# The cross-polytope, the ball, and the cube

Jordan Bell
March 28, 2016

## 1 lq norms and volume of the unit ball

For $x,y\in\mathbb{R}^{n}$,

 $\left\langle x,y\right\rangle=\sum_{j=1}^{n}x_{j}y_{j}.$

Let $e_{1},\ldots,e_{n}$ be the standard basis for $\mathbb{R}^{n}$.

 $x=\sum_{j=1}^{n}x_{j}e_{j}=\sum_{j=1}^{n}\left\langle x,e_{j}\right\rangle e_{% j}.$

For $1\leq q<\infty$ let

 $|x|_{q}=\left(\sum_{j=1}^{n}|x_{j}|^{q}\right)^{1/q}$

and for $q=\infty$ let

 $|x|_{\infty}=\max_{1\leq j\leq n}|x_{j}|.$

Then for for $1\leq q\leq\infty$ let

 $B_{q}^{n}=\{x\in\mathbb{R}^{n}:|x|_{q}\leq 1\}.$

For $0\leq k\leq n$ let $\lambda_{k}$ be $k$-dimensional Lebesgue measure on $\mathbb{R}^{n}$. We calculate the volume of the unit ball with the $\ell^{q}$ norm for $1\leq q<\infty$.

###### Theorem 1.

For $n\geq 1$ and for $1\leq q<\infty$,

 $\lambda_{n}(B_{q}^{n})=\frac{(2\Gamma(\frac{1}{q}+1))^{n}}{\Gamma(\frac{n}{q}+% 1)}.$
###### Proof.

For $R\geq 0$ let $V_{q}^{n}(R)=\lambda_{n}(R\cdot B_{q}^{n})$. For $n=1$,

 $V_{q}^{1}(R)=\lambda_{1}(R\cdot B_{q}^{1})=\int_{-R\leq x_{1}\leq R}d\lambda_{% 1}(x_{1})=2R.$

By induction, suppose for some $n$ that

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

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

 $\displaystyle V_{q}^{n+1}(R)$ $\displaystyle=\int_{|x_{1}|^{q}+\cdots+|x_{n}|^{q}+|x_{n+1}|^{q}\leq R^{q}}d% \lambda_{n+1}(x)$ $\displaystyle=\int_{-R\leq x_{n+1}\leq R}\left(\int_{|x_{1}|^{q}+\cdots+|x_{n}% |_{q}\leq R^{q}-|x_{n+1}|^{q}}d\lambda_{n}(x_{1},\ldots,x_{n})\right)d\lambda_% {1}(x_{n+1})$ $\displaystyle=\int_{-R\leq x_{n+1}\leq R}V_{q}^{n}((R^{q}-|x_{n+1}|^{q})^{1/q}% )d\lambda_{1}(x_{n+1})$ $\displaystyle=\int_{-R\leq x_{n+1}\leq R}\frac{(2(R^{q}-|x_{n+1}|^{q})^{1/q}% \Gamma(\frac{1}{q}+1))^{n}}{\Gamma(\frac{n}{q}+1)}d\lambda_{1}(x_{n+1})$ $\displaystyle=\frac{(2\Gamma(\frac{1}{q}+1))^{n}}{\Gamma(\frac{n}{q}+1)}\int_{% -1\leq t\leq 1}(R^{q}-|Rt|^{q})^{n/q}\cdot Rd\lambda_{1}(t)$ $\displaystyle=\frac{(2\Gamma(\frac{1}{q}+1))^{n}}{\Gamma(\frac{n}{q}+1)}\cdot R% ^{n+1}\cdot 2\int_{0\leq t\leq 1}(1-t^{q})^{n/q}d\lambda_{1}(t).$

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 $B(a,b)=\int_{0}^{1}u^{a-1}(1-u)^{b-1}d\lambda_{1}(u)$,

 $\displaystyle\int_{0\leq t\leq 1}(1-t^{q})^{n/q}d\lambda_{1}(t)$ $\displaystyle=\int_{0\leq u\leq 1}(1-u)^{n/q}\cdot\frac{1}{q}u^{\frac{1}{q}-1}% d\lambda_{1}(u)$ $\displaystyle=\frac{1}{q}B\left(\frac{1}{q},\frac{n}{q}+1\right).$

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

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

Therefore

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

which proves the claim. ∎

$B_{1}^{n}$ is an $n$-dimensional cross-polytope, $B_{2}^{n}$ is an $n$-dimensional Euclidean ball, and $B_{\infty}^{n}$ is an $n$-dimensional cube.

