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A Banach space $X$ is a Grothendieck space when every weak-star convergent sequence in $X^{\ast }$ is weakly convergent. Every reflexive Banach space has this property, and every separable Grothendieck space is reflexive.

A nonempty norm-closed convex subset $C$ of a reflexive Banach space attains its distance to zero. For $d={\inf }_C\Vert x\Vert$, the sets $C\cap (d+1/n)B_X$ are nonempty weakly closed subsets of one weakly compact ball, and form a decreasing family. Compactness gives a point in their intersection, of norm $d$. Mazur theorem supplies weak closedness of $C$ and the balls; no sequential-compactness theorem is required.

For $K\ge 1$, define the normalized ${\ell }^1$ treeThe upper bound follows from normalization. This is a closed tree: it suffices to impose the lower inequalities for a countable dense set of scalar tuples. The Bourgain l1-index iswhere $o$ is its well-founded tree order, and the index is ${\omega }_1$ if a tree is ill-founded. An infinite branch is exactly a sequence equivalent to the unit vector basis of ${\ell }^1$.

If $X$ is separable and contains no ${\ell }^1$, its unit sphere is Polish and every $T_K(X)$ is closed and well-founded. Part (a) makes each order countable, and their countable supremum is countable. In particular every separable reflexive Banach space has countable index, since its closed subspaces are reflexive whereas ${\ell }^1$ is not.

We now construct separable reflexive spaces with arbitrarily high index. Take any countable well-founded rooted tree $T$. Recursively, at each node $t$, defineAt a terminal node this is just $\mathbb{R}$. The recursion is legitimate by well-foundedness. Each $X_t$ is separable and reflexive: a countable ${\ell }^2$ sum of separable reflexive spaces is separable and reflexive, and adjoining one dimension in a finite direct sum preserves reflexivity. Let $X_T$ be the root space, and let $e_t$ be the unit vector at each node coordinate.

Along every finite branch $t_1\prec \cdots \prec t_n$, the recursive norm is exactlyIndeed, at every step only one child component is occupied, so its ${\ell }^2$ norm equals that component's norm, while the node coordinate adds by the ${\ell }^1$ sum. Thus $T$ embeds by extension into $T_1(X_T)$, and its rank is a lower bound for the index, up to the harmless empty-root convention. Countable well-founded trees have arbitrarily high countable ranks: add a root for a successor stage and attach countably many trees of cofinal ranks at a limit stage.

Suppose now a separable reflexive space $U$ were universal. Put $\beta =I_1(U)<{\omega }_1$ and choose $T$ of height larger than $\beta$. An isomorphic embedding $J:X_T\to U$ has bounds $c\Vert x\Vert \le \Vert Jx\Vert \le C\Vert x\Vert$. The normalized images $J{e}_t/\Vert J{e}_t\Vert$ along each branch satisfy the lower ${\ell }^1$ estimate $c/C$, so the tree embeds into $T_K(U)$ for some integer $K\ge C/c$. It would have order greater than $\beta$, contradicting its definition. Therefore no separable reflexive Banach space is universal for all separable reflexive Banach spaces. The embedding constant need not be uniform across spaces; the supremum over all $K$ handles this.

Symmetry implies that all eigenvalues of $A(x)$ lie between $\lambda$ and $\Lambda$, so $\Vert A(x){\Vert }_{\mathrm{op}}\le \Lambda$. In particular the bilinear form $B$ from part (i) is bounded and symmetric under the printed assumptions. The symmetric elliptic Dirichlet energy with a nonpositive potential isAll its terms are finite. For an admissible variation $\phi \in H_0^1$, its first variation is $B(u,\phi )+\int f\phi$. Thus stationarity is exactly the weak formulation of $Lu=f$.

