A linear operator $T:X\to Y$ is a compact operator if the image of the closed unit ball has compact closure in $Y$. Equivalently, every bounded sequence $(x_n)$ in $X$ has a subsequence whose images converge in the norm of $Y$. This implies boundedness of $T$.

The Schauder theorem for compact operators says that for a bounded linear operator between Banach spaces,where the dual map of bounded linear operator is $T^{\ast }f=f\circ T$. The forward implication also holds for normed spaces; completeness of the target gives the usual converse. The compactness definition requires the image subsequence's limit to belong to $Y$, so that completeness qualification matters when stating the theorem beyond Banach spaces.

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 $\mathrm{Re}x(1_A)=1$.

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.

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 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.

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 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.

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

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 $y$.

An element is a Positive element of a C-star algebra if it is self-adjoint and its spectrum lies in $[0,\infty )$, equivalently if $a=b^{\ast }b$ for some $b$.

For a normal $a$, the continuous functional calculus is the unique unital isometric *-isomorphism $C(\sigma (a))\to C^{\ast }(1,a)$ sending the coordinate function $z$ to $a$. It respects sums, products, conjugation and the unit, satisfies $\Vert f(a)\Vert ={\max }_{\sigma (a)}|f|$, and has $\sigma (f(a))=f(\sigma (a))$. For self-adjoint $a$ it preserves the order of real-valued functions.

The piecewise definition of $g_n$ is consistent for $n\ge 2$. At $n=1$ the printed branches conflict, even assigning both zero and one at $t=1/2$. All the requested limit and face arguments use only $n\ge 2$; discarding this first index resolves that harmless initial-index defect without altering any limit.

For $n\ge 2$, interpolate linearly from $(1/(2n),0)$ to $(1/n,g(1/n))$, and from $(1-1/n,g(1-1/n))$ to $(1-1/(2n),1)$. On the middle interval the error is zero. On the left both $g$ and $g_n$ lie between zero and $g(1/n)$; on the right both lie between $g(1-1/n)$ and one. Thusby continuity at the endpoints.

Put $\delta =1/(2n)$. To prove the bounds on $h_n$, it suffices to show $0\le t-\delta g_n(t)\le 1-\delta$. For $t\le \delta$, $g_n=0$; for $t\ge 1-\delta$, $g_n=1$. On the remaining interval, $\delta \le t\le 1-\delta$ and $0\le g_n\le 1$, so both inequalities follow. Consequently $g_n$ and $h_n$ are positive contractions.

For the face assertion, write $r=\Vert a\Vert$. Since $a\in [0,1_A]$ and $0\notin F$, $0<r\le 1$. If $r<1$, then $a=r(a/r)+(1-r)0$, with both endpoints in $[0,1_A]$. The face property would force $0\in F$, a contradiction. Thus $\Vert a\Vert =1$. Its spectrum contains one by positivity and the norm formula, and contains zero because it is noninvertible.

Applying functional calculus to the scalar identity givesSince $0<\delta <1$ and $a\in F$, the face property gives $g_n(a)\in F$. Isometry of functional calculus gives $g_n(a)\to g(a)$ in norm; closedness of $F$ therefore proves

An element of a C*-algebra is a Normal element of a C-star algebra if

(vi) implies (iv) without assuming beforehand that any coordinate functional is continuous. Define the everywhere-defined linear mