Functional Analysis and Banach and C*-Algebras course of the University of Oxford (C4.1)

An isomorphic embedding of normed vector spaces is a bounded linear operator that is injective and whose inverse function on its image is bounded. Equivalently, $T$ is bounded below: there is $c>0$ such thatIndeed this inequality gives injectivity and $\Vert T^{-1}z\Vert \le c^{-1}\Vert z\Vert$ for $z\in T(X)$; the converse follows by applying the boundedness of $T^{-1}$ to $Tx$. Together with $\Vert Tx\Vert \le \Vert T\Vert \Vert x\Vert$, it makes $T$ an isomorphic embedding onto its image with the inherited norm. The image is necessarily closed when $X$ is a Banach space; that extra conclusion requires completeness.

A quotient operator is a surjective bounded linear operator that is an open map. For a linear map, being an open map is equivalent to the existence of $c>0$ such thatThe inclusion gives a neighbourhood of $Tx$ inside the image of every sufficiently small ball about $x$, by translation and scaling. Conversely, an open map sends $B_X(0,1)$ to a neighbourhood of zero. A linear open map onto $Y$ is automatically surjective, since its image contains a ball and is a linear subspace.

The Banach-space adjoint is $T^{\ast }g=g\circ T$. For $g\in Y^{\ast }$, the ball inclusion givesAlso $\Vert T^{\ast }g\Vert \le \Vert T\Vert \Vert g\Vert$. Thus the Banach-space adjoint is an isomorphic embedding, with lower bound $c$.

We use successive approximation for quotient operators. Fix $y$ with $\Vert y\Vert <1$. The hypothesis gives $x_0$ with $\Vert x_0\Vert <1$ and $\Vert y-T{x}_0\Vert <\delta$. Inductively, ifapply the hypothesis to $r_n/{\delta }^n$, obtaining $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and $\Vert r_n-T{x}_n\Vert <{\delta }^{n+1}$. If a residual is zero, take all remaining terms zero.

The series $\sum x_n$ converges in the Banach space $X$ because $\sum \Vert x_n\Vert <\sum {\delta }^n=(1-\delta {)}^{-1}$. Write its sum as $x$. The continuity of $T$ and the residual estimate giveScaling now yieldsThis ball inclusion proves that $T$ is a quotient operator. Completeness of $Y$ was unnecessary: the convergent series is constructed in $X$ and its image has limit $y$.

The real Hahn-Banach separation theorem says that if $C$ is a nonempty open convex set in a real normed vector space and $y\notin C$, there is a nonzero continuous linear functional $g$ such that $g(y)\ge {\sup }_{z\in C}g(z)$.

Put $\delta =1-1/M\in (0,1)$. The prescribed set is open because it is a union of translates of an open ball, and it is convex because both summands are convex. If $\Vert y\Vert <1$ and $y\notin C$, Hahn-Banach separation gives a nonzero $g\in Y^{\ast }$ withThe displayed supremum is a sum of independent suprema; the symmetry of the unit balls identifies them with the respective operator norms. But $g(y)\le \Vert g\Vert \Vert y\Vert <\Vert g\Vert$, a contradiction. Hence $y=Tx+r$ for some $\Vert x\Vert \le 1$ and $\Vert r\Vert <\delta$.

To obtain the strict bound on $x$ required in the approximation hypothesis, choose $t<1$ sufficiently close to $1$. Then $\Vert tx\Vert <1$ andThus the approximation condition holds with exactly $\delta =1-1/M$. When $X$ is a Banach space, the preceding successive approximation for quotient operators proves $T$ is a quotient operator. Combining this with the Banach-space adjoint estimate proves

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets; equivalently, every countable intersection of open dense sets is dense.

Let $E_n=\mathrm{span}\{e_1,\dots ,e_n\}$ in the space of finitely supported sequences. Each $E_n$ is a closed linear subspace for every norm, since a finite-dimensional subspace is closed. Henceis a closed linear subspace of the Banach space $X$. Every finitely supported sequence lies in some $E_n$, so $X={\bigcup }_{n\ge 1}{F}_n$. By the Baire category theorem, some $F_N$ has nonempty interior.

