Relative norm-preserving Hahn–Banach extension over the real and complex fields

Put $C=\mathrm{\parallel }g\mathrm{\parallel }\ge 0$. For $m\mathrm{\ne }0$, the vector $m\mathrm{/}\mathrm{\parallel }m\mathrm{\parallel }$ is in the unit ball of $M$, so $\mathrm{\mid }g(m)\mathrm{\mid }=\mathrm{\parallel }m\mathrm{\parallel }\mathrm{\mid }g(m\mathrm{/}\mathrm{\parallel }m\mathrm{\parallel })\mathrm{\mid }\le C\mathrm{\parallel }m\mathrm{\parallel }$; for $m=0$ the same inequality holds because $g(0)=0$. Set $p(x)=C\mathrm{\parallel }x\mathrm{\parallel }$. Then $p(tx)=tp(x)$ for $t\ge 0$ and $p(x+y)\le p(x)+p(y)$ by the norm axioms.

Over $\mathbb{R}$, $g\le p{\mathrm{\mid }}_M$ by the preceding estimate. The real extension theorem gives $F{\mathrm{\mid }}_M=g$ and $-C\mathrm{\parallel }x\mathrm{\parallel }\le F(x)\le C\mathrm{\parallel }x\mathrm{\parallel }$. Thus $\mathrm{\mid }F(x)\mathrm{\mid }\le C\mathrm{\parallel }x\mathrm{\parallel }$, so $F\in X^{\ast }$ and $\mathrm{\parallel }F\mathrm{\parallel }\le C$.

Over $\mathbb{C}$, the underlying real space of $M$ is a real linear subspace of the underlying real $X$, because closure under complex scalars includes closure under real scalars. Let $u=\mathrm{Re}g$. It is real linear and $u(m)\le \mathrm{\mid }g(m)\mathrm{\mid }\le p(m)$. The real extension theorem gives real-linear $U:X\to \mathbb{R}$ with $U{\mathrm{\mid }}_M=u$ and $U\le p$.

Define $F(x)=U(x)-iU(ix)$. The reconstruction lemma gives complex linearity and $\mathrm{Re}F=U$. For $m\in M$, also $im\in M$, whence $F(m)=u(m)-iu(im)=g(m)$ by the same lemma applied to $g$.

If $F(x)=0$ then $\mathrm{\mid }F(x)\mathrm{\mid }\le C\mathrm{\parallel }x\mathrm{\parallel }$. Otherwise set $a=\overline{F(x)}\mathrm{/}\mathrm{\mid }F(x)\mathrm{\mid }$. Then $\mathrm{\mid }a\mathrm{\mid }=1$ and $F(ax)=aF(x)=\mathrm{\mid }F(x)\mathrm{\mid }$ is real, so $\mathrm{\mid }F(x)\mathrm{\mid }=U(ax)\le C\mathrm{\parallel }ax\mathrm{\parallel }=C\mathrm{\parallel }x\mathrm{\parallel }$. Consequently the complex extension is bounded and $\mathrm{\parallel }F\mathrm{\parallel }\le C$.

In either field, $F$ extends $g$. For every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$, $\mathrm{\mid }g(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$; taking the supremum gives $C\le \mathrm{\parallel }F\mathrm{\parallel }$. Together with the upper bounds this yields $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }g\mathrm{\parallel }$. If $C=0$, the bound forces $F=0$; in particular this covers $M=\{0\}$ and the zero space.