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11.44.4 A norm defined by the critical scalarA20C27AA8420

Mixed products turn the comparison of critical scalars into a triangle inequality for $n$th roots of densities, and the subsolution estimate associated with the strict barrier gives continuity in the corresponding Lebesgue norm.

Fix $1<q<p$ and set $\ell =nq$. On real $L^\ell (\mathrm{d}V)$ define

\[ N(a)= \begin{cases} c(|a|^ n)^{-1/n},& a\ne 0,\\ 0,& a=0. \end{cases} \]

Proposition 11.44.5.91FDE4AB5C22 The function $N$ is a lattice norm on $L^\ell (\mathrm{d}V)$, that is, a norm satisfying $N(a)\leq N(b)$ whenever $|a|\leq |b|$. There is $C_ q<\infty $, depending only on $X,\theta ,\mathrm{d}V$ and $q$, such that

8267EBDA38A9 11.44.6
\begin{equation} N(a)\leq C_ q\| a\| _\ell ,\qquad |N(a)-N(b)|\leq C_ q\| a-b\| _\ell . \end{equation}

Consequently, for nonzero $g,h\geq 0$ in $L^ q$,

\[ |c(g)^{-1/n}-c(h)^{-1/n}| \leq C_ q\| g-h\| _ q^{1/n}. \]

Proof. If a potential $v$ of minimal type satisfies $\mathrm{MA}_\theta (v)\geq d g\, \mathrm{d}V$, then $d\leq c(g)$, since otherwise comparison with a critical potential for $g$ contradicts part (1) of corollary 15.13.3. Hence $g\leq h$ implies $c(g)\geq c(h)$. Uniqueness of the critical scalar also gives $c(tg)=c(g)/t$ for $t>0$. These facts prove that $N$ is monotone and absolutely homogeneous, and positivity follows from $0<c(g)<\infty $.

For the triangle inequality, let $a,b\geq 0$ be nonzero, put $A=N(a)$ and $B=N(b)$, and choose critical potentials $v_ a,v_ b$ with measures $(a/A)^ n\mathrm{d}V$ and $(b/B)^ n\mathrm{d}V$. On every relatively compact subset of $\Omega $, (5.28.9) gives

\[ \mathrm{MA}_\theta \left(\frac{A v_ a+B v_ b}{A+B}\right) \geq \left(\frac{a+b}{A+B}\right)^ n\mathrm{d}V, \]

where both potentials are locally bounded. Exhausting $\Omega $ gives the inequality on $X$, since $X\setminus \Omega $ has zero smooth volume. The averaged potential has minimal type, so the first paragraph gives $N(a+b)\leq A+B$. The cases where $a$ or $b$ vanishes are trivial, and for real functions we use $|a+b|\leq |a|+|b|$ and monotonicity.

Next, apply lemma 15.14.1 with $\eta =V_\theta $ to nonnegative densities of $L^ q$ norm at most one. It gives a uniform $m>0$ and subsolutions of minimal type, hence $c(g)\geq m$ for each such nonzero density. For $a\ne 0$, take $g=|a|^ n/\| a\| _\ell ^ n$, which has $L^ q$ norm one. Homogeneity gives $N(a)\leq m^{-1/n}\| a\| _\ell $, which is the first bound in (11.44.6), and the reverse triangle inequality gives the second. Applying the second bound to $g^{1/n},h^{1/n}$ and using $|g^{1/n}-h^{1/n}|^ n\leq |g-h|$ proves the last assertion. $\square$

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