$H^p$ equals $L^p$ with equivalent norms for $1<p<\mathrm{\infty }$

Given: Countable Choice and the ultrafilter lemma, $n\ge 1$, $1<p<\mathrm{\infty }$, an admissible kernel $\phi$, and $f\in {\mathcal{S}}^{\mathrm{\prime }}$.

If $0\le \mathrm{\Phi }\in L^1({\mathbb{R}}^n)$ is radially nonincreasing, then the associated maximal operator is dominated by the centered Hardy-Littlewood maximal operator: $(\mathrm{\Phi }\ast \mathrm{\mid }u\mathrm{\mid })(x)\le \mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{1}Mu(x)$ (Radially decreasing kernels are dominated by the maximal function).

The centered Hardy-Littlewood maximal operator satisfies the strong $L^p$ bound $\mathrm{\parallel }Mu{\mathrm{\parallel }}_{L^p}\le C_{n,p}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$, $1<p<\mathrm{\infty }$ (The centered maximal operator is bounded on $L^p({\mathbb{R}}^n)$ for $1<p<\mathrm{\infty }$, The centered and uncentered Hardy-Littlewood maximal functions).

Let $p_{\alpha ,0}(\phi )={\sup }_x\mathrm{\mid }{x}^{\alpha }\phi (x)\mathrm{\mid }$ be the Schwartz seminorms of Schwartz space and its seminorms, and set $C_{\phi }:=2^n(1+\sqrt{n}{)}^{n+1}(p_{0,0}(\phi )+{\sum }_{i=1}^n{p}_{(n+1)e_i,0}(\phi ))<\mathrm{\infty }$. Since $\mathrm{\mid }x\mathrm{\mid }\le \sqrt{n}{\max }_i\mathrm{\mid }{x}_i\mathrm{\mid }$, the elementary inequality $(1+r{)}^{n+1}\le 2^n(1+r^{n+1})$ shows $(1+\mathrm{\mid }x\mathrm{\mid }{)}^{n+1}\mathrm{\mid }\phi (x)\mathrm{\mid }\le C_{\phi }$. Thus $G(x):=C_{\phi }(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-n-1}$ is a radially nonincreasing integrable pointwise majorant of $\mathrm{\mid }\phi \mathrm{\mid }$. If $u\in L^p$, then $\mathrm{\mid }(u\ast {\phi }_t)(x)\mathrm{\mid }\le (\mathrm{\mid }u\mathrm{\mid }\ast \mathrm{\mid }\phi {\mathrm{\mid }}_t)(x)$; normalised dilations $G_t$ are radially nonincreasing with $\mathrm{\parallel }{G}_t{\mathrm{\parallel }}_1=\mathrm{\parallel }G{\mathrm{\parallel }}_1$, so $\mathrm{\mid }u\mathrm{\mid }\ast \mathrm{\mid }\phi {\mathrm{\mid }}_t\le \mathrm{\mid }u\mathrm{\mid }\ast G_t\le \mathrm{\parallel }G{\mathrm{\parallel }}_{1}Mu$ (Radially decreasing kernels are dominated by the maximal function).

For $1<p<\mathrm{\infty }$ complex $L^p$ is the dual of complex $L^{p^{\mathrm{\prime }}}$ by Complex Lp duality from real Lp duality. Separability first applies to the Borel restriction: rational boxes countably generate it (For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n), and bounded cubes give sigma-finiteness. Completion does not change $L^{p^{\mathrm{\prime }}}$: $\mathcal{L}({\mathbb{R}}^n)$ is exactly the completion of the restriction of ${\lambda }_n$ to the Borel sets replaces each measurable set in a sequence of simple approximants by a Borel set modulo a null set; the countable union of these exceptions is null, yielding a Borel representative. Separability on the Borel restriction therefore gives separability of Lebesgue $L^{p^{\mathrm{\prime }}}$; hence the unit ball of $L^p$ is weak-star sequentially compact, the weak-star topology on norm-bounded sets is metrised by a countable dense set, and the dual norm is weak-star lower semicontinuous (For $1<p<\mathrm{\infty }$, the same representation theorem holds on arbitrary measure spaces, If $\mu$ is sigma-finite and $\mathcal{A}$ is countably generated, then $L^p(\mu )$ is separable for $1\le p<\mathrm{\infty }$, A separable predual has weak-star sequentially compact dual ball, Conjugate exponents, including the endpoint conventions).

Proof technique: direct domination by the Hardy-Littlewood maximal function, then weak-star sequential compactness for the reverse inclusion.