Here is the direct method in the calculus of variations. Put $u=\psi +w$. Expanding the symmetric quadratic form givesThe Poincaré inequality controls $\Vert w{\Vert }_2$ by $\Vert Dw{\Vert }_2$. Therefore $J$ is bounded below on the admissible class, and a minimizing sequence is bounded in $H^1$. By weak compactness in this reflexive Banach space, a subsequence converges weakly to $u$. The affine closed subspace $\psi +H_0^1$ is weakly closed, so $u$ remains admissible.

The symmetric square root $A^{1/2}(x)$ is measurable and bounded. Since $-q\ge 0$, the two bounded linear mapstake weak $H^1$ convergence to weak $L^2$ convergence. The weak lower semicontinuity of the Hilbert norm shows that $B(u,u)$ is weakly lower semicontinuous. The term $\int fu$ is weakly continuous. Hence $J(u)\le \mathrm{liminf}J(u_k)$ and $u$ attains the minimum. Varying in either sign yields the weak equation.

The sign $q\le 0$ was used both to make the potential energy nonnegative, hence weakly lower semicontinuous by this argument, and to obtain coercivity. Finally $J(u+z)-J(u)=B(z,z)/2\ge \lambda \Vert Dz{\Vert }_2^2/2$ for $z\in H_0^1$, since the first variation at $u$ vanishes. Therefore

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Weak compactness of the adjoint forces weak compactness of the original operator. Assume $T^{\ast }:Y^{\ast }\to X^{\ast }$ is weakly compact. Apply the already proved $(i)⟹(ii)$ to this operator, rather than assuming the adjoint equivalence in advance. It givesTake $\psi \in Y^{\ast \ast \ast }$ annihilating $J_{Y}Y$, so $\psi (J_{Y}y)=0$ for every $y\in Y$. Write $T^{\ast \ast \ast }\psi =J_{X^{\ast }}f$ for some $f\in X^{\ast }$. For every $x\in X$,Hence $f=0$ and $T^{\ast \ast \ast }\psi =0$. It follows thatfor every $\psi$ annihilating $J_{Y}Y$.

The vector subspace $J_{Y}Y$ is closed in the norm topology in $Y^{\ast \ast }$ because $Y$ is complete and $J_Y$ is an isometry. The Hahn-Banach theorem says that any point outside a closed linear subspace can be separated from it by a bounded linear functional vanishing on that subspace. Applying this in $Y^{\ast \ast }$ shows that an element annihilated by all such $\psi \in Y^{\ast \ast \ast }$ must belong to $J_{Y}Y$. Therefore $T^{\ast \ast }F\in J_{Y}Y$ for every $F$, proving $(iv)⟹(ii)$. The reverse implications above now establish all four equivalences, including weak compactness of an operator and its adjoint:

Finally, if $X$ is a reflexive Banach space, every $F\in X^{\ast \ast }$ is $J_{X}x$ for some $x$, and $T^{\ast \ast }F=J_{Y}Tx\in J_{Y}Y$. If $Y$ is a reflexive Banach space, $Y^{\ast \ast }=J_{Y}Y$, so the same range condition holds automatically. In either case, $(ii)⟹(i)$ proves

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every bounded linear functional continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<ϵ,1\le j\le m\}$ for finite families in $X^{\ast }$.

Mazur theorem says that for any convex subset $C$ of a real or complex normed vector space,The norm closure is convex. If $x$ is outside it, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $a$ with $\mathrm{Re}f(x)>a\ge {\sup }_{y\in C}\mathrm{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the norm closure. The other inclusion follows because the weak topology is weaker than the norm topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.

The sequential formulation, Mazur lemma, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the norm closure of its convex hull. Select a finite convex combination of the $n$th tail at norm distance less than $1/n$ from $x$.

A weakly bounded set $D$ satisfies ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^{\ast }$. The completeness of the dual space holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the unit ball. The Uniform boundedness principle therefore gives ${\sup }_{x\in D}\Vert Jx\Vert <\infty$, and $\Vert Jx\Vert =\Vert x\Vert$ proves norm boundedness.