 $\lambda_{n}(B_{1}^{n})=\frac{2^{n}}{n!},\quad\lambda_{n}(B_{2}^{n})=\frac{\pi^% {n/2}}{\Gamma(\frac{n}{2}+1)},\quad\lambda_{n}(B_{\infty}^{n})=2^{n},$

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

## 2 Intersection of a hyperplane and the cube

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

 $P_{\xi,t}=\{x\in\mathbb{R}^{n}:\left\langle x,\xi\right\rangle=t\}.$

In particular,

 $\xi^{\perp}=P_{\xi,0}.$

Let

 $A_{\xi}(t)=\lambda_{n-1}(P_{\xi,t}\cap B_{\infty}^{n})=\int_{P_{\xi,t}}1_{B_{% \infty}^{n}}(x)d\lambda_{n-1}(x).$
###### Theorem 2.

For $\xi\in S^{n-1}$ and $t\in\mathbb{R}$,

 $A_{\xi}(t)=\frac{2^{n}}{\pi}\int_{0}^{\infty}\cos tr\cdot\prod_{i=1}^{n}\frac{% 2}{\xi_{i}r}\sin\xi_{i}rdr.$
###### Proof.

Then by Fubini’s theorem,

 $\displaystyle\widehat{A}_{\xi}(\tau)$ $\displaystyle=\int_{\mathbb{R}}A_{\xi}(t)e^{-2\pi it\tau}d\lambda_{1}(t)$ $\displaystyle=\int_{\mathbb{R}}\left(\int_{P_{\xi,t}}1_{B_{\infty}^{n}}(x)e^{-% 2\pi i\left\langle x,\xi\right\rangle\tau}d\lambda_{n-1}(x)\right)d\lambda_{1}% (t)$ $\displaystyle=\int_{\mathbb{R}^{n}}1_{B_{\infty}^{n}}(x)e^{-2\pi i\left\langle x% ,\xi\right\rangle\tau}d\lambda_{n}(x).$

Now,

 $1_{B_{\infty}^{n}}(x)=(1_{B_{\infty}^{1}}\otimes\cdots\otimes 1_{B_{\infty}^{1% }})(x)=\prod_{i=1}^{n}1_{B_{\infty}^{1}}(x_{i}),$

whence, by Fubini’s theorem,

 $\displaystyle\int_{\mathbb{R}^{n}}1_{B_{\infty}^{n}}(x)e^{-2\pi i\left\langle x% ,\xi\right\rangle\tau}d\lambda_{n}(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(\prod_{i=1}^{n}1_{B_{\infty}^{1}}(x_{% i})e^{-2\pi ix_{i}\xi_{i}\tau}\right)d\lambda_{n}(x)$ $\displaystyle=\prod_{i=1}^{n}\int_{\mathbb{R}}1_{B_{\infty}^{1}}(x_{i})e^{-2% \pi ix_{i}\xi_{i}\tau}d\lambda_{1}(x_{i}).$

But, when $\xi_{i}\tau\neq 0$,

 $\int_{\mathbb{R}}1_{B_{\infty}^{1}}(x_{i})e^{-2\pi ix_{i}\xi_{i}\tau}d\lambda_% {1}(x_{i})=\int_{-1}^{1}e^{-2\pi ix_{i}\xi_{i}\tau}d\lambda_{1}(x_{i})=\frac{1% }{\pi\xi_{i}\tau}\sin 2\pi\xi_{i}\tau,$

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,

 $\displaystyle A_{\xi}(t)$ $\displaystyle=\int_{\mathbb{R}}\widehat{A}_{\xi}(\tau)e^{2\pi it\tau}d\lambda_% {1}(\tau)$ $\displaystyle=\int_{\mathbb{R}}\widehat{A}_{\xi}(\tau)\cos 2\pi t\tau d\lambda% _{1}(\tau)$ $\displaystyle=2\int_{0}^{\infty}\widehat{A}_{\xi}(\tau)\cos 2\pi t\tau d\tau$ $\displaystyle=2\int_{0}^{\infty}\cos 2\pi t\tau\cdot\prod_{i=1}^{n}\frac{1}{% \pi\xi_{i}\tau}\sin 2\pi\xi_{i}\tau d\tau$ $\displaystyle=\frac{2^{n}}{\pi}\int_{0}^{\infty}\cos tr\cdot\prod_{i=1}^{n}% \frac{2}{\xi_{i}r}\sin\xi_{i}rdr.$

## 3 Schwartz functions

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

 $\int_{|x|_{2}\leq R}|f(x)|d\lambda_{n}(x)=O(R^{N}),\qquad R\to\infty,$

it is a fact that

 $\phi\mapsto\left\langle f,\phi\right\rangle=\int_{\mathbb{R}^{n}}f(x)\phi(x)d% \lambda_{n}(x),\qquad\phi\in\mathscr{S},$

is a tempered distribution.

