Quantitative Bishop–Phelps support functional construction

Choose a real number $\eta$ with $0<\eta <\epsilon$. The restriction $F=f{\mathrm{\mid }}_C$ is continuous and bounded above because $C$ is bounded. By [F1], $C$ with its norm metric is complete. Fix one $x_0\in C$, possible because $C$ is nonempty, and for $x\in C$ define $S(x)=\{y\in C:F(y)\ge F(x)+\eta \mathrm{\parallel }y-x\mathrm{\parallel }\}\mathrm{.}$ Each $S(x)$ is nonempty because it contains $x$, and it is closed because $F(y)-F(x)-\eta \mathrm{\parallel }y-x\mathrm{\parallel }$ is continuous in $y$.

If $y\in S(x)$ and $z\in S(y)$, then $F(z)\ge F(y)+\eta \mathrm{\parallel }z-y\mathrm{\parallel }\ge F(x)+\eta (\mathrm{\parallel }y-x\mathrm{\parallel }+\mathrm{\parallel }z-y\mathrm{\parallel })\ge F(x)+\eta \mathrm{\parallel }z-x\mathrm{\parallel }\mathrm{.}$ Thus $z\in S(x)$ and $S(y)\subseteq S(x)$. Since $F$ is bounded above and $S(x)$ is nonempty, its supremum is a real number. For every $n$ and every $x\in C$, the defining approximation property of the supremum supplies $y\in S(x)$ with $F(y)>{\sup }_{z\in S(x)}F(z)-2^{-n}\mathrm{.}$

Let a state be a nonempty finite sequence $(x_0,\dots ,x_n)$ in $C$ which starts at the fixed $x_0$ and, at each earlier index $k<n$, has $x_{k+1}\in S(x_k)$ and $F(x_{k+1})>{\sup }_{z\in S(x_k)}F(z)-2^{-k}\mathrm{.}$ Relate each state to every valid one-term extension. Step 2.1 proves that this relation is entire, so one application of DC gives a compatible infinite sequence $(x_n)$ with both displayed properties for every $n$.

The ascent condition gives $\eta \mathrm{\parallel }{x}_{n+1}-x_n\mathrm{\parallel }\le F(x_{n+1})-F(x_n)\mathrm{.}$ Consequently the partial sums of ${\sum }_n\mathrm{\parallel }{x}_{n+1}-x_n\mathrm{\parallel }$ are nondecreasing and bounded above by $({\sup }_{C}F-F(x_0))\mathrm{/}\eta$. By [F3] the series converges, hence its tails tend to zero and $(x_n)$ is Cauchy. Completeness gives a limit $v\in C$.

Transitivity in step 2.1 makes every tail point $x_k$, $k\ge n$, belong to $S(x_n)$; closedness gives $v\in S(x_n)$ for every $n$. If $z\in S(v)$, then transitivity also puts $z$ in every $S(x_n)$, and the approximate-supremum condition gives $F(z)<F(x_{n+1})+2^{-n}\mathrm{.}$ Letting $n\to \mathrm{\infty }$ and using continuity yields $F(z)\le F(v)$. But $z\in S(v)$ also gives $F(z)\ge F(v)+\eta \mathrm{\parallel }z-v\mathrm{\parallel }$, so $z=v$. Therefore $F(c)<F(v)+\eta \mathrm{\parallel }c-v\mathrm{\parallel }(c\in C,c\mathrm{\ne }v),$ and the corresponding non-strict inequality holds for every $c\in C$.

Put $D=\{t(c-v):t\ge 0,c\in C\}\mathrm{.}$ Because $C-v$ is convex and contains zero, $D$ is a convex cone: for $t,s\ge 0$ the sum $t(c-v)+s(c^{\mathrm{\prime }}-v)$ is zero if $t+s=0$, and otherwise equals $(t+s)(\frac{t}{t+s}c+\frac{s}{t+s}{c}^{\mathrm{\prime }}-v)$. Step 5.1 and real linearity give $f(d)\le \eta \mathrm{\parallel }d\mathrm{\parallel }(d\in D)\mathrm{.}$ This includes $d=0$.

For $x\in X$ define $p(x)={\inf }_{d\in D}(\eta \mathrm{\parallel }x+d\mathrm{\parallel }-f(d))\mathrm{.}$ The set being infimized is nonempty because $0\in D$. Step 6.1 and the reverse triangle inequality give every one of its terms at least $-\eta \mathrm{\parallel }x\mathrm{\parallel }$, while $d=0$ gives a term equal to $\eta \mathrm{\parallel }x\mathrm{\parallel }$. Thus [F3] makes $p(x)$ a finite real and -\eta\|x\|\le p(x)\le\eta\|x\|.\tag{1} Both bounds are uniform in the choice of $d$.

The cone identities imply $p(tx)=tp(x)$ for $t>0$, and (1) gives $p(0)=0$. If $d_1,d_2\in D$, then $d_1+d_2\in D$ and $\eta \mathrm{\parallel }x+y+d_1+d_2\mathrm{\parallel }-f(d_1+d_2)\le \eta \mathrm{\parallel }x+d_1\mathrm{\parallel }-f(d_1)+\eta \mathrm{\parallel }y+d_2\mathrm{\parallel }-f(d_2)\mathrm{.}$ Taking infima first over $d_1,d_2$ and then over $D$ proves $p(x+y)\le p(x)+p(y)$. Hence $p$ is sublinear.

Apply [F4] to the zero functional on $\{0\}$, dominated by $p$, to obtain a real-linear $h:X\to \mathbb{R}$ with $h(x)\le p(x)$. Applying this at $x$ and $-x$ and using (1) gives $\mathrm{\mid }h(x)\mathrm{\mid }\le \eta \mathrm{\parallel }x\mathrm{\parallel }$, so $h\in X^{\ast }$ and $\mathrm{\parallel }h\mathrm{\parallel }\le \eta$ by [F5]. For $d\in D$, the candidate $d$ in the infimum gives $p(-d)\le -f(d)$, whence $-h(d)=h(-d)\le p(-d)\le -f(d)$ and $h(d)\ge f(d)$. Thus the extension dominates $f$ on the support cone.

Set $g=f-h$. Then $g\in X^{\ast }$ and $\mathrm{\parallel }g-f\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }\le \eta <\epsilon$. For every $c\in C$, the vector $c-v$ lies in $D$, so step 9.1 gives $g(c)-g(v)=f(c-v)-h(c-v)\le 0.$ Thus $g$ attains its supremum on $C$ at $v$. The proof permits a singleton $C$, $f=0$, $X=\{0\}$ and the closed-boundary cases; DC is used only in step 3.1 and HB only in step 9.1.