One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^{\ast }:{\sup }_{x\in D}|f(x)|\le n\}$ cover the Banach space $X^{\ast }$. The Baire category theorem makes some $F_n$ contain a ball $f_0+r{B}_{X^{\ast }}$, after shrinking the radius. For $\Vert g\Vert \le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so ${\sup }_D|g(x)|\le 2n$. Scaling and the dual norm formula yield ${\sup }_D\Vert x\Vert \le 2n/r$. A weakly compact set is weakly bounded because each functional has compact, hence bounded, image. Consequently it is norm bounded.

For a Banach space, let $J:X\to X^{\ast \ast }$ be its canonical isometry. We use two explicitly stated weak-star facts: Banach-Alaoglu theorem makes a dual unit ball weak-star compact, and Goldstine theorem makes $J(B_X)$ weak-star dense in $B_{X^{\ast \ast }}$. The weak topology on $X$ is carried by $J$ to $\sigma (X^{\ast \ast },X^{\ast })$ on its image, since their coordinates are the same evaluations $f(x)$.

If $X$ is a reflexive Banach space, $J(B_X)=B_{X^{\ast \ast }}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the Hausdorff space $X^{\ast \ast }$. Goldstine density then forces $J(B_X)=B_{X^{\ast \ast }}$, which implies $JX=X^{\ast \ast }$. This proves the weak compactness characterization of reflexivity:Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^{\ast }}$ separate points. If $K$ is empty there is nothing to prove; otherwise defineThe summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Thus $d$ induces precisely the weak topology on $K$, the countable separating family metrizes a weakly compact set result. No norm-density of the separating family in $X^{\ast }$ is claimed or required.

The direct method in the calculus of variations gives the following existence theorem. Suppose bounded sequences in the Banach space $\mathcal{U}$ have $\tau$-convergent subsequences, and $E$ is a proper extended-real function with coercivity and is $\tau$-sequentially lower semicontinuous. Then $E$ has a finite-valued global minimizer.

To prove this, properness makes $m=\inf E<+\infty$. First rule out $m=-\infty$: a sequence with $E(u_n)\to -\infty$ eventually lies in a fixed sublevel set, which is bounded by coercivity. A $\tau$-convergent subsequence would then have a limit $u$ with $E(u)\le -\infty$, contradicting the codomain. Hence $m$ is finite.

Choose a minimizing sequence with $E(u_n)\to m$. It eventually belongs to the bounded sublevel set $\{E\le m+1\}$. Extract $u_{n_j}\stackrel{\tau }{\to }\bar{u}$. By sequential lower semicontinuity,Thus $E(\bar{u})=m$. In particular, a reflexive Banach space with the weak topology supplies the required subsequence property by weak sequential compactness of bounded sequences in a reflexive Banach space. A strictly convex function has at most one minimizer; this is an additional property, not part of the existence theorem.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.

For a Banach space $X$, weak convergence $u_k\rightharpoonup u$ meansFor its continuous dual space $X^{\ast }$, convergence in the weak-star topology meansWeak convergence on $X^{\ast }$ tests against all of $X^{\ast \ast }$, whereas weak-star convergence tests only against the canonical image of $X$. Thus weak convergence always implies weak-star convergence. If $X$ is a reflexive Banach space, that image is all of $X^{\ast \ast }$, so the weak and weak-star topologies on $X^{\ast }$ coincide. Weak-star topology is defined relative to a chosen predual; it is not an extra intrinsic topology on an arbitrary Banach space without such an identification.

Write $h(t)=t/(1+t)$ for $t\ge 0$. Every summand in $d$ is nonnegative and at most $2^{-m}$, so the series converges. Symmetry is immediate. If $d(x,y)=0$, then ${\varphi }_m(x-y)=0$ for every $m$. For arbitrary $\varphi \in ({\ell }^p{)}^{\ast }$, choose a sequence of ${\varphi }_m$ converging to $\varphi$ in operator norm. Since $x-y$ is fixed, $\varphi (x-y)=0$. The continuous dual space separates points, so $x=y$.

