Reflexivity and Eberlein Smulian — Examples

By [F2], the induced function is a norm. The assumed completeness and [F3] therefore make $H$ a real or complex Banach space. This also covers the zero Hilbert space.

For arbitrary $x,y\in H$, expand with [F1]: $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+⟨x,y⟩+⟨y,x⟩+\mathrm{\parallel }y{\mathrm{\parallel }}^2,$ while $\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2-⟨x,y⟩-⟨y,x⟩+\mathrm{\parallel }y{\mathrm{\parallel }}^2\mathrm{.}$ Adding cancels the two cross terms, over both scalar fields, and gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2+\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=2\mathrm{\parallel }x{\mathrm{\parallel }}^2+2\mathrm{\parallel }y{\mathrm{\parallel }}^2\mathrm{.}$

Now let $x,y$ lie in the closed unit ball and suppose $\mathrm{\parallel }x-y\mathrm{\parallel }\ge \epsilon$. By [F2] and step 1.2, ${\parallel \frac{x+y}{2}\parallel }^2=\frac{\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2}{2}-\frac{\mathrm{\parallel }x-y{\mathrm{\parallel }}^2}{4}\le 1-\frac{{\epsilon }^2}{4}\mathrm{.}$ The radicand $r=1-{\epsilon }^2\mathrm{/}4$ belongs to $[0,1)$ because $0<\epsilon \le 2$. Let $s=\sqrt{r}\ge 0$ as in [F4]. If $s\ge 1$, then either $s=1$ or strict monotonicity of squaring gives $s^2>1$, whereas $s^2=r<1$; hence $s<1$. Since both $\mathrm{\parallel }(x+y)\mathrm{/}2\mathrm{\parallel }$ and $s$ are nonnegative, their squared inequality and strict monotonicity of squaring give $\parallel \frac{x+y}{2}\parallel \le s=1-\delta (\epsilon )\mathrm{.}$ Thus $\delta (\epsilon )=1-s>0$. At $\epsilon =2$, $r=s=0$ and $\delta (2)=1$.

Step 2.1 applies to every closed-unit-ball pair satisfying the separation hypothesis and supplies a positive number depending only on $\epsilon$. Therefore [F3] proves that $H$ is uniformly convex. No choice principle is used: the square root in the displayed formula is unique by [F4].

The first assertion is exactly [F1]: under ${\mathrm{A}\mathrm{C}}_{\omega }$, both scalar versions of $L^p(\mu )$ are reflexive for every measure space and every $1<p<\mathrm{\infty }$. Empty and zero measure spaces are included; their $L^p$ spaces are zero and the cited theorem still applies.

Under the three additional principles stated for the endpoint clause, [F3] gives nonreflexivity of both real and complex ${\mathrm{\ell }}^1$. Those principles are used here only through that cited corollary.

The $c_0$ conclusion needs none of those additional principles. Fix $\mathbb{K}$. By [F4], $c_0(\mathbb{K})$ is a Banach space. Let $T:{\mathrm{\ell }}^1(\mathbb{K})\to c_0(\mathbb{K}{)}^{\ast }$ be the isometric bijection in [F4], so $T(a)(x)={\sum }_n{a}_n{x}_n$. Identify $({\mathrm{\ell }}^1(\mathbb{K}){)}^{\ast }$ with ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{K})$ by [F4], using the same bilinear series pairing. For $x\in c_0$ and $a\in {\mathrm{\ell }}^1$, $J_{c_0}(x)(T(a))=T(a)(x)={\sum }_n{a}_n{x}_n\mathrm{.}$ Thus, under the composite identification $c_0^{\ast \ast }\cong {\mathrm{\ell }}^{\mathrm{\infty }}$, the canonical image $J_{c_0}(x)$ is exactly the bounded sequence $x$. The constant sequence $\mathbf{1}=(1,1,\dots )$ belongs to ${\mathrm{\ell }}^{\mathrm{\infty }}$ but not to $c_0$ by [F4], so it is a concrete bidual element outside $J_{c_0}(c_0)$. Hence $J_{c_0}$ is not onto and [F4] proves that real and complex $c_0$ are not reflexive.

Give $\mathbb{N}$ counting measure. By [F2], the real or complex $L^p$ space in step 1.1 is isometrically the corresponding ${\mathrm{\ell }}^p$. Reflexivity therefore gives the asserted sequence-space specialization.

Steps 1.1 and 2.1 prove the positive result in the entire open exponent range, while steps 1.2, 1.3, and 2.1 provide the promised failures outside it. Countable choice is used through the arbitrary-measure $L^p$ theorem. The ultrafilter lemma, DC, and relative HB are additionally used only for the selected ${\mathrm{\ell }}^1$ proof; the direct canonical-image proof for $c_0$ is choice-free. No assertion about reflexivity of $L^1$ or $L^{\mathrm{\infty }}$ on every measure space is made.

Let $T:{\mathrm{\ell }}^1(\mathbb{K})\to c_0(\mathbb{K}{)}^{\ast }$ be the isometric bijection from [F2], so $T(a)(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$. Identify $({\mathrm{\ell }}^1(\mathbb{K}){)}^{\ast }$ with ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{K})$ through [F2]. Both identifications use this bilinear series pairing, including over $\mathbb{C}$.

