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The Hahn-Banach theorem for bounded linear functionals says that a bounded linear functional on any linear subspace of a normed vector space extends to the whole space with its norm unchanged. No closedness or completeness of the subspace is required. Question 1 uses real-valued ${\ell }_{\infty }(\Gamma )$, so its operator assertions are read over the real scalar field. The analogous complex statements use complex-valued indexed functions.

For $x\ne 0$, define $g(tx)=t\Vert x\Vert$ on its one-dimensional span. This is a norm-one functional, so Hahn-Banach theorem supplies an extension $f\in X^{\ast }$ withThe canonical embedding into the bidual is $Jx(f)=f(x)$. It is linear and $\Vert Jx\Vert \le \Vert x\Vert$. The bounded linear functional just constructed gives the reverse inequality for nonzero $x$, and the zero case is immediate. Thus $\boxed{\Vert Jx\Vert =\Vert x\Vert }$ and $J$ is injective.

For the coordinate functional representation of an operator into bounded indexed functions, let $e_{\gamma }$ evaluate a coordinate and put $f_{\gamma }=e_{\gamma }T$. If $T$ is bounded and linear, then $f_{\gamma }\in X^{\ast }$ and $\Vert f_{\gamma }\Vert \le \Vert T\Vert$. Conversely, if $M={\sup }_{\gamma }\Vert f_{\gamma }\Vert <\infty$, the formula $(Tx)(\gamma )=f_{\gamma }(x)$ defines a bounded scalar function for each $x$, is linear, and satisfies $\Vert Tx{\Vert }_{\infty }\le M\Vert x\Vert$. Combining the two estimates givesFor an empty index set both spaces/families have norm bound zero; the supremum of the empty nonnegative family is taken as zero.

Choose $\Gamma =B_{X^{\ast }}$ and $Tx=(f(x){)}_{f\in B_{X^{\ast }}}$. The Hahn-Banach theorem makes $\Vert Tx{\Vert }_{\infty }=\Vert x\Vert$, so this is a linear isometric embedding into bounded scalar functions on an index set.

If $X$ is nonzero and separable, choose a dense sequence $u_n$ in its unit sphere and supporting functionals $f_n$ with $\Vert f_n\Vert =f_n(u_n)=1$. For a unit vector $u$ and $ϵ>0$, some $u_n$ satisfies $\Vert u-u_n\Vert <ϵ$, whence $|f_n(u)|\ge 1-ϵ$. Thus $(f_n)$ is a countable norming family and $x\mapsto (f_n(x))$ is an isometry into the l-infinity sequence space. Completeness of $X$ is not needed.

If instead $X=E^{\ast }$ for a separable normed vector space $E$, choose a dense sequence $v_n$ in $B_E$. The evaluations $f_n(x)=x(v_n)$ lie in $B_{X^{\ast }}$ and continuity of $x\in E^{\ast }$ gives ${\sup }_n|x(v_n)|=\Vert x\Vert$. This again gives $\Gamma =\mathbb{N}$, even when $E^{\ast }$ is not separable. If $X=\{0\}$, use zero coordinates throughout.

To prove 1-injectivity, take $T:Y\to {\ell }_{\infty }(\Gamma )$ on a subspace of $Z$. Extend every coordinate $f_{\gamma }\in Y^{\ast }$ to ${\tilde{f}}_{\gamma }\in Z^{\ast }$ by Hahn-Banach theorem, keeping its norm. Their uniform bound defines $\tilde{T}z=({\tilde{f}}_{\gamma }(z))$, and the coordinate norm identity yieldsSelection of these extensions uses the usual choice convention for arbitrary index sets. This verifies the lambda-injective normed space definition with $\lambda =1$.

For the final retraction characterization of lambda-injectivity, suppose first that $X$ is lambda-injective and $i:X\to Z$ is a linear isometry. The inverse $i(X)\to X$ has norm one when $X\ne 0$; extend it to $P:Z\to X$ with $\Vert P\Vert \le \lambda$. Then $Pi=I_X$, and $iP$ is a bounded projection onto $i(X)$. For the zero space take $P=0$.

Conversely, suppose every linear isometry out of $X$ has such a left inverse. Fix an isometry $j:X\to {\ell }_{\infty }(\Gamma )$ as above and a left inverse $P$ with $\Vert P\Vert \le \lambda$. For any $T:Y\to X$, extend $jT$ to $S:Z\to {\ell }_{\infty }(\Gamma )$ by 1-injectivity. Then $\tilde{T}=PS$ extends $T$, and $\Vert \tilde{T}\Vert \le \lambda \Vert S\Vert =\lambda \Vert T\Vert$. Therefore $X$ is lambda-injective exactly when every isometric embedding $i:X\to Z$ admits a bounded map $P:Z\to X$ satisfying

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

A Banach space is separable when it has a countable dense subset for its norm topology. It then has a countable base of norm balls and a countable norming family. Its continuous dual space need not be norm separable.

The weak topology and norm topology of a separable Banach space generate the same Borel sigma-algebra. A countable norming family makes every norm ball weakly Borel measurable, and separability makes every norm-open set a countable union of such balls. The reverse inclusion follows because the weak topology is coarser.