

Let . For ,

$${\widehat{f}}_{q,h}(\xi )=\frac{q}{\mathrm{\Gamma }(h/q)}{\int }_0^{\mathrm{\infty }}{t}^{n-h-1}\prod _{j=1}^n{\gamma }_q(t{\xi }_j)dt,$$

and for $q=\mathrm{\infty }$,

$${\widehat{f}}_{\mathrm{\infty },h}(\xi )=2^{n}h{\int }_0^{\mathrm{\infty }}{t}^{n-h-1}\prod _{j=1}^n\frac{\sin t{\xi }_j}{t{\xi }_j}dt$$

Proof.

Suppose that . For $x\ne 0$, doing the change of variable $z=t^{1/q}{|x|}_q^{-1}$,

${\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1}{e}^{-z^q{|x|}_q^q}𝑑z$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{(t^{1/q}{|x|}_q^{-1})}^{h-1}{e}^{-t}{|x|}_q^{-1}{\displaystyle \frac{1}{q}}{t}^{\frac{1}{q}-1}𝑑t$

$={\displaystyle \frac{{|x|}_q^{-h}}{q}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{\frac{h}{q}-1}{e}^{-t}𝑑t$

$={\displaystyle \frac{{|x|}_q^{-h}}{q}}\cdot \mathrm{\Gamma }(h/q),$

i.e. $f_{q,h}(x)=\frac{q}{\mathrm{\Gamma }(h/q)}{\int }_0^{\mathrm{\infty }}{z}^{h-1}{e}^{-z^q{|x|}_q^q}𝑑z$.

For $z>0$ define $F_{q,z}:{ℝ}^n\to ℝ$ by

$$F_{q,z}(x)=e^{-{|zx|}_q^q},x\in {ℝ}^n,$$

which is a Schwartz function. Doing the change of variable $y=z\cdot x$ and using Fubini’s theorem,

${\widehat{F}}_{q,z}(\xi )$ $={\displaystyle {\int }_{{ℝ}^n}}{e}^{-{|zx|}_q^q}{e}^{-2\pi i⟨x,\xi ⟩}𝑑{\lambda }_n(x)$

$={\displaystyle {\int }_{{ℝ}^n}}{e}^{-{|y_1|}^q-\mathrm{\cdots }-{|y_n|}^q}{e}^{-2\pi i⟨y,z^{-1}\xi ⟩}\cdot z^{-n}𝑑{\lambda }_n(y)$

$=z^{-n}{\displaystyle \prod _{j=1}^n}{\displaystyle {\int }_{ℝ}}{e}^{-{|y_j|}^q}{e}^{-2\pi i{y}_j\cdot z^{-1}{\xi }_j}𝑑{\lambda }_1(y_j)$

$=z^{-n}{\displaystyle \prod _{j=1}^n}{\gamma }_q(z^{-1}{\xi }_j).$

Then for $\varphi \in 𝒮$,

$⟨{\widehat{f}}_{q,h},\varphi ⟩$ $={\displaystyle {\int }_{{ℝ}^n}}{f}_{q,h}(\xi )\widehat{\varphi }(\xi )𝑑{\lambda }_n(\xi )$

$={\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1}{e}^{-z^q{|\xi |}_q^q}𝑑z\right)\widehat{\varphi }(\xi )𝑑{\lambda }_n(\xi )$

$={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1}\left({\displaystyle {\int }_{{ℝ}^n}}{e}^{-{|z\xi |}_q^q}\widehat{\varphi }(\xi )𝑑{\lambda }_n(\xi )\right)𝑑z$

$={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1}⟨F_{q,z},\widehat{\varphi }⟩𝑑z$

$={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1}⟨{\widehat{F}}_{q,z},\varphi ⟩𝑑z$

$={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1-n}\left({\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \prod _{j=1}^n}{\gamma }_q(z^{-1}{\xi }_j)\cdot \varphi (\xi )d{\lambda }_n(\xi )\right)𝑑z$

$={\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1-n}{\displaystyle \prod _{j=1}^n}{\gamma }_q(z^{-1}{\xi }_j)dz\right)\varphi (\xi )𝑑{\lambda }_n(\xi ).$

This implies, doing the change of variable $z=t^{-1}$,

${\widehat{f}}_{q,h}(\xi )$ $={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{z}^{h-1-n}{\displaystyle \prod _{j=1}^n}{\gamma }_q(z^{-1}{\xi }_j)dz$

$={\displaystyle \frac{q}{\mathrm{\Gamma }(h/q)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{n-h-1}{\displaystyle \prod _{j=1}^n}{\gamma }_q(t{\xi }_j)dt.$

∎

4 Fourier transform

We remind ourselves that $c_q(z)=e^{-{|z|}^q}$, $z\in ℝ$, and ${\gamma }_q={\widehat{c}}_q$. We prove that ${\gamma }_q$ is positive and logconvex.

Theorem 5.

For $1\le q\le 2$, ${\gamma }_q(\sqrt{z})>0$, and $z\mapsto \log {\gamma }_q(\sqrt{z})$ is convex on ${ℝ}_{\ge 0}$.

Proof.