The function $h$ is increasing and subadditive:Combining this with the triangle inequality for each $|{\varphi }_m(x-y)|$ and summing proves the triangle inequality for $d$. Hence $d$ is a metric on $B$.

Suppose first that $x^{(n)}$ converges weakly to $x$, meaning thatFor each fixed $m$, the $m$th summand of $d(x^{(n)},x)$ tends to zero. A finite-head and geometric-tail estimate, using that every summand is bounded by $2^{-m}$, then gives $d(x^{(n)},x)\to 0$.

Conversely, if $d(x^{(n)},x)\to 0$, each nonnegative summand tends to zero, and thereforefor every $m$. Given $\varphi \in ({\ell }^p{)}^{\ast }$ and $\epsilon >0$, choose $m$ with $\Vert \varphi -{\varphi }_m\Vert <\epsilon$. Since $x^{(n)},x\in B$,Taking the limit superior and then letting $\epsilon \downarrow 0$ proves weak convergence. ThusThe same approximation argument for neighbourhoods shows that $d$ induces the weak topology on $B$, as in the metrization of the weak topology on a bounded set.

If $q$ is conjugate to $p$, the given duality of sequence spaces applied twice givesso ${\ell }^p$ is a reflexive Banach space. Its closed unit ball is weakly compact by the weak compactness characterization of reflexivity. Since $d$ induces that topology on $B$,

On $C_c^0({\mathbb{R}}^n)$ equipped with the $L^{\infty }$ norm, point evaluation is linear and satisfieswith equality for a compactly supported function equal to one near zero. Thus $\Vert \Lambda \Vert =1$. The Hahn-Banach theorem extends it, without increasing its norm, to a continuous functional $\tilde{\Lambda }$ on $L^{\infty }({\mathbb{R}}^n)$.

This extension cannot be represented by an $L^1$ function. If $h\in L^1$ satisfied $\tilde{\Lambda }(f)=\int fh$, choose continuous functions $0\le f_k\le 1$ with $f_k(0)=1$ and support in the ball of radius $1/k$. Thenby the absolute continuity of the Lebesgue integral, a contradiction. Therefore $(L^{\infty }{)}^{\prime }$ strictly contains the functionals represented by $L^1$; this is a singular functional on L infinity.

Finally, $L^p({\mathbb{R}}^n)$ is a reflexive Banach space exactly whenThe endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive.

The relation that $X$ is finitely representable in $Y$ means that every finite-dimensional subspace $E\subseteq X$ has, for every $\epsilon >0$, a linear copy in $Y$ of distortion below $1+\epsilon$. A superreflexive Banach space is one for which every Banach space finitely representable in it is a reflexive Banach space.

Suppose first that $X$ is reflexive and $(x_i)\subseteq B_X$. By weak compactness, the sequence has a weak cluster point $x\in B_X$. The point $x$ belongs to the weak closure of every tail convex hull, and the weak and norm closures of a convex set agree by the Hahn-Banach separation theorem. Choose a convex combinationwithin $\theta /2$ of $x$, enlarging $n$ beyond all indices used. Then choosewithin $\theta /2$ of $x$. It follows that $\Vert y-z\Vert <\theta$.

Conversely, the James reflexivity criterion in convex-block form says that a nonreflexive Banach space has a number $\theta >0$ and a sequence $(x_i)\subseteq B_X$ such thatfor every $n$. One obtains this form from the usual bidual separation proof by applying the principle of local reflexivity to each finite-dimensional stage. Such a sequence contradicts the asserted property. This proves the convex-block separation criterion for reflexivity.

