BMO, John-Nirenberg, and H1 Duality — Examples

Linearity of the integral over the cube $Q$ gives $(b+c{)}_Q=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q(b+c)=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_{Q}b+\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_{Q}c=b_Q+c$.

By step 1.1, $\mathrm{\mid }(b+c)-(b+c{)}_Q\mathrm{\mid }=\mathrm{\mid }b-b_Q\mathrm{\mid }$ pointwise on $Q$, so $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }(b+c)-(b+c{)}_Q\mathrm{\mid }=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q\mathrm{\mid }$ for every cube; taking the supremum over all cubes gives $\mathrm{\parallel }b+c{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$, including the value $+\mathrm{\infty }$.

The identity of step 2.1 shows that the seminorm is constant on each equivalence class modulo constants, so it descends to $\mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)\mathrm{/}\mathbb{C}$; on classes it is a norm because $\mathrm{\parallel }[b]\mathrm{\parallel }=0$ holds exactly when $b$ is almost everywhere constant by [L1], that is exactly for the zero class. No choice principle is used.

Local integrability and the scaling reduction. For every $R<\mathrm{\infty }$ the shell bound of [L2] gives ${\int }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }dz<\mathrm{\infty }$, because $\{\mathrm{\mid }z\mathrm{\mid }\le 1\}$ is covered by the shells $j\ge 0$ with total ${\sum }_j{2}^{(1-j)n}(j+1)\log 2<\mathrm{\infty }$, while on $\{1<\mathrm{\mid }z\mathrm{\mid }\le R\}$ one has $\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }\le \log R$ and the enclosing cube $[-R,R{]}^n$ has volume $(2R{)}^n$; hence $b\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. For a cube $Q$ write $x=x_Q+ry$ with $y\in [-\frac{1}{2},\frac{1}{2}{]}^n$, so that $\mathrm{\mid }Q\mathrm{\mid }=r^n$ and, by [L2], $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-c\mathrm{\mid }={\int }_{[-\frac{1}{2},\frac{1}{2}{]}^n}\mid \log r+\log \mathrm{\mid }\xi +y\mathrm{\mid }-c\mid dy$ with $\xi :=x_Q\mathrm{/}r$; choosing $c=\log r+c_{\xi }$ shows that it suffices to bound $A(\xi ):={\int }_{[-\frac{1}{2},\frac{1}{2}{]}^n}\mathrm{\mid }\log \mathrm{\mid }\xi +y\mathrm{\mid }-c_{\xi }\mathrm{\mid }dy$ uniformly in $\xi$, and then [L1] gives $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q\mathrm{\mid }\le 2A(\xi )$.

The regular case $\mathrm{\mid }\xi \mathrm{\mid }>2\sqrt{n}$. With $c_{\xi }=\log \mathrm{\mid }\xi \mathrm{\mid }$, for $y\in [-\frac{1}{2},\frac{1}{2}{]}^n$ and $t\in [0,1]$ one has $\mathrm{\mid }\xi +ty\mathrm{\mid }\ge \mathrm{\mid }\xi \mathrm{\mid }-\frac{1}{2}\sqrt{n}>\frac{3}{2}\sqrt{n}$, so $\log \mathrm{\mid }z\mathrm{\mid }$ is differentiable along the segment and $\mid \log \mathrm{\mid }\xi +y\mathrm{\mid }-\log \mathrm{\mid }\xi \mathrm{\mid }\mid \le {\int }_0^1\frac{\mathrm{\mid }y\mathrm{\mid }}{\mathrm{\mid }\xi +ty\mathrm{\mid }}dt\le \frac{\sqrt{n}\mathrm{/}2}{3\sqrt{n}\mathrm{/}2}=\frac{1}{3}$; hence $A(\xi )\le \frac{1}{3}$.

The singular case $\mathrm{\mid }\xi \mathrm{\mid }\le 2\sqrt{n}$. With $c_{\xi }=0$ the shifted cube lies in $\{\mathrm{\mid }z\mathrm{\mid }\le 2\sqrt{n}+\frac{1}{2}\sqrt{n}\}\subseteq \{\mathrm{\mid }z\mathrm{\mid }\le 3\sqrt{n}\}$, so $A(\xi )\le {\int }_{\mathrm{\mid }z\mathrm{\mid }\le 3\sqrt{n}}\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }dz\le C_n$ by the estimates of step 1.1 with $R=3\sqrt{n}$.

Combining steps 2.1 and 1.2, ${\sup }_{\xi }A(\xi )\le C_n<\mathrm{\infty }$, so [L1] and step 1.1 give $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\le 2C_n$: the seminorm depends only on $n$ and $b\in \mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)$.