Let , and for $z\in [0,\mathrm{\infty })$ let $f(z)=\exp z$ and $g(z)=-z^{\alpha }$. Then for $k\in {\mathbb{Z}}_{\ge 0}$ and $z\in (0,\mathrm{\infty })$,

$$g^{(k)}(z)=-k!\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)z^{\alpha -k},\mathrm{sgn}{g}^{(k)}(z)={(-1)}^k.$$

For $n\ge 1$, Faà di Bruno’s formula tells us

${(f\circ g)}^{(n)}(z)$ $={\displaystyle \sum _{(m_1,\mathrm{\dots },m_n),1\cdot m_1+\mathrm{\cdots }+n\cdot m_n=n}}{\displaystyle \frac{n!}{m_1!\mathrm{\cdots }{m}_n!}}(f^{(m_1+\mathrm{\cdots }+m_n)}\circ g)(z)$

$\cdot {\displaystyle \prod _{k=1}^n}{\left({\displaystyle \frac{g^{(k)}(z)}{k!}}\right)}^{m_k}.$

Then

${(-1)}^n{(f\circ g)}^{(n)}(z)$ $={\displaystyle \sum _{(m_1,\mathrm{\dots },m_n),1\cdot m_1+\mathrm{\cdots }+n\cdot m_n=n}}{\displaystyle \frac{n!}{m_1!\mathrm{\cdots }{m}_n!}}(\exp \circ g)(z)$

$\cdot {\displaystyle \prod _{k=1}^n}{\left({(-1)}^k{\displaystyle \frac{g^{(k)}(z)}{k!}}\right)}^{m_k}$

$\ge 0.$

This shows that $f\circ g$ is completely monotone. Furthermore, $(f\circ g)(0)=1$, so by the Bernstein-Widder theorem there is a Borel probability measure $\mu$ on $[0,\mathrm{\infty })$ such that

$$(f\circ g)(z)={\int }_{[0,\mathrm{\infty })}{e}^{-zt}𝑑\mu (t),z\in [0,\mathrm{\infty }).$$

With $\alpha =\frac{q}{2}$, there is thus a Borel probability measure ${\mu }_q$ on $[0,\mathrm{\infty })$ such that

$$\exp (-z^{q/2})={\int }_{[0,\mathrm{\infty })}{e}^{-zt}𝑑{\mu }_q(t),z\in [0,\mathrm{\infty }).$$

Then for $z\in ℝ$,

$$c_q(z)=\exp (-{|z|}^q)=\exp (-{(z^2)}^{q/2})={\int }_{[0,\mathrm{\infty })}{e}^{-z^{2}t}𝑑{\mu }_{q/2}(t).$$

For $w\in ℝ$ we calculate, using the Fourier transform of a Gaussian,

${\gamma }_q(w)$ $={\displaystyle {\int }_{ℝ}}{e}^{-2\pi iwz}{c}_q(z)𝑑{\lambda }_1(z)$

$={\displaystyle {\int }_{ℝ}}{e}^{-2\pi iwz}\left({\displaystyle {\int }_{[0,\mathrm{\infty })}}{e}^{-z^{2}t}𝑑{\mu }_{q/2}(t)\right)𝑑{\lambda }_1(z)$

$={\displaystyle {\int }_{[0,\mathrm{\infty })}}\left({\displaystyle {\int }_{ℝ}}{e}^{-t{z}^2}{e}^{-2\pi iwz}𝑑{\lambda }_1(z)\right)𝑑{\mu }_{q/2}(t)$

$={\displaystyle {\int }_{[0,\mathrm{\infty })}}\sqrt{{\displaystyle \frac{\pi }{t}}}\exp \left(-{\displaystyle \frac{{(\pi w)}^2}{t}}\right)𝑑{\mu }_{q/2}(t)$

$={\pi }^{1/2}{\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/2}{e}^{-{\pi }^2{w}^2/t}𝑑{\mu }_{q/2}(t).$

From the final expression it is evident that ${\gamma }_k(w)>0$. Furthermore, for $w_1,w_2\in (0,\mathrm{\infty })$, using the Cauchy-Schwarz inequality,

$\log {\gamma }_q\left(\sqrt{{\displaystyle \frac{w_1+w_2}{2}}}\right)$ $={\displaystyle \frac{1}{2}}\log {({\pi }^{1/2}{\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/4}{e}^{-{\pi }^2\cdot \frac{w_1}{2t}}\cdot t^{-1/4}{e}^{-{\pi }^2\cdot \frac{w_2}{2t}}d{\mu }_{q/2}(t))}^2$

$\le {\displaystyle \frac{1}{2}}\log (\pi {\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/2}{e}^{-{\pi }^2\cdot \frac{w_1}{t}}d{\mu }_{q/2}(t)$

$\cdot {\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/2}{e}^{-{\pi }^2\cdot \frac{w_2}{t}}d{\mu }_{q/2}(t))$

$={\displaystyle \frac{1}{2}}\log \left({\pi }^{1/2}{\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/2}{e}^{-{\pi }^2\cdot \frac{w_1}{t}}𝑑{\mu }_{q/2}(t)\right)$

$+{\displaystyle \frac{1}{2}}\log \left({\pi }^{1/2}{\displaystyle {\int }_{[0,\mathrm{\infty })}}{t}^{-1/2}{e}^{-{\pi }^2\cdot \frac{w_2}{t}}𝑑{\mu }_{q/2}(t)\right)$

$={\displaystyle \frac{1}{2}}\log {\gamma }_q(\sqrt{w_1})+{\displaystyle \frac{1}{2}}\log {\gamma }_q(\sqrt{w_2}).$

Because $w\mapsto \log {\gamma }_q(\sqrt{w})$ is continuous, this suffices to prove that it is convex. ∎