Consider now the uniform finite version. If it fails for some $\theta >0$, choose for every $N$ a sequence $(x_i^{(N)}{)}_{i=1}^N\subseteq B_X$ for which every cut has convex-hull distance at least $\theta$. In a free ultrapower $X_{\mathcal{U}}$, letfilling the finitely many missing coordinates arbitrarily. Every initial-tail pair of convex hulls of $({\xi }_i)$ remains at distance at least $\theta$, so the first criterion makes $X_{\mathcal{U}}$ nonreflexive. Since an ultrapower is finitely representable in $X$, this contradicts superreflexivity.

Conversely, if $X$ is not superreflexive, choose a nonreflexive space $Y$ finitely representable in $X$. The first criterion supplies a sequence in $B_Y$ whose convex blocks are separated by some $\theta >0$. For each $N$, transfer the span of its first $N$ vectors to $X$ with distortion arbitrarily close to one and normalize. The resulting $N$-term sequence violates the uniform condition, with separation at least, say, $\theta /2$. This proves the uniform finite convex-block criterion for superreflexivity.

A purely metric equivalent is the diamond-graph characterization of superreflexivity:For sufficiency, argue contrapositively. If $X$ is not superreflexive, the preceding uniform criterion supplies arbitrarily long unit-ball sequences whose convex hulls stay a fixed distance apart across every cut. At each replacement step in the diamond graph, map the two new branches to convex combinations on opposite sides of the corresponding cut. The upper bound follows from convexity and the lower bound from the fixed separation. The standard recursive diamond construction therefore embeds every $D_m$ into $X$ with one distortion constant independent of $m$. Thus divergence of the diamond distortions forces superreflexivity.

For $p\ge 2$, integrate the stated scalar inequality to obtain the Clarkson inequalityIf $f,g$ belong to the unit ball and $\Vert f-g{\Vert }_p\ge \epsilon$, thenThus $L^p[0,1]$ is uniformly convex.

Now let $X$ be uniformly convex. It is enough to show that every $\Phi \in S_{X^{\ast \ast }}$ lies in the canonical image of $X$. Given $\epsilon >0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^{\ast }}$ with $\Phi (f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, thenso $\Vert x-y\Vert <\epsilon$. By Goldstine theorem, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_{\alpha })$; completeness gives $x_{\alpha }\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is reflexive.

For $p\ge 2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p{)}^{\ast }=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.

A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the reflexive Banach space $X$, which is weakly compact. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.

The compact operator $T$ is weak-to-weak continuous, so $T{x}_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\Vert T{x}_{k_n}-Tz\Vert \ge \epsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence

Take a minimizing sequence $(X_k,Y_k)$ in the affine Sobolev class $W$. The standing bounds $C^{-1}\le X_k\le C$ make the energy uniformly equivalent toThe fixed boundary values and the Poincaré inequality therefore bound the sequence in $H^1(\Omega ;{\mathbb{R}}^2)$. By weak compactness in a reflexive Banach space, a subsequence converges weakly to $(X,Y)$. The assumed weak closedness keeps the limit in $W$, and the assumed weak lower semicontinuity givesThus the direct method in the calculus of variations produces a minimizer. Its first variation vanishes in every compactly supported direction, so part (b) makes it a weak solution of the system.

If $X$ is a reflexive Banach space, its closed unit ball is weakly compact. A bounded linear map is weak-to-weak continuous, so $T(B_X)$ is weakly compact in $Y$. Compact subsets of a Hausdorff space are closed, hence $T(B_X)$ is weakly closed and therefore norm closed.

Now suppose the two norms on $V$ have the same continuous dual as a set. Each $X_j^{\ast }$ is a Banach space in its dual norm. The identityhas closed graph: if $f_n\to f$ in the first dual norm and $f_n\to g$ in the second, evaluating at each $x\in V$ gives $f(x)=g(x)$. The closed graph theorem makes $I$ bounded, and the same argument for $I^{-1}$ makes the dual norms equivalent. Thus there are $c,C>0$ withThe dual formulafrom the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent.