Complex Lq norm recovery from finite simple dual tests

Put $a=\mathrm{\mid }g\mathrm{\mid }$ and $\theta =\bar{g}\mathrm{/}a$ on $\{a>0\}$, zero elsewhere. The function $b=1\mathrm{/}a$ on $\{a>0\}$, zero elsewhere, is measurable: for $t\ge 0$, its strict upper level set is $\{0<a<1\mathrm{/}t\}$ when $t>0$ and $\{a>0\}$ when $t=0$; negative upper thresholds give $X$. Real positive powers of $a$ have upper sets $\{a>t^{1\mathrm{/}\gamma }\}$ for $\gamma >0,t\ge 0$. Thus F9–F10 and F1 show the phase, the powers and all ensuing products are measurable. F8 gives $\mathrm{\mid }\theta \mathrm{\mid }\le 1$ and $g\theta =a$.

Let $E$ be measurable of finite measure and let $h$ be a bounded measurable function vanishing outside $E$, with $\mathrm{\parallel }h{\mathrm{\parallel }}_r\le 1$ and ${\int }_E\mathrm{\mid }g\mathrm{\mid }<\mathrm{\infty }$. Round each coordinate of $h$ down to an integer multiple of $1\mathrm{/}m$ on $E$ and set the result $t_m$ to zero off $E$. This has finite range, measurable fibers, and $\mathrm{\mid }{t}_m-h\mathrm{\mid }\le \sqrt{2}\mathrm{/}m$. When $r=\mathrm{\infty }$ and $\mathrm{\mid }h\mathrm{\mid }\le 1$, replace each of its finitely many values $z$ by $z\mathrm{/}\max (1,\mathrm{\mid }z\mathrm{\mid })$; these values lie in the closed unit disk and the error is at most $2\sqrt{2}\mathrm{/}m$, because the displacement of $z$ is at most $\mathrm{\mid }z-h\mathrm{\mid }$. For finite $r$, the uniform error $e_m$ gives $\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_r\le e_m\mu (E{)}^{1\mathrm{/}r}\to 0$; for infinity it gives $\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_{\mathrm{\infty }}\le e_m\to 0$. Put $d_m=\max (1,\mathrm{\parallel }{t}_m{\mathrm{\parallel }}_r)$ and $s_m=t_m\mathrm{/}{d}_m$. F2 gives $1\le d_m\le 1+\mathrm{\parallel }{t}_m-h{\mathrm{\parallel }}_r$, hence $d_m\to 1$ and $s_m\to h$ uniformly, since $h$ is bounded. Each $s_m$ is an admissible finite simple test, and F7 gives $\mathrm{\mid }\int g(s_m-h)\mathrm{\mid }\le {\sup }_E\mathrm{\mid }{s}_m-h\mathrm{\mid }{\int }_E\mathrm{\mid }g\mathrm{\mid }\to 0$. Therefore $S(g)\ge \mathrm{\mid }\int gh\mathrm{\mid }$. If $\mu (E)=0$, the integral is zero and the zero test already suffices.

Let $E$ have finite measure with $a\le B<\mathrm{\infty }$ on $E$, and put $A=\mathrm{\parallel }g{\mathbf{1}}_E{\mathrm{\parallel }}_q$ for finite $q$. If $A=0$, $S(g)\ge A$ from the zero test. If $A>0$ and $1<q<\mathrm{\infty }$, set $h={\mathbf{1}}_E\theta a^{q-1}\mathrm{/}{A}^{q-1}$. This is bounded, and $(q-1)r=q$ gives $\int \mathrm{\mid }h{\mathrm{\mid }}^r=A^{-q}{\int }_E{a}^q=1$. Moreover $\int gh=A^{-(q-1)}{\int }_E{a}^q=A$. For $q=1$ set $h={\mathbf{1}}_E\theta$ instead; it is bounded by one and $\int gh={\int }_{E}a=A$. In both cases ${\int }_E\mathrm{\mid }g\mathrm{\mid }\le B\mu (E)<\mathrm{\infty }$, so step 1.2 applies and gives $S(g)\ge \mathrm{\parallel }g{\mathbf{1}}_E{\mathrm{\parallel }}_q$.

Under sigma-finiteness take a covering $H_j$ of finite measure and put $E_n=({\bigcup }_{j\le n}{H}_j)\cap \{1\mathrm{/}n\le a\le n\}$ for $n\ge 1$. These finite-measure sets increase and cover $\{a>0\}$. For finite $q$, $a^q{\mathbf{1}}_{E_n}\uparrow a^q$ everywhere, so F6 yields $\mathrm{\parallel }g{\mathbf{1}}_{E_n}{\mathrm{\parallel }}_q\uparrow \mathrm{\parallel }g{\mathrm{\parallel }}_q$, possibly infinitely. Step 2.1 proves $S(g)\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. If that norm is finite, F2 bounds every admissible test by $\mathrm{\parallel }g{\mathrm{\parallel }}_q$; if it is infinite the lower bound already gives equality at infinity. The assumed test integrability ensures every integral in the defining supremum is meaningful.

If instead $g\in L^q$ with $q<\mathrm{\infty }$ on an arbitrary measure space, use $E_n=\{1\mathrm{/}n\le a\le n\}$. F11 gives $n^{-q}\mu (E_n)\le \int a^q<\mathrm{\infty }$, hence $\mu (E_n)<\mathrm{\infty }$. The same increasing limit and step 2.1 give $S(g)\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. Hölder gives the reverse bound and integrability of every $gs$: a finite simple function of finite-measure support belongs to every finite-exponent Lr, and is bounded when $r=\mathrm{\infty }$. No sigma-finiteness of $X$ is needed.

Now let $q=\mathrm{\infty }$ and $M=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}>0$, possibly infinite under the sigma-finite hypothesis. For any $0<t<M$, the measurable set $D=\{a>t\}$ has positive measure, since otherwise $t$ would be an essential bound. In the sigma-finite case, $D={\bigcup }_n(D\cap E_n)$ with the sets from step 3.1. F12 implies some $B=D\cap E_n$ has positive measure, and $0<\mu (B)<\mathrm{\infty }$ with $a\le n$ there. In the semifinite already-L-infinity case, $M<\mathrm{\infty }$; the set $\{a>M+1\}$ is null by F5. Semifiniteness applied to $D\cap \{a\le M+1\}$ gives a measurable $B$ of finite positive measure. In either case $h=\theta {\mathbf{1}}_B\mathrm{/}\mu (B)$ is bounded, $\mathrm{\parallel }h{\mathrm{\parallel }}_1=1$, and $\int gh=\mu (B{)}^{-1}{\int }_{B}a\ge t$. Step 1.2 now yields $S(g)\ge t$. Letting $t\uparrow M$, or taking arbitrarily large $t$ if $M=\mathrm{\infty }$, proves $S(g)\ge M$. For finite $M$, F2 gives $S(g)\le M$. If $M=0$, F2 and the zero test give $S(g)=0$.

The measure on $X=\{x\}$ with $\mu (\mathrm{\varnothing })=0$, $\mu (X)=\mathrm{\infty }$ is countably additive: a disjoint family contains at most one nonempty member. It is not semifinite. For $g(x)=1$, its essential norm is one, yet the only finite-measure-supported simple function is zero, so $S(g)=0$. This verifies the necessity of a measure hypothesis at the infinity endpoint. Finally the zero test makes every stated supremum nonempty; on zero measure spaces it and every other integral have value zero. The previous steps prove all the claimed identities and therefore the finite-bound membership conclusion.