Bounded BMO functions dualise H1 boundedly

By [F1] fix a representation $f={\sum }_j{\lambda }_j{a}_j$ with ${\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }\le C_{n,\tilde{N},\phi }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}$. The partial sums are $L^1$ functions with $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1}\le {\sum }_{j\le N}\mathrm{\mid }{\lambda }_j\mathrm{\mid }\mathrm{\parallel }{a}_j{\mathrm{\parallel }}_{L^1}\le {\sum }_{j\le N}\mathrm{\mid }{\lambda }_j\mathrm{\mid }$ by [F2]; they therefore form a Cauchy sequence in $L^1$, and by completeness [F3] converge in $L^1$ to some $g\in L^1$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^1}\le {\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$. For every test function $\psi$ one has $⟨f,\psi ⟩={\lim }_N⟨S_N,\psi ⟩={\lim }_N\int S_N\psi =\int g\psi$ by [F3] and the ${\mathcal{S}}^{\mathrm{\prime }}$-convergence of the partial sums; hence $f$ is represented by the $L^1$ function $g$, and $\int fb=\int gb$ converges absolutely with $\int \mathrm{\mid }gb\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^1}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}<\mathrm{\infty }$.

Since $S_N\to g$ in $L^1$ and $b\in L^{\mathrm{\infty }}$, [F3] gives $\int S_{N}b\to \int gb$; and $\mid \int S_{N}b\mid =\mid {\sum }_{j\le N}{\lambda }_j\int a_{j}b\mid \le {\sum }_{j\le N}\mathrm{\mid }{\lambda }_j\mathrm{\mid }\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\le C_{n,\tilde{N},\phi }\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}$ by [F2]. Passing to the limit gives $\mid \int fb\mid =\mid \int gb\mid \le C_{n,\tilde{N},\phi }\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}$.

Step 2.1 is the asserted bound with constant $C_{n,\tilde{N},\phi }$, and step 1.1 is the asserted absolute convergence. Countable Choice is inherited from the suppliers.