Milman–Pettis theorem

We first transfer uniform convexity to $X^{\ast \ast }$. Fix $\epsilon \in (0,2]$ and take from [F1] a number $\delta >0$ for the separation threshold $\epsilon \mathrm{/}2$. Let $U,V\in B_{X^{\ast \ast }}$ satisfy $\mathrm{\parallel }U-V\mathrm{\parallel }\ge \epsilon$. By [F3], choose $f_0\in B_{X^{\ast }}$ with $\mathrm{\mid }(U-V)(f_0)\mathrm{\mid }>3\epsilon \mathrm{/}4$. In the complex case multiply $f_0$ by a scalar of modulus one, and in the real case change its sign if necessary, to obtain $f\in B_{X^{\ast }}$ with $\mathrm{Re}(U-V)(f)>3\epsilon \mathrm{/}4$.

Fix an arbitrary $g\in B_{X^{\ast }}$ and $\eta >0$, and put $m:=\mathrm{Re}(U-V)(f)-\epsilon \mathrm{/}2>0$ and $\tau :=\min \{m\mathrm{/}4,\eta \}>0$. Apply [F2] separately to $U$ and $V$, each time with the two tests $f,g$ and tolerance $\tau$, obtaining $x,y\in B_X$. Then $\mathrm{Re}f(x-y)>\mathrm{Re}(U-V)(f)-2\tau >\epsilon \mathrm{/}2$, so $\mathrm{\parallel }x-y\mathrm{\parallel }>\epsilon \mathrm{/}2$. By [F1], $\mathrm{\parallel }(x+y)\mathrm{/}2\mathrm{\parallel }\le 1-\delta$. Approximation at $g$ therefore gives $\mid \frac{(U+V)(g)}{2}\mid <\mid \frac{g(x)+g(y)}{2}\mid +\tau \le 1-\delta +\eta \mathrm{.}$ If the left side exceeded $1-\delta$, taking $\eta$ to be half that positive gap would contradict this inequality. Hence $\mathrm{\mid }(U+V)(g)\mathrm{/}2\mathrm{\mid }\le 1-\delta$. Taking the supremum over $g\in B_{X^{\ast }}$ by [F3] yields $\mathrm{\parallel }(U+V)\mathrm{/}2\mathrm{\parallel }\le 1-\delta$.

Thus $X^{\ast \ast }$, with its given dual norm, is uniformly convex: the modulus at $\epsilon$ may be taken to be any modulus of $X$ at $\epsilon \mathrm{/}2$. Notice that step 2.1 used two finite-data witnesses only after $U,V,f,g,\eta$ were fixed; it selected no sequence or family of witnesses.

Let $z\in X^{\ast \ast }$ have norm one and let $\rho >0$. Put $\epsilon :=\min \{1,\rho \}\in (0,1]$ and let $\delta >0$ be the bidual modulus established in step 3.1. Set $\gamma :=\min \{\delta \mathrm{/}4,1\mathrm{/}4\}$. By [F3] choose $f_0\in B_{X^{\ast }}$ with $\mathrm{\mid }z(f_0)\mathrm{\mid }>1-\gamma$, and rotate or change its sign to get $f\in B_{X^{\ast }}$ with $\mathrm{Re}z(f)>1-\gamma$. By [F2], choose $x\in B_X$ with $\mathrm{\mid }f(x)-z(f)\mathrm{\mid }<\gamma$. Hence $\mathrm{Re}f(x)>1-2\gamma$, and $\parallel \frac{z+J_{X}x}{2}\parallel \ge \mathrm{Re}\frac{z(f)+f(x)}{2}>1-\frac{3\gamma }{2}>1-\delta \mathrm{.}$ The contrapositive of the bidual uniform-convexity estimate gives $\mathrm{\parallel }z-J_{X}x\mathrm{\parallel }<\epsilon \le \rho$.

It follows that $J_X(B_X)$ is norm dense in $B_{X^{\ast \ast }}$. Indeed, the zero vector is $J_{X}0$. For nonzero $w\in B_{X^{\ast \ast }}$ and a prescribed $\rho >0$, apply step 4.1 to $z=w\mathrm{/}\mathrm{\parallel }w\mathrm{\parallel }$ with tolerance $\rho \mathrm{/}\mathrm{\parallel }w\mathrm{\parallel }$, obtaining $x\in B_X$; then $\mathrm{\parallel }w-J_X(\mathrm{\parallel }w\mathrm{\parallel }x)\mathrm{\parallel }<\rho$ and $\mathrm{\parallel }w\mathrm{\parallel }x\in B_X$.

By [F4], $J_X$ is an isometry, so its range $J_X(X)$ is a normed subspace isometric to the complete space $X$. Under the assumed ${\mathrm{A}\mathrm{C}}_{\omega }$, [F5] makes this range norm closed in $X^{\ast \ast }$. Step 5.1 puts every element of $B_{X^{\ast \ast }}$ in its norm closure and hence in the range. Scaling then gives $J_X(X)=X^{\ast \ast }$: the zero element is already in the range, and a nonzero element is its norm times an element of the bidual unit sphere. Thus $J_X$ is surjective, and [F6] says that $X$ is reflexive.

HB is used exactly in [F2] for Goldstine finite-data approximation and in [F4] for the canonical isometry. Countable Choice is used exactly through the complete-subspace closedness statement [F5]. No compactness theorem and no ultrafilter principle occurs. If $X=\{0\}$, then $X^{\ast \ast }=\{0\}$ and step 6.1 is immediate. All scalar inequalities use real parts, so the proof covers both real and complex scalars.