A linear subspace with nonempty interior is the whole space: translate a ball in it to zero and scale any vector into that ball. Thus $F_N=X$, giving $T(X)\subseteq E_N$ andThe proof uses no particular choice of norm on the space of finitely supported sequences.

Under duality of sequence spaces, $a=(a_n)\in {\ell }_1$ represents $f_a(x)=\sum a_n{x}_n$ on $c_0$, with operator norm $\Vert f_a\Vert =\sum |a_n|$. Here $c_0$ is the space of sequences converging to zero, equipped with its supremum norm.

If $\Vert x{\Vert }_{\infty }\le 1$ and $f_a(x)=\sum |a_n|$, thenEvery summand is nonnegative, so $x_n=\mathrm{sgn}(a_n)$ whenever $a_n\ne 0$. Infinitely many such coordinates would give infinitely many $|x_n|=1$, contradicting $x_n\to 0$. Therefore $a$ has finite support.

Conversely, if $a$ has finite support, choose $x_n=\mathrm{sgn}(a_n)$ on that support and $x_n=0$ elsewhere. Then $x\in c_0$, $\Vert x{\Vert }_{\infty }\le 1$, and $f_a(x)=\Vert f_a\Vert$. The zero functional is included by taking $x=0$. ThusThis is norm attainment on the vanishing sequence space.

The compact set $S(B_{c_0})$ is bounded, so $S$ is a bounded linear operator. For each $g\in Y^{\ast }$, the continuous linear functional $g$ attains its maximum on that compact set. Symmetry of $S(B_{c_0})$ implies that maximum is $\Vert S^{\ast }g\Vert$, so $S^{\ast }g$ attains its norm on $B_{c_0}$. By norm attainment on the vanishing sequence space,Regard the Banach-space adjoint $S^{\ast }$ as a bounded linear operator from the Banach space $Y^{\ast }$ into $c_{00}$ with its inherited ${\ell }_1$ norm. The completeness of the dual space holds even if $Y$ is incomplete. The preceding Baire category theorem argument shows that $S^{\ast }$ has finite rank.

If $S^{\ast }(Y^{\ast })=\mathrm{span}\{f_1,\dots ,f_r\}$, set $N={\bigcap }_{j=1}^r\ker f_j$. For $x\in N$ and every $g\in Y^{\ast }$, $g(Sx)=(S^{\ast }g)(x)=0$. The Hahn-Banach theorem implies that continuous linear functionals separate points of $Y$, so $Sx=0$. Thus $N\subseteq \ker S$, andThis proves compact closed-ball image forces finite rank on c0.

Define a diagonal operator on sequence space byIt is bounded because $\Vert Sx{\Vert }_2^2\le \Vert x{\Vert }_{\infty }^2{\sum }_{n\ge 1}{4}^{-n}=\Vert x{\Vert }_{\infty }^2/3$. Its truncation $S_{m}x=(2^{-1}{x}_1,\dots ,2^{-m}{x}_m,0,\dots )$ is a finite-rank operator, andEquality follows by using finitely supported sequences equal to $1$ on longer and longer finite tails. An operator norm limit of compact operators into the Banach space ${\ell }_2$ is compact; therefore $S$ is a compact operator. The vectors $S{e}_n=2^{-n}{e}_n$ are linearly independent, so its rank is infinite.

There is no conflict with the preceding result about a compact closed-ball image. Indeed, take $x^{(m)}=(1,\dots ,1,0,\dots )$ with $m$ initial ones. Then $S{x}^{(m)}\to (2^{-n}{)}_{n\ge 1}$ in ${\ell }_2$, but that limit has no preimage in $c_0$: its only possible preimage is the constant sequence $1$. Thus $S(B_{c_0})$ is relatively compact but is not closed and hence is not compact.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every $f\in X^{\ast }$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\epsilon$. The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.

The canonical embedding into the bidual isIt is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. For $x\ne 0$, the Hahn-Banach theorem supplies a norming functional $f$ with $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, proving equality. Thus $J_X$ is an isometric embedding. The coordinates defining the weak-star topology on $X^{\ast \ast }$, restricted to $J_X(X)$, are precisely $J_{X}x\mapsto f(x)$ for $f\in X^{\ast }$. These are exactly the coordinates defining the weak topology on $X$, so $J_X$ is also a topological embedding for these topologies.

A normed vector space is a reflexive space if $J_X$ is onto $X^{\ast \ast }$. Such a space is automatically a Banach space, since it is isometric to the complete continuous dual space $X^{\ast \ast }$. The Banach-Alaoglu theorem states that the closed unit ball of a continuous dual space is compact for its weak-star topology. The Goldstine theorem states that $J_X(B_X)$ is dense in the weak-star topology in $B_{X^{\ast \ast }}$. ConsequentlyIf $J_X$ is onto, its weak-to-weak-star homeomorphism carries $B_X$ onto the Banach-Alaoglu theorem's compact ball $B_{X^{\ast \ast }}$. Conversely, if $B_X$ is weakly compact, $J_X(B_X)$ is compact and hence closed in the Hausdorff weak-star topology. Its density from the Goldstine theorem makes it all of $B_{X^{\ast \ast }}$, and scaling gives surjectivity of $J_X$.

Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the Hahn-Banach theorem, choose norming functionals $f_n$ with $\Vert f_n\Vert =1$ and $f_n(u_n)=1$. These continuous linear functionals separate points: for $x\ne 0$, choose $\Vert u_n-x/\Vert x\Vert \Vert <1/2$ to get $f_n(x)>\Vert x\Vert /2$. DefineThe series converges uniformly, separates distinct points, and obeys the triangle inequality because $\min (c,a+b)\le \min (c,a)+\min (c,b)$. Thus $d$ is a metric. Its topology is coarser than the weak topology: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$. On a weakly compact set $K$, the identity $(K,\text{weak})\to (K,d)$ is therefore a continuous bijection from a compact space into a Hausdorff space, hence a homeomorphism. This proves that $K$ is metrizable in its weak topology, an instance of countable separating family metrizes a weakly compact set.

If $X$ is also a reflexive space, its closed unit ball is weakly compact and therefore metrizable by this construction. Compactness in a metric space implies sequential compactness, so every sequence in $B_X$ has a weakly convergent subsequence. Scaling proves the claim for every bounded sequence.

For the converse, assume every bounded sequence in $X$ has a weakly convergent subsequence. Use the same metric $d$ on $B_X$. A weak limit of points of $B_X$ remains in $B_X$: $|f(x)|\le 1$ for every $\Vert f\Vert \le 1$, and the Hahn-Banach theorem recovers $\Vert x\Vert \le 1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.

We must still recover the entire weak topology, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the weak topology $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of metric limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies weak convergence. Every $f\in X^{\ast }$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a metric space. The weak topology and $d$-topology on $B_X$ coincide. The ball is weakly compact, so the criterion above provesThe zero space satisfies all statements directly. The converse did not assume completeness, nor use the full Eberlein-Šmulian theorem.

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.

For a self-adjoint C-star element $a$, the following are three equivalent characterizations of a Positive element of a C-star algebra:If $a=b^2$ with $b=b^{\ast }$, the spectral mapping theorem gives nonnegative spectrum. Conversely, the continuous functional calculus applied to $\sqrt{\lambda }$ on $\sigma (a)$ gives a self-adjoint C-star element $b$ with $b^2=a$. For real $\sigma (a)$,Thus the third condition means $\sigma (a)\subseteq [0,2t]$, while the first implies the second by taking the interval $[0,\Vert a\Vert ]$. The second implies the third with $t=\Vert a\Vert$ if $a\ne 0$; if $a=0$, any $t>0$ works. This proves every equivalence.

The positive cone of a C-star algebra contains zero and is closed under multiplication by nonnegative real scalars. To verify addition without assuming that positive elements commute, put $r=\Vert a\Vert$, $s=\Vert b\Vert$ for $a,b\ge 0$. ThenIf $r+s>0$, the third characterization makes $a+b$ positive; otherwise both are zero. Hence it is a convex cone. If $a$ and $-a$ are positive, $\sigma (a)=\{0\}$ and $a=0$ by the self-adjoint spectral norm formula; therefore the cone is pointed and defines a partial order on $A_{\mathrm{sa}}$ by $h\le k$ if $k-h\in A_{+}$. It is generating because the continuous functional calculus givesFor $a\ge 0$, the spectral mapping theorem yieldsFinally, let $a\in M_2(\mathbb{C})$ be Hermitian, and write $\tau =\mathrm{tr}a$, $d=\det a$. Its two eigenvalues are real. Both are nonnegative exactly when their sum $\tau$ and product $d$ are nonnegative. The corresponding sum and product for $I-a$ are $2-\tau$ and $1-\tau +d$. Therefore the necessary and sufficient conditions for a positive contraction areIndeed, nonnegative sum and product force two real numbers to be nonnegative, including the zero-product cases. Applying this to $a$ and $I-a$ places both eigenvalues in $[0,1]$.

Let $c=x(1_A)\ge 0$. Positivity on $a^{\ast }a$ implies positivity on every Positive element of a C-star algebra, by taking its positive square root. For a self-adjoint C-star element $h$, the positive cone of a C-star algebra givesBoth $x(\Vert h\Vert 1_A+h)$ and $x(\Vert h\Vert 1_A-h)$ are nonnegative real numbers. Their difference shows $x(h)$ is real, and the same inequalities give $|x(h)|\le c\Vert h\Vert$.

For an arbitrary $a\in A$, write $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint. Since the involution is isometric, $\Vert h\Vert ,\Vert k\Vert \le \Vert a\Vert$. HenceThus every positive functional on a C-star algebra is bounded, with the preliminary bound $\Vert x\Vert \le 2x(1_A)$. This proof does not presuppose the Cauchy–Schwarz inequality for positive C-star functionals. It also gives $x(a^{\ast })=\overline{x(a)}$, since $x(h)$ and $x(k)$ are real.

Set $A_0=x(a^{\ast }a)$, $B_0=x(b^{\ast }b)$ and $c=x(a^{\ast }b)$. The preceding reality argument gives $x(b^{\ast }a)=\bar{c}$. Positivity therefore implies, for every $\lambda \in \mathbb{C}$,If $A_0>0$, choose $\lambda =-c/A_0$ to obtain $0\le B_0-|c{|}^2/A_0$. If $A_0=0$ and $c\ne 0$, choosing $\lambda =-tc$ and letting $t\to \infty$ contradicts positivity. Thus $c=0$ in that case too. This proves the Cauchy–Schwarz inequality for positive C-star functionals:Apply it with $a=1_A$, $b=z$. Since $0\le z^{\ast }z\le \Vert z{\Vert }^2{1}_A$, positivity givesThus $\Vert x\Vert \le x(1_A)$, while evaluation on $1_A$ gives the reverse inequality. ConsequentlyThe zero functional is covered as well. In the trivial zero algebra all assertions are immediate, and its state space is empty.

For a nonzero unital C-star algebra, the set $S_A$ of state on a C-star algebra functionals is convex, since both positivity and evaluation $x(1_A)=1$ are preserved by convex combinations. The norm identity places it in the closed dual unit ball. Suppose a state satisfies $x=ty+(1-t)z$, with $0<t<1$ and $\Vert y\Vert ,\Vert z\Vert \le 1$. Evaluating at the unit and taking real parts givesSince both weights are strictly positive, both real parts equal $1$. Also $|y(1_A)|,|z(1_A)|\le 1$, so $y(1_A)=z(1_A)=1$. HenceThe stipulated converse to the norm identity makes $y,z$ positive, so both belong to $S_A$. Therefore $S_A$ is a face of a convex set, namely the closed dual unit ball. More precisely it is the exposed face of the dual ball defined by the unit, since it is cut out by $\mathrm{Re}x(1_A)=1$.

The quotient norm isChanging the representative does not change the infimum. The triangle inequality and absolute homogeneity follow from those of the original norm. Vanishing means $x$ belongs to the closure of $Y$, hence to $Y$ because the subspace is closed. Thus this is a norm.