For $x\in c_0$ and $a\in {\mathrm{\ell }}^1$, $J_{c_0}(x)(T(a))=T(a)(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n\mathrm{.}$ The functional on the right is represented, under the second identification in step 1.1, by the bounded sequence $x$ itself. Hence the composite $c_0\stackrel{J_{c_0}}{\to }{c}_0^{\ast \ast }\cong {\mathrm{\ell }}^{\mathrm{\infty }}$ is precisely the canonical inclusion $x\mapsto x$.

The constant sequence $\mathbf{1}=(1,1,\dots )$ lies in ${\mathrm{\ell }}^{\mathrm{\infty }}$, has norm one, and does not tend to zero. Thus [F3] gives $\mathbf{1}\mathrm{\notin }{c}_0$, so step 2.1 exhibits a concrete member of $c_0^{\ast \ast }$ outside the range of $J_{c_0}$.

The canonical map is not onto. Since $c_0(\mathbb{K})$ is Banach, [F1] therefore proves that it is not reflexive. The calculation covers both scalar fields, including their identical bilinear convention, and uses no choice principle.

Let $B=\{x\in X:\mathrm{\parallel }x{\mathrm{\parallel }}_1<1\}$. It is norm-open by [F2]. Suppose for contradiction that it is weakly open. Since $0\in B$, [F1] supplies a finite basic weak neighborhood $U={\bigcap }_{j=1}^m{f}_j^{-1}(V_j)$ with $0\in U\subseteq B$, where $f_j\in X^{\ast }$ and each scalar-open $V_j$ contains zero. The case $m=0$ means $U=X$ and already contradicts $U\subseteq B$, since $2e_0\mathrm{\notin }B$.

Assume $m\ge 1$. Let $E=\mathrm{span}\{e_0,\dots ,e_m\}$ and define the linear map $A:E⟶{\mathbb{K}}^m,A(x)=(f_1(x),\dots ,f_m(x))\mathrm{.}$ The coordinate vectors are linearly independent by their explicit coordinates, so $\dim E=m+1$, whereas $\mathrm{rank}A\le m$. Rank--nullity [F4] therefore gives a nonzero $v\in \ker A$.

For every scalar $\lambda$, each $f_j(\lambda v)=0\in V_j$, so $\lambda v\in U$. Since $v\mathrm{\ne }0$, choose the positive real scalar $\lambda =2\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_1$. Absolute homogeneity gives $\mathrm{\parallel }\lambda v{\mathrm{\parallel }}_1=2$, hence $\lambda v\mathrm{\notin }B$. This contradicts $U\subseteq B$.

Thus the norm-open ball $B$ is not weakly open. Since [F1] says the weak topology is contained in the norm topology, the containment is strict over both scalar fields.

Norm convergence implies weak convergence because the weak topology is coarser by [F1]. Conversely, [F5] turns every weakly convergent sequence in $X$ into a norm-convergent sequence. The two topologies therefore have exactly the same convergent sequences even though step 4.1 proves that they are different.

The witness uses $m+1$ explicit coordinate vectors for an arbitrary finite list of $m$ functionals, so it also covers one functional and a list containing zero or repeated functionals. The empty list was handled in step 1.1. Only finite-dimensional rank--nullity and one formula-defined rescaling are used; there is no choice principle, and no assertion about nets having the same convergence behavior is made.

If $f=0$, take $y=0$. Then $f(x)=⟨x,0⟩$ and $\mathrm{\parallel }f\mathrm{\parallel }=0=\mathrm{\mid }f(x)\mathrm{\mid }$ for every $x$ in the closed unit ball, including $x=0$ when $H=\{0\}$. Hence assume from now on that $f\mathrm{\ne }0$; in particular $H\mathrm{\ne }\{0\}$ and $\mathrm{\parallel }f\mathrm{\parallel }>0$.

Choose $w\in H$ with $f(w)\mathrm{\ne }0$ and put $v=w\mathrm{/}f(w)$, so $f(v)=1$. Let $M=\ker f$. It is a linear subspace, and the following estimate proves that it is closed and hence Banach.

The set $D=\{\mathrm{\parallel }v-u\mathrm{\parallel }:u\in M\}$ is nonempty (take $u=0$) and bounded below by $0$, so [F3] gives $d=\inf D$; the functional estimate below also proves that this infimum is positive.

For each natural $n$, define the following set of approximate minimizers and use Countable Choice to select from all of them.

Put $a_n=v-m_n$. Expanding squared norms and using that the kernel contains midpoints gives the estimate below.

The sequence $(m_n)$ is Cauchy by the following explicit use of the reciprocal bound in the estimate from step 4.1.

Since $M$ is Banach, $m_n\to m$ for some $m\in M$; the minimizing bounds and triangle inequality show that this limit realizes the infimum.

Set $z=v-m$. Then $f(z)=1$, and $z\mathrm{\ne }0$ because $\mathrm{\parallel }z\mathrm{\parallel }=d>0$; real variations, and then the $iu$ variation over the complex field, prove that $z$ is orthogonal to the kernel.

For arbitrary $x\in H$, subtracting $f(x)z$ puts the remainder in the kernel and yields the unique representing vector.

Cauchy–Schwarz supplies the upper bound, and evaluation at the normalized representing vector supplies equality and norm attainment.