Unboundedness. As $x\to 0$ one has $\log \mathrm{\mid }x\mathrm{\mid }\to -\mathrm{\infty }$, so $b$ is unbounded on every ball $B(0,\epsilon )$, hence on every neighbourhood of $0$. If $g\in L^{\mathrm{\infty }}({\mathbb{R}}^n)$ satisfied $g=b$ almost everywhere, then for every $M>\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ the set $\{x:\mathrm{\mid }b(x)\mathrm{\mid }>M\}$ would have positive measure (it contains $B(0,\epsilon )\setminus \{0\}$ for $\epsilon =e^{-M}$, which has positive measure because the ball contains a nondegenerate cube and a singleton has measure zero) while $\{\mathrm{\mid }g\mathrm{\mid }>M\}$ is null, contradicting almost-everywhere equality; so no bounded function represents the class of $b$. If the class of $b$ were the image of $[g]$ with $g\in L^{\mathrm{\infty }}$, then $b=g+c$ almost everywhere for a constant $c$, and $g+c$ would be a bounded representative, which is impossible. Thus the class of $b$ is not in the image of $L^{\mathrm{\infty }}\mathrm{/}\mathbb{C}$, and the continuous injection is not surjective.

For every $m\ge 1$ the cube $C_m:=[3,R_m\mathrm{/}\sqrt{n}{]}^n$ is contained in $E_m$: every $x\in C_m$ has $\mathrm{\mid }x\mathrm{\mid }\ge 3\sqrt{n}>2$ and $\mathrm{\mid }x\mathrm{\mid }\le \sqrt{n}(R_m\mathrm{/}\sqrt{n})=R_m$, and its volume is $\mathrm{\mid }{C}_m\mathrm{\mid }=(R_m\mathrm{/}\sqrt{n}-3{)}^n$. On $C_m$ one has $b(x)=\log \mathrm{\mid }x\mathrm{\mid }\ge \log 3>0$, so $\mathrm{\mid }b\mathrm{\mid }=b\ge \log 3$ there.

Therefore ${\int }_{E_m}\mathrm{\mid }b\mathrm{\mid }\ge {\int }_{C_m}\mathrm{\mid }b\mathrm{\mid }\ge (\log 3)\mathrm{\mid }{C}_m\mathrm{\mid }=(\log 3)(R_m\mathrm{/}\sqrt{n}-3{)}^n$, which tends to $+\mathrm{\infty }$ as $m\to \mathrm{\infty }$.

By [F2] and step 2.1, ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }b\mathrm{\mid }={\lim }_{m\to \mathrm{\infty }}{\int }_{E_m}\mathrm{\mid }b\mathrm{\mid }=+\mathrm{\infty }$; since $b\in \mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)$ by [F1], the inclusion $\mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)\subseteq L^1({\mathbb{R}}^n)$ is refuted.

Writing $A(\lambda ):=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\mathrm{\mid }\{x\in Q:\mathrm{\mid }b-b_Q\mathrm{\mid }>\lambda \}\mathrm{\mid }$, [L1] divided by $\mathrm{\mid }Q\mathrm{\mid }$ gives the identity $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q{\mathrm{\mid }}^q=q{\int }_0^{\mathrm{\infty }}{\lambda }^{q-1}A(\lambda )d\lambda$.

If $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>0$, [L2] bounds $A(\lambda )\le C_n{e}^{-c_n\lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}$ for every $\lambda >0$, so the integral of step 1.1 is at most $q{C}_n{\int }_0^{\mathrm{\infty }}{\lambda }^{q-1}{e}^{-c_n\lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}d\lambda =q{C}_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}^q{\int }_0^{\mathrm{\infty }}{\mu }^{q-1}{e}^{-c_n\mu }d\mu$ after the substitution $\lambda =\mu \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$; the last integral is finite, so with $C_{n,q}:=q{C}_n{\int }_0^{\mathrm{\infty }}{\mu }^{q-1}{e}^{-c_n\mu }d\mu$ the asserted bound follows.

If $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=0$, then $b$ is constant almost everywhere by [L2], so $A(\lambda )=0$ for every $\lambda >0$ and the integral of step 1.1 vanishes; in particular the bound of step 2.1 holds with both sides $0$ for any finite $C_{n,q}$.

Steps 2.1 and 3.1 give $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q{\mathrm{\mid }}^q\le C_{n,q}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}^q$ for every cube, which is exactly the per-cube upper bound used in [L3]: taking $q$-th roots and the supremum over $Q$ yields the equivalence of the $L^q$ oscillation seminorms with the BMO seminorm.

Low frequencies. For every $n<n_0$, $\mathrm{\mid }{e}^{2\pi i{\xi }_{n}x}-e^{2\pi i{\xi }_n{x}_I}\mathrm{\mid }\le 2\pi \mathrm{\mid }{\xi }_n\mathrm{\mid }\mathrm{\mid }x-x_I\mathrm{\mid }$. After $x=x_I+Ly$, ${(\int {\varphi }_I(x)\mathrm{\mid }{e}^{2\pi i{\xi }_{n}x}-e^{2\pi i{\xi }_n{x}_I}{\mathrm{\mid }}^{2}dx)}^{1\mathrm{/}2}\le C_{\varphi }\mathrm{\mid }{\xi }_n\mathrm{\mid }{L}^{3\mathrm{/}2}\mathrm{.}$ The triangle inequality in $L^2({\varphi }_{I}dx)$, Cauchy--Schwarz, and the upper frequency bound therefore give $\begin{array}{rl}{\displaystyle \mathrm{\parallel }{f}_{<}-c_I{\mathrm{\parallel }}_{L^2({\varphi }_{I}dx)}}& {\displaystyle \le C_{\varphi }{L}^{3\mathrm{/}2}\sum _{n<n_0}\mathrm{\mid }{c}_n\mathrm{\mid }\mathrm{\mid }{\xi }_n\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle \le C_{b,\varphi }{L}^{3\mathrm{/}2}\left(\sum _n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\right(\sum _{n<n_0}{4}^n{)}^{1\mathrm{/}2}}\\ {\displaystyle }& {\displaystyle \le C_{b,\varphi }{L}^{1\mathrm{/}2}(\sum _n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2},}\end{array}$ because $2^{n_0}<2L^{-1}$.

