The Analytic Hahn Banach Theorem

If $m+t{x}_0=m^{\mathrm{\prime }}+t^{\mathrm{\prime }}{x}_0$ with $m,m^{\mathrm{\prime }}\in M$ and $t,t^{\mathrm{\prime }}\in \mathbb{R}$, then $m-m^{\mathrm{\prime }}=(t^{\mathrm{\prime }}-t)x_0\in M$. If $t\mathrm{\ne }{t}^{\mathrm{\prime }}$, closure under scalar multiplication from [L3] gives $x_0\in M$, contradicting the hypothesis. So $t=t^{\mathrm{\prime }}$ and then $m=m^{\mathrm{\prime }}$. Therefore every element of $N$ has a unique representation $m+t{x}_0$, and $F_c$ is well defined.

For $y,z\in M$, one has $F_c(y-x_0)=f(y)-c\text{and}{F}_c(z+x_0)=f(z)+c\mathrm{.}$ Therefore $F_c(y-x_0)\le p(y-x_0)⟺f(y)-p(y-x_0)\le c,$ and $F_c(z+x_0)\le p(z+x_0)⟺c\le p(z+x_0)-f(z)\mathrm{.}$

Let $y,z\in M$. Since $y+z\in M$ by [L3], linearity and domination on $M$ give $f(y)+f(z)=f(y+z)\le p(y+z)\mathrm{.}$ Also $y+z=(y-x_0)+(z+x_0),$ so subadditivity from [L1] yields $p(y+z)\le p(y-x_0)+p(z+x_0)\mathrm{.}$ Combining these inequalities gives $f(y)-p(y-x_0)\le p(z+x_0)-f(z)\mathrm{.}$ Hence every lower endpoint is at most every upper endpoint.

Suppose first that the two inequalities from step 1.2 hold for every $y,z\in M$. Let $m+t{x}_0\in N$. If $t>0$, then $m+t{x}_0=t(\frac{m}{t}+x_0),$ so by linearity and positive homogeneity, $F_c(m+t{x}_0)=t{F}_c(\frac{m}{t}+x_0)\le tp(\frac{m}{t}+x_0)=p(m+t{x}_0)\mathrm{.}$ If $t<0$, then $m+t{x}_0=(-t)(\frac{-m}{t}-x_0),$ so the lower-bound half of step 1.2 applied to $y:=-m\mathrm{/}t$ gives $F_c(m+t{x}_0)=(-t)F_c(\frac{-m}{t}-x_0)\le (-t)p(\frac{-m}{t}-x_0)=p(m+t{x}_0)\mathrm{.}$ If $t=0$, then $m+t{x}_0=m\in M$, so the hypothesis $f\le p$ gives $F_c(m)=f(m)\le p(m)$. Thus $F_c\le p$ on $N$. Conversely, if $F_c\le p$ on $N$, then applying that inequality to $y-x_0$ and $z+x_0$ yields the two inequalities in step 1.2.

Step 1.3 shows that the set of lower endpoints is bounded above by every upper endpoint, and the set of upper endpoints is bounded below by every lower endpoint. Completeness of $\mathbb{R}$ therefore gives real numbers $\alpha \le \beta$ with the displayed formulas in the statement. By step 2.1, a real number $c$ is admissible exactly when it lies between every lower endpoint and every upper endpoint, that is, exactly when $\alpha \le c\le \beta$. Therefore the admissible values form the nonempty interval $[\alpha ,\beta ]$.

If $x\in N_1\cap N_2$ for $(N_1,g_1),(N_2,g_2)\in \mathcal{C}$, then [L2] gives comparability. Suppose $N_1\subseteq N_2$. Since the chain order is extension, $g_2{\mathrm{\mid }}_{N_1}=g_1$, so $g_1(x)=g_2(x)$. The other inclusion case is the same. Therefore $G(x)$ is well defined on overlaps.

Let $u,v\in U$ and $a\in \mathbb{R}$. Choose $(N_1,g_1),(N_2,g_2)\in \mathcal{C}$ with $u\in N_1$ and $v\in N_2$. By [L2], one of the domains contains the other; after relabeling, assume $N_1\subseteq N_2$. Then $u,v\in N_2$, so $u+v\in N_2$ and $au\in N_2$ because $N_2$ is a linear subspace. Hence $u+v,au\in U$. Thus $U$ is a linear subspace.

With the same choice of $N_2$ as in step 1.2, one has $G(u+v)=g_2(u+v)=g_2(u)+g_2(v)=G(u)+G(v),$ and $G(au)=g_2(au)=a{g}_2(u)=aG(u),$ by [L1]. Therefore $G$ is linear.

If $u\in U$, choose $(N,g)\in \mathcal{C}$ with $u\in N$. Then $G(u)=g(u)\le p(u)$ by the defining property of the chain element, so $G$ is dominated by $p$. If every chain element extends the same $f:M\to \mathbb{R}$, then every $m\in M$ lies in each domain and all values there equal $f(m)$, so $G{\mathrm{\mid }}_M=f$.

Let $\mathcal{E}$ be the set of all pairs $(N,g)$ such that $M\subseteq N\subseteq X$, the set $N$ is a linear subspace of $X$, the map $g:N\to \mathbb{R}$ is linear, $g{\mathrm{\mid }}_M=f$, and $g\le p$ on $N$. Order $\mathcal{E}$ by extension: $(N_1,g_1)⪯(N_2,g_2)⟺N_1\subseteq N_2\text{ and }{g}_2{\mathrm{\mid }}_{N_1}=g_1\mathrm{.}$ The pair $(M,f)$ lies in $\mathcal{E}$, so this poset is nonempty.

Let $\mathcal{C}\subseteq \mathcal{E}$ be a chain. If $\mathcal{C}=\mathrm{\varnothing }$, then $(M,f)$ is an upper bound for it. If $\mathcal{C}\mathrm{\ne }\mathrm{\varnothing }$, [L2] applies to the union of its domains and yields a well-defined linear functional dominated by $p$; because every chain element extends $f$, that union functional still extends $f$. Hence every chain in $\mathcal{E}$ has an upper bound in $\mathcal{E}$.

By [L3], choose a maximal element $(N,F)$ of $\mathcal{E}$. If $N\mathrm{\ne }X$, choose $x_0\in X\setminus N$. Applying [L1] to the dominated functional $F$ on the subspace $N$ produces a dominated linear extension $\tilde{F}$ on $N+\mathbb{R}{x}_0$. Then $(N,F)\prec (N+\mathbb{R}{x}_0,\tilde{F})$, contradicting maximality. Therefore $N=X$.

Since the maximal domain is all of $X$, the corresponding functional $F$ is the required dominated extension of $f$.

Define $p:X\to \mathbb{R}$ by $p(x):=\mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ The triangle inequality and real homogeneity of the norm make $p$ sublinear. For $m\in M$, [L3] and [L4] give $f_0(m)\le \mathrm{\mid }{f}_0(m)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }m\mathrm{\parallel }=p(m),$ so $f_0\le p$ on $M$.

By [L1], there exists a linear functional $F:X\to \mathbb{R}$ extending $f_0$ and satisfying $F(x)\le p(x)$ for every $x\in X$.

Apply step 2.1 to $x$ and to $-x$. Since $p(-x)=p(x)$, one gets $-\mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le F(x)\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel },$ hence $\mathrm{\mid }F(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }(x\in X)\mathrm{.}$ So $F$ is bounded in the sense of [L2], and [L3] gives $\mathrm{\parallel }F\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Since $F{\mathrm{\mid }}_M=f_0$, for every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$ one has $\mathrm{\mid }{f}_0(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$. Taking the supremum over the unit ball of the normed subspace $M$ and using [L3] and [L4] yields $\mathrm{\parallel }{f}_0\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$.

Steps 3.1 and 3.2 give $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }$, so $F$ is the required norm-preserving extension.

Let $u:=\mathrm{Re}f$. For $x,y\in X$ and $a\in \mathbb{R}$, $u(x+y)=\mathrm{Re}(f(x+y))=\mathrm{Re}f(x)+\mathrm{Re}f(y),$ and $u(ax)=\mathrm{Re}(f(ax))=\mathrm{Re}(af(x))=au(x)\mathrm{.}$ So $u$ is real linear.

Write $f(x)=a+ib$ with $a,b\in \mathbb{R}$. Since $f$ is complex linear, $f(ix)=if(x)=ia-b,$ so $\mathrm{Re}f(x)=a$ and $\mathrm{Re}f(ix)=-b$. Therefore $u(x)-iu(ix)=a-i(-b)=a+ib=f(x)\mathrm{.}$

Conversely, let $u$ be real linear and define $g(x):=u(x)-iu(ix)$. Additivity is immediate from real linearity of $u$. Also $g(ix)=u(ix)-iu(i^{2}x)=u(ix)+iu(x)=i(u(x)-iu(ix))=ig(x)\mathrm{.}$ Now for $\lambda =a+ib\in \mathbb{C}$ with $a,b\in \mathbb{R}$, $g(\lambda x)=g(ax+bix)=ag(x)+bg(ix)=ag(x)+big(x)=\lambda g(x)\mathrm{.}$ Hence $g$ is complex linear.

Taking real parts in the definition of $g$ gives $\mathrm{Re}g(x)=u(x)$ for every $x$. Together with step 1.2, this shows that a complex linear functional is uniquely determined by its real part.

Let $u:=\mathrm{Re}{f}_0$. By [L2], $u$ is real linear on the underlying real subspace $M$. Also $\mathrm{\mid }u(m)\mathrm{\mid }\le \mathrm{\mid }{f}_0(m)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }m\mathrm{\parallel }(m\in M),$ so $u$ is bounded with $\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Apply [L1] to the underlying real normed spaces. This yields a bounded real linear functional $U:X\to \mathbb{R}$ extending $u$ and satisfying $\mathrm{\parallel }U\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$. Define $F(x):=U(x)-iU(ix)(x\in X)\mathrm{.}$ By [L2], $F$ is complex linear and $\mathrm{Re}F=U$.

For $m\in M$, the equality $U{\mathrm{\mid }}_M=u$ and [L2] give $F(m)=u(m)-iu(im)=f_0(m)\mathrm{.}$ So $F$ extends $f_0$.

Fix $x\in X$. Choose $\theta \in \mathbb{R}$ so that $e^{i\theta }F(x)=\mathrm{\mid }F(x)\mathrm{\mid }$ is a nonnegative real number. Since $\mathrm{Re}F=U$ and $F$ is complex linear, $\mathrm{\mid }F(x)\mathrm{\mid }=\mathrm{Re}(e^{i\theta }F(x))=\mathrm{Re}F(e^{i\theta }x)=U(e^{i\theta }x)\mathrm{.}$ Therefore $\mathrm{\mid }F(x)\mathrm{\mid }\le \mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }{e}^{i\theta }x\mathrm{\parallel }=\mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel },$ where the last inequality uses step 1.1 and $\mathrm{\parallel }U\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$. Hence $\mathrm{\parallel }F\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Since $F{\mathrm{\mid }}_M=f_0$, every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$ satisfies $\mathrm{\mid }{f}_0(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$. Taking the supremum over the unit ball of $M$ gives $\mathrm{\parallel }{f}_0\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$. Combined with step 3.2, this yields $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Let $M:=\mathrm{span}\{x\}$. Define $f_0:M\to \mathbb{R}$ or $f_0:M\to \mathbb{C}$ by $f_0(\lambda x):=\lambda \mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ Because $x\mathrm{\ne }0$, each vector of $M$ has a unique representation $\lambda x$. Moreover, $\mathrm{\mid }{f}_0(\lambda x)\mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }\lambda x\mathrm{\parallel },$ so $f_0$ is bounded and $\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$.

If the scalar field is $\mathbb{R}$, [L2] extends $f_0$ to a bounded real linear functional $f$ on $X$ with $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$. Since $x=1\cdot x$, $f(x)=f_0(x)=\mathrm{\parallel }x\mathrm{\parallel }$.

If the scalar field is $\mathbb{C}$, [L3] extends $f_0$ to a bounded complex linear functional $f$ on $X$ with $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$. Again $f(x)=f_0(x)=\mathrm{\parallel }x\mathrm{\parallel }$.

In either scalar case, the extension produced in step 2.1 or step 2.2 is a bounded linear functional on $X$, hence an element of $X^{\ast }$ by [L1], with norm $1$ and value $\mathrm{\parallel }x\mathrm{\parallel }$ at $x$.

Since $x\mathrm{\ne }y$, the vector $x-y$ is nonzero.

By [L1], there exists $f\in X^{\ast }$ with $f(x-y)=\mathrm{\parallel }x-y\mathrm{\parallel }\mathrm{.}$ In particular $f(x-y)\mathrm{\ne }0$.

If $f(x)=f(y)$, then linearity would give $f(x-y)=0$, contradicting step 2.1. Therefore $f(x)\mathrm{\ne }f(y)$, so the dual space separates $x$ and $y$.

If $f\in X^{\ast }$ and $\mathrm{\parallel }f\mathrm{\parallel }\le 1$, then [L1] gives $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ So the displayed supremum is at most $\mathrm{\parallel }x\mathrm{\parallel }$.

If $x=0$, step 1.1 already shows that the supremum is $0$, so the formula holds. Assume now that $x\mathrm{\ne }0$. By [L2], choose $g\in X^{\ast }$ with $\mathrm{\parallel }g\mathrm{\parallel }=1$ and $g(x)=\mathrm{\parallel }x\mathrm{\parallel }$. Then the displayed supremum is at least $\mathrm{\mid }g(x)\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }$.

Step 1.1 gives an upper bound of $\mathrm{\parallel }x\mathrm{\parallel }$, and step 2.1 gives equality. So $\mathrm{\parallel }x\mathrm{\parallel }=\sup \{\mathrm{\mid }f(x)\mathrm{\mid }:f\in X^{\ast },\mathrm{\parallel }f\mathrm{\parallel }\le 1\}\mathrm{.}$

If the scalar field is $\mathbb{R}$, then [L1] applies exactly as stated to the given subspace $M$ and produces a norm-preserving extension of $f_0$ to $X$.

If the scalar field is $\mathbb{C}$, then [L2] applies exactly as stated to the given subspace $M$ and produces a norm-preserving extension of $f_0$ to $X$.

Neither step 1.1 nor step 1.2 uses or requires that $M$ be closed; each cited theorem assumes only that $M$ is a linear subspace. Therefore every bounded linear functional on an arbitrary subspace extends with the same norm.