###### Lemma 3.

For $1\leq q\leq\infty$ and for $0, $|x|_{q}^{-h}$ is a tempered distribution.

###### Proof.

For $1\leq q\leq 2$,

 $|x|_{2}\leq|x|_{q}\leq n^{\frac{1}{q}-\frac{1}{2}}|x|_{2},$

and for $2\leq q\leq\infty$,

 $|x|_{q}\leq|x|_{2}\leq n^{\frac{1}{2}-\frac{1}{q}}|x|_{q}.$

Then for $1\leq q\leq 2$ and for $0, using polar coordinates and as $\sigma(S^{n-1})=\frac{2\pi^{(n+1)/2}}{\Gamma(\frac{n+1}{2})}$,

 $\displaystyle\int_{|x|_{2}\leq R}|x|_{q}^{-h}d\lambda_{n}(x)$ $\displaystyle\leq\int_{|x|_{2}\leq R}|x|_{2}^{-h}d\lambda_{n}(x)$ $\displaystyle=\int_{S^{n-1}}\left(\int_{0}^{\infty}r^{-h}\cdot r^{n-1}dr\right% )d\sigma$ $\displaystyle=\frac{2\pi^{(n+1)/2}}{\Gamma(\frac{n+1}{2})}\cdot\int_{0}^{R}r^{% -h+n-1}dr$ $\displaystyle=\frac{2\pi^{(n+1)/2}}{\Gamma(\frac{n+1}{2})}\cdot\frac{r^{-h+n}}% {-h+n}\bigg{|}_{0}^{R}$ $\displaystyle=\frac{2\pi^{(n+1)/2}}{\Gamma(\frac{n+1}{2})}\cdot\frac{R^{-h+n}}% {-h+n}$ $\displaystyle=O(R^{-h+n}).$

For $2\leq q\leq\infty$ and for $0,

 $\displaystyle\int_{|x|_{2}\leq R}|x|_{q}^{-h}d\lambda_{n}(x)$ $\displaystyle\leq\int_{|x|_{2}\leq R}(n^{-\frac{1}{2}+\frac{1}{q}}|x|_{2})^{-h% }d\lambda_{n}(x)$ $\displaystyle=n^{\frac{h}{2}-\frac{h}{q}}\int_{|x|_{2}\leq R}|x|_{2}^{-h}d% \lambda_{n}(x)$ $\displaystyle=n^{\frac{h}{2}-\frac{h}{q}}\cdot\frac{2\pi^{(n+1)/2}}{\Gamma(% \frac{n+1}{2})}\cdot\frac{R^{-h+n}}{-h+n}$ $\displaystyle=O(R^{-h+n}).$

For $\phi\in\mathscr{S}$ let

 $\widehat{\phi}(\xi)=\int_{\mathbb{R}^{n}}\phi(x)e^{-2\pi i\left\langle x,\xi% \right\rangle}d\lambda_{n}(x).$

For $1\leq q<\infty$ define $c_{q}:\mathbb{R}\to\mathbb{R}$ by

 $c_{q}(z)=e^{-|z|^{q}},\qquad z\in\mathbb{R},$

which belongs to $\mathscr{S}(\mathbb{R})$, and let $\gamma_{q}=\widehat{c}_{q}$.

For a tempered distribution $T$,

 $\left\langle\widehat{T},\phi\right\rangle=\left\langle T,\widehat{\phi}\right% \rangle,\qquad\phi\in\mathscr{S}.$

Define $f_{q,h}(x)=|x|_{q}^{-h}$. We calculate the Fourier transform of the tempered distribution $f_{q,h}$.11 1 Alexander Koldobsky and Vladyslav Yaskin, The Interface between Convex Geometry and Harmonic Analysis, p. 9, Lemma 2.1.

###### Theorem 4.

Let $0. For $1\leq q<\infty$,

 $\widehat{f}_{q,h}(\xi)=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}t^{n-h-1}\prod_{j% =1}^{n}\gamma_{q}(t\xi_{j})dt,$

and for $q=\infty$,

 $\widehat{f}_{\infty,h}(\xi)=2^{n}h\int_{0}^{\infty}t^{n-h-1}\prod_{j=1}^{n}% \frac{\sin t\xi_{j}}{t\xi_{j}}dt$
###### Proof.

