[proofplan] We reduce the desired moment comparison to a one-dimensional tail estimate for the measurable seminorm variable $Y(x)=s(x)$ on the probability space $(\mathbb R^n,\mathcal B(\mathbb R^n),\mu)$. The log-concavity of $\mu$ and the seminorm property imply the standard Borell tail bound: after normalising by the $p$-moment, the upper tail of $Y$ decays exponentially with constants depending only on $p$. Integrating this tail estimate by the [layer-cake formula](/theorems/2956) gives the $q$-moment bound, and the constant is computed explicitly from the resulting Gamma integral. [/proofplan]

[step:Declare the seminorm random variable and normalise its $p$-moment] Let $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\mu)$ denote the probability space in the theorem statement, and define the measurable map \begin{align*} Y:(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))&\to([0,\infty),\mathcal{B}([0,\infty)))\\ x&\mapsto s(x). \end{align*} Since $s$ is a seminorm, $s$ is convex and continuous, hence Borel measurable. Define \begin{align*} M_p:=\left(\int_{\mathbb{R}^n}s(x)^p\,d\mu(x)\right)^{1/p}=\left(\int_{\mathbb{R}^n}Y(x)^p\,d\mu(x)\right)^{1/p}. \end{align*} If $M_p=0$, then $Y=0$ $\mu$-a.e., so the desired inequality is immediate. Assume from now on that $M_p>0$, and define the normalised random variable \begin{align*} Z:(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))&\to([0,\infty),\mathcal{B}([0,\infty)))\\ x&\mapsto \frac{s(x)}{M_p}. \end{align*} Then \begin{align*} \int_{\mathbb{R}^n}Z(x)^p\,d\mu(x)=1. \end{align*} [/step]

[step:Apply Borell's log-concave tail estimate to the normalised seminorm]We use the following one-dimensional form of Borell's tail estimate for seminorms under log-concave measures: if $\nu$ is a log-concave probability measure on $\mathbb{R}^n$, $r: \mathbb{R}^n\to[0,\infty)$ is a seminorm, $1\le p<\infty$, and \begin{align*} \int_{\mathbb{R}^n}r(x)^p\,d\nu(x)=1, \end{align*} then for every $t\ge 2$, \begin{align*} \nu(\{x\in\mathbb{R}^n:r(x)>t\})\le \exp\left(-\frac{t}{2e}\right). \end{align*} The hypotheses are satisfied with $\nu=\mu$ and $r=Z$: the measure $\mu$ is log-concave by assumption, $Z$ is a positive scalar multiple of the seminorm $s$ and therefore is again a seminorm, and the preceding step gives $\int Z^p\,d\mu=1$. Hence, for every $t\ge 2$, \begin{align*} \mu(\{x\in\mathbb{R}^n:Z(x)>t\})\le \exp\left(-\frac{t}{2e}\right). \end{align*}[/step]

[guided]The point of the normalisation is that Borell's tail estimate is scale invariant once the $p$-moment is fixed. We apply it to the map \begin{align*} Z:(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))&\to([0,\infty),\mathcal{B}([0,\infty)))\\ x&\mapsto \frac{s(x)}{M_p}. \end{align*} We verify each hypothesis. First, $\mu$ is log-concave by the theorem statement. Second, because $M_p>0$ and $s$ is a seminorm, $Z=M_p^{-1}s$ is also a seminorm: it is non-negative, positively homogeneous, and subadditive. Third, the previous step proved \begin{align*} \int_{\mathbb{R}^n}Z(x)^p\,d\mu(x)=1. \end{align*} Borell's tail estimate therefore gives, for every $t\ge 2$, \begin{align*} \mu(\{x\in\mathbb{R}^n:Z(x)>t\})\le \exp\left(-\frac{t}{2e}\right). \end{align*} This is where both structural hypotheses are used: log-concavity supplies the exponential decay, and the seminorm property is the convex homogeneity condition needed by the tail estimate.[/guided]

[step:Convert the tail estimate into a $q$-moment bound]Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $\mathbb R$. For $q>0$, define the Gamma value \begin{align*} \Gamma(q):=\int_0^\infty u^{q-1}e^{-u}\,d\mathcal L^1(u). \end{align*} By the layer-cake formula for the non-negative measurable function $Z^q$, with respect to the probability measure $\mu$, \begin{align*} \int_{\mathbb{R}^n}Z(x)^q\,d\mu(x) &=q\int_0^\infty t^{q-1}\mu(\{x\in\mathbb{R}^n:Z(x)>t\})\,d\mathcal{L}^1(t). \end{align*} Split the integral at $t=2$. Since $\mu$ is a probability measure, $\mu(\{Z>t\})\le 1$ for $0<t<2$, and the tail estimate applies for $t\ge 2$. Therefore \begin{align*} \int_{\mathbb{R}^n}Z(x)^q\,d\mu(x) &\le q\int_0^2 t^{q-1}\,d\mathcal{L}^1(t)+q\int_2^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t)\\ &\le 2^q+q\int_0^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t). \end{align*} Using the substitution $u=t/(2e)$, so that $t=2eu$ and $d\mathcal{L}^1(t)=2e\,d\mathcal{L}^1(u)$, gives \begin{align*} q\int_0^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t) &=q(2e)^q\int_0^\infty u^{q-1}e^{-u}\,d\mathcal{L}^1(u)\\ &=q(2e)^q\Gamma(q). \end{align*} Thus \begin{align*} \int_{\mathbb{R}^n}Z(x)^q\,d\mu(x)\le 2^q+q(2e)^q\Gamma(q). \end{align*}[/step]

[guided]We now turn tail decay into a moment estimate. The correct tool is the layer-cake formula, applied to the non-negative measurable function $Z^q$ on the probability space $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\mu)$. It gives \begin{align*} \int_{\mathbb{R}^n}Z(x)^q\,d\mu(x) &=q\int_0^\infty t^{q-1}\mu(\{x\in\mathbb{R}^n:Z(x)>t\})\,d\mathcal{L}^1(t). \end{align*} The tail estimate is only stated for $t\ge 2$, so we split the one-dimensional [Lebesgue integral](/page/Lebesgue%20Integral) into the intervals $(0,2)$ and $[2,\infty)$. On $(0,2)$ we use only that $\mu$ is a probability measure: \begin{align*} q\int_0^2 t^{q-1}\mu(\{x\in\mathbb{R}^n:Z(x)>t\})\,d\mathcal{L}^1(t) \le q\int_0^2 t^{q-1}\,d\mathcal{L}^1(t)=2^q. \end{align*} On $[2,\infty)$ we use Borell's tail bound: \begin{align*} q\int_2^\infty t^{q-1}\mu(\{x\in\mathbb{R}^n:Z(x)>t\})\,d\mathcal{L}^1(t) \le q\int_2^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t). \end{align*} We enlarge the integration domain from $[2,\infty)$ to $[0,\infty)$ because the integrand is non-negative. This gives \begin{align*} q\int_2^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t) \le q\int_0^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t). \end{align*} Now substitute $u=t/(2e)$. Under this substitution $t=2eu$ and $d\mathcal{L}^1(t)=2e\,d\mathcal{L}^1(u)$, while the domain $[0,\infty)$ is mapped onto $[0,\infty)$. Hence \begin{align*} q\int_0^\infty t^{q-1}\exp\left(-\frac{t}{2e}\right)\,d\mathcal{L}^1(t) &=q(2e)^q\int_0^\infty u^{q-1}e^{-u}\,d\mathcal{L}^1(u)\\ &=q(2e)^q\Gamma(q). \end{align*} Combining the small-tail and large-tail pieces yields \begin{align*} \int_{\mathbb{R}^n}Z(x)^q\,d\mu(x)\le 2^q+q(2e)^q\Gamma(q). \end{align*}[/guided]

[step:Rescale to obtain the stated moment comparison] Define \begin{align*} C_{p,q}:=\left(2^q+q(2e)^q\Gamma(q)\right)^{1/q}. \end{align*} This constant is positive and depends only on $q$, hence only on $p$ and $q$. Since $Z=s/M_p$, the preceding estimate gives \begin{align*} \left(\int_{\mathbb{R}^n}s(x)^q\,d\mu(x)\right)^{1/q} &=M_p\left(\int_{\mathbb{R}^n}Z(x)^q\,d\mu(x)\right)^{1/q}\\ &\le C_{p,q}M_p\\ &=C_{p,q}\left(\int_{\mathbb{R}^n}s(x)^p\,d\mu(x)\right)^{1/p}. \end{align*} This is the asserted inequality. [/step]