Use the absolutely convergent series criterion for completeness. If $\sum \Vert {\xi }_n{\Vert }_{X/Y}<\infty$, choose representatives $z_n$ of ${\xi }_n$ satisfying $\Vert z_n\Vert \le \Vert {\xi }_n{\Vert }_{X/Y}+2^{-n}$. The series $\sum z_n$ converges in the Banach space $X$ to some $z$. Since the quotient map is contractive, $\sum {\xi }_n$ converges to $z+Y$. Consequently the quotient is a Banach space.

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ is a unit vector at distance greater than $1/2$ from $Y$. Starting with one unit vector, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular

The natural canonical embedding into the bidual is $J_{X}x(f)=f(x)$. It is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. A norming functional from the Hahn-Banach theorem gives the reverse inequality, so it is an isometry. A reflexive Banach space is one for which this canonical embedding is onto.

A nonreflexive example is $c_0$. Its dual is ${\ell }^1$ and its bidual is ${\ell }^{\infty }$; the canonical map is the natural inclusion, which misses the constant sequence one. This also shows why merely having an isometric embedding in the bidual is insufficient.

An infinite-dimensional reflexive example is ${\ell }^2$, by the Riesz representation theorem.

The weak topology on $X$ is the coarsest topology making every $f\in X^{\ast }$ continuous. The weak-star topology on $X^{\ast }$ is the coarsest making every evaluation $f\mapsto f(x)$, $x\in X$, continuous. In each case basic neighborhoods impose finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of a dual space is compact for its weak-star topology. The separation theorem needed here says: if $K$ is a nonempty weak-star compact convex subset of a dual space $E^{\ast }$ and $z\notin K$, then some $e\in E$ strictly separates them,The real-part convention covers complex spaces; for real spaces omit it.

Let $K$ be the weak-star closure of $J_X(B_X)$ inside $B_{X^{\ast \ast }}$. This is compact by Banach-Alaoglu, convex, and nonempty. If $z\in B_{X^{\ast \ast }}\setminus K$, separation in the dual pair $(X^{\ast \ast },X^{\ast })$ gives $f\in X^{\ast }$ withBut $|z(f)|\le \Vert z\Vert \Vert f\Vert \le \Vert f\Vert$, a contradiction. Therefore $K=B_{X^{\ast \ast }}$, proving the required Goldstine theorem density.

If $X$ is reflexive, $J_X$ identifies its weak topology with the weak-star topology of $X^{\ast \ast }$ and carries $B_X$ onto $B_{X^{\ast \ast }}$, so Banach-Alaoglu gives weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact, hence closed in the Hausdorff bidual topology. Density then forces $J_X(B_X)=B_{X^{\ast \ast }}$. Scaling gives surjectivity on the entire bidual. Thus

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).

(ii) implies (i): together with the preliminary estimate, (ii) gives equivalent norms,An equivalent norm on a Banach space is complete: a primed-norm Cauchy sequence has an original-norm limit, and the upper bound makes convergence hold also in the primed norm. (ii) implies (iii) immediately because $\Vert P_k\Vert \le C$ for every $k$.

(iii) implies (ii): the preliminary bounds show ${\sup }_k\Vert P_{k}x\Vert <\infty$ for every $x$. Applying the Uniform boundedness principle to the bounded maps $P_k$ gives ${\sup }_k\Vert P_k\Vert <\infty$.

(iii) implies (iv): the normalization $\Vert x_n\Vert =1$ givesHence every coordinate functional is bounded.

(iv) implies (iii): $P_k={\sum }_{n=1}^k{x}_n{\phi }_n$ is a finite sum of bounded rank-one operators.

(iv) implies (v): continuity gives ${\phi }_n(y_r)\to 0$ whenever $y_r\to 0$, so any scalar limit $\beta$ must vanish.

(iv) implies (vi): fix $n$. Continuity gives ${\phi }_n(y_r)\to {\phi }_n(x)$, while the assumed uniform coordinate convergence gives ${\phi }_n(y_r)\to {\beta }_n$. Uniqueness of the scalar limit yields ${\beta }_n={\phi }_n(x)$ for every $n$.