High frequencies. Define $G_{nm}:=L^{-1}{\int }_{\mathbb{R}}{\varphi }_I(x)e^{2\pi i({\xi }_n-{\xi }_m)x}dx(n,m\ge n_0)\mathrm{.}$ The change of variables $x=x_I+Ly$ and [L2] give, for every fixed integer $M\ge 1$, $\mathrm{\mid }{G}_{nm}\mathrm{\mid }\le C_{M,\varphi }(1+L\mathrm{\mid }{\xi }_n-{\xi }_m\mathrm{\mid }{)}^{-M}\mathrm{.}$ Choose an integer $K\ge 1$ such that $b{2}^{-K}\le a\mathrm{/}2$. If $\mathrm{\mid }n-m\mathrm{\mid }>K$, the frequency comparability implies $\mathrm{\mid }{\xi }_n-{\xi }_m\mathrm{\mid }\ge \mid \mathrm{\mid }{\xi }_n\mathrm{\mid }-\mathrm{\mid }{\xi }_m\mathrm{\mid }\mid \ge (a\mathrm{/}2)2^{\max (n,m)}\mathrm{.}$ For each fixed $n\ge n_0$, the terms with $\mathrm{\mid }m-n\mathrm{\mid }\le K$ contribute at most a constant to ${\sum }_{m\ge n_0}\mathrm{\mid }{G}_{nm}\mathrm{\mid }$. For $m>n+K$ their sum is bounded by $C{\sum }_{r\ge 1}(1+L{2}^{n+r}{)}^{-M}\le C{\sum }_{r\ge 1}{2}^{-Mr}<\mathrm{\infty },$ using $L{2}^n\ge 1$. For $n_0\le m<n-K$, there are at most $n-n_0$ terms, each bounded by $C(1+L{2}^n{)}^{-M}$; their total is bounded uniformly because $L{2}^n\ge 2^{n-n_0}$ and ${\sup }_{r\ge 0}r(1+2^r{)}^{-M}<\mathrm{\infty }$. Hence the absolute row sums of $(G_{nm}{)}_{n,m\ge n_0}$ are uniformly bounded. Since $\mathrm{\mid }{G}_{nm}\mathrm{\mid }=\mathrm{\mid }{G}_{mn}\mathrm{\mid }$, the inequality $2\mathrm{\mid }{c}_n\mathrm{\mid }\mathrm{\mid }{c}_m\mathrm{\mid }\le \mathrm{\mid }{c}_n{\mathrm{\mid }}^2+\mathrm{\mid }{c}_m{\mathrm{\mid }}^2$ yields $\begin{array}{rl}{\displaystyle \int {\varphi }_I\mathrm{\mid }{f}_{\ge }{\mathrm{\mid }}^2}& {\displaystyle \le L\sum _{n,m\ge n_0}\mathrm{\mid }{c}_n\mathrm{\mid }\mathrm{\mid }{c}_m\mathrm{\mid }\mathrm{\mid }{G}_{nm}\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle \le C_{a,b,\varphi }L\sum _n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}}\end{array}$

Combining the two pieces with $\mathrm{\mid }f-c_I{\mathrm{\mid }}^2\le 2\mathrm{\mid }{f}_{<}-c_I{\mathrm{\mid }}^2+2\mathrm{\mid }{f}_{\ge }{\mathrm{\mid }}^2$ gives ${\int }_{\mathbb{R}}{\varphi }_I\mathrm{\mid }f-c_I{\mathrm{\mid }}^2\le C_{a,b,\varphi }L{\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}$ Since ${\varphi }_I\ge 1$ on $I$, Cauchy--Schwarz and [L1] imply $\begin{array}{rl}{\displaystyle L^{-1}{\int }_I\mathrm{\mid }f-f_I\mathrm{\mid }}& {\displaystyle \le 2L^{-1}{\int }_I\mathrm{\mid }f-c_I\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle \le 2L^{-1\mathrm{/}2}{({\int }_I\mathrm{\mid }f-c_I{\mathrm{\mid }}^2)}^{1\mathrm{/}2}}\\ {\displaystyle }& {\displaystyle \le C_{a,b,\varphi }(\sum _n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\mathrm{.}}\end{array}$ Taking the supremum over all nondegenerate intervals proves the BMO bound. The proof uses finite sums only and no choice principle.