Suppose that $1\leq q<\infty$. For $x\neq 0$, doing the change of variable $z=t^{1/q}|x|_{q}^{-1}$,

 $\displaystyle\int_{0}^{\infty}z^{h-1}e^{-z^{q}|x|_{q}^{q}}dz$ $\displaystyle=\int_{0}^{\infty}(t^{1/q}|x|_{q}^{-1})^{h-1}e^{-t}|x|_{q}^{-1}% \frac{1}{q}t^{\frac{1}{q}-1}dt$ $\displaystyle=\frac{|x|_{q}^{-h}}{q}\int_{0}^{\infty}t^{\frac{h}{q}-1}e^{-t}dt$ $\displaystyle=\frac{|x|_{q}^{-h}}{q}\cdot\Gamma(h/q),$

i.e. $f_{q,h}(x)=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1}e^{-z^{q}|x|_{q}^{q}}dz$.

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

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

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

 $\displaystyle\widehat{F}_{q,z}(\xi)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-|zx|_{q}^{q}}e^{-2\pi i\left\langle x,% \xi\right\rangle}d\lambda_{n}(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-|y_{1}|^{q}-\cdots-|y_{n}|^{q}}e^{-2\pi i% \left\langle y,z^{-1}\xi\right\rangle}\cdot z^{-n}d\lambda_{n}(y)$ $\displaystyle=z^{-n}\prod_{j=1}^{n}\int_{\mathbb{R}}e^{-|y_{j}|^{q}}e^{-2\pi iy% _{j}\cdot z^{-1}\xi_{j}}d\lambda_{1}(y_{j})$ $\displaystyle=z^{-n}\prod_{j=1}^{n}\gamma_{q}(z^{-1}\xi_{j}).$

Then for $\phi\in\mathscr{S}$,

 $\displaystyle\left\langle\widehat{f}_{q,h},\phi\right\rangle$ $\displaystyle=\int_{\mathbb{R}^{n}}f_{q,h}(\xi)\widehat{\phi}(\xi)d\lambda_{n}% (\xi)$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(\frac{q}{\Gamma(h/q)}\int_{0}^{\infty% }z^{h-1}e^{-z^{q}|\xi|_{q}^{q}}dz\right)\widehat{\phi}(\xi)d\lambda_{n}(\xi)$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1}\left(\int_{\mathbb% {R}^{n}}e^{-|z\xi|_{q}^{q}}\widehat{\phi}(\xi)d\lambda_{n}(\xi)\right)dz$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1}\left\langle F_{q,z% },\widehat{\phi}\right\rangle dz$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1}\left\langle% \widehat{F}_{q,z},\phi\right\rangle dz$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1-n}\left(\int_{% \mathbb{R}^{n}}\prod_{j=1}^{n}\gamma_{q}(z^{-1}\xi_{j})\cdot\phi(\xi)d\lambda_% {n}(\xi)\right)dz$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(\frac{q}{\Gamma(h/q)}\int_{0}^{\infty% }z^{h-1-n}\prod_{j=1}^{n}\gamma_{q}(z^{-1}\xi_{j})dz\right)\phi(\xi)d\lambda_{% n}(\xi).$

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

 $\displaystyle\widehat{f}_{q,h}(\xi)$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}z^{h-1-n}\prod_{j=1}^{n}% \gamma_{q}(z^{-1}\xi_{j})dz$ $\displaystyle=\frac{q}{\Gamma(h/q)}\int_{0}^{\infty}t^{n-h-1}\prod_{j=1}^{n}% \gamma_{q}(t\xi_{j})dt.$

## 4 Fourier transform

We remind ourselves that $c_{q}(z)=e^{-|z|^{q}}$, $z\in\mathbb{R}$, and $\gamma_{q}=\widehat{c}_{q}$. We prove that $\gamma_{q}$ is positive and logconvex.22 2 Alexander Koldobsky and Vladyslav Yaskin, The Interface between Convex Geometry and Harmonic Analysis, p. 4, Lemma 1.4.

###### Theorem 5.

For $1\leq q\leq 2$, $\gamma_{q}(\sqrt{z})>0$, and $z\mapsto\log\gamma_{q}(\sqrt{z})$ is convex on $\mathbb{R}_{\geq 0}$.

###### Proof.

Let $0<\alpha\leq 1$, and for $z\in[0,\infty)$ let $f(z)=\exp z$ and $g(z)=-z^{\alpha}$. Then for $k\in\mathbb{Z}_{\geq 0}$ and $z\in(0,\infty)$,

 $g^{(k)}(z)=-k!\binom{\alpha}{k}z^{\alpha-k},\qquad\mathrm{sgn}\,g^{(k)}(z)=(-1% )^{k}.$

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

 $\displaystyle(f\circ g)^{(n)}(z)$ $\displaystyle=\sum_{(m_{1},\ldots,m_{n}),1\cdot m_{1}+\cdots+n\cdot m_{n}=n}% \frac{n!}{m_{1}!\cdots m_{n}!}(f^{(m_{1}+\cdots+m_{n})}\circ g)(z)$ $\displaystyle\cdot\prod_{k=1}^{n}\left(\frac{g^{(k)}(z)}{k!}\right)^{m_{k}}.$

Then

 $\displaystyle(-1)^{n}(f\circ g)^{(n)}(z)$ $\displaystyle=\sum_{(m_{1},\ldots,m_{n}),1\cdot m_{1}+\cdots+n\cdot m_{n}=n}% \frac{n!}{m_{1}!\cdots m_{n}!}(\exp\circ g)(z)$ $\displaystyle\cdot\prod_{k=1}^{n}\left((-1)^{k}\frac{g^{(k)}(z)}{k!}\right)^{m% _{k}}$ $\displaystyle\geq 0.$

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 $[0,\infty)$ such that

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

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

 $\exp(-z^{q/2})=\int_{[0,\infty)}e^{-zt}d\mu_{q}(t),\qquad z\in[0,\infty).$

Then for $z\in\mathbb{R}$,

 $c_{q}(z)=\exp(-|z|^{q})=\exp(-(z^{2})^{q/2})=\int_{[0,\infty)}e^{-z^{2}t}d\mu_% {q/2}(t).$

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

 $\displaystyle\gamma_{q}(w)$ $\displaystyle=\int_{\mathbb{R}}e^{-2\pi iwz}c_{q}(z)d\lambda_{1}(z)$ $\displaystyle=\int_{\mathbb{R}}e^{-2\pi iwz}\left(\int_{[0,\infty)}e^{-z^{2}t}% d\mu_{q/2}(t)\right)d\lambda_{1}(z)$ $\displaystyle=\int_{[0,\infty)}\left(\int_{\mathbb{R}}e^{-tz^{2}}e^{-2\pi iwz}% d\lambda_{1}(z)\right)d\mu_{q/2}(t)$ $\displaystyle=\int_{[0,\infty)}\sqrt{\frac{\pi}{t}}\exp\left(-\frac{(\pi w)^{2% }}{t}\right)d\mu_{q/2}(t)$ $\displaystyle=\pi^{1/2}\int_{[0,\infty)}t^{-1/2}e^{-\pi^{2}w^{2}/t}d\mu_{q/2}(% t).$

From the final expression it is evident that $\gamma_{k}(w)>0$. Furthermore, for $w_{1},w_{2}\in(0,\infty)$, using the Cauchy-Schwarz inequality,

 $\displaystyle\log\gamma_{q}\left(\sqrt{\frac{w_{1}+w_{2}}{2}}\right)$ $\displaystyle=\frac{1}{2}\log\left(\pi^{1/2}\int_{[0,\infty)}t^{-1/4}e^{-\pi^{% 2}\cdot\frac{w_{1}}{2t}}\cdot t^{-1/4}e^{-\pi^{2}\cdot\frac{w_{2}}{2t}}d\mu_{q% /2}(t)\right)^{2}$ $\displaystyle\leq\frac{1}{2}\log\bigg{(}\pi\int_{[0,\infty)}t^{-1/2}e^{-\pi^{2% }\cdot\frac{w_{1}}{t}}d\mu_{q/2}(t)$ $\displaystyle\cdot\int_{[0,\infty)}t^{-1/2}e^{-\pi^{2}\cdot\frac{w_{2}}{t}}d% \mu_{q/2}(t)\bigg{)}$ $\displaystyle=\frac{1}{2}\log\left(\pi^{1/2}\int_{[0,\infty)}t^{-1/2}e^{-\pi^{% 2}\cdot\frac{w_{1}}{t}}d\mu_{q/2}(t)\right)$ $\displaystyle+\frac{1}{2}\log\left(\pi^{1/2}\int_{[0,\infty)}t^{-1/2}e^{-\pi^{% 2}\cdot\frac{w_{2}}{t}}d\mu_{q/2}(t)\right)$ $\displaystyle=\frac{1}{2}\log\gamma_{q}(\sqrt{w_{1}})+\frac{1}{2}\log\gamma_{q% }(\sqrt{w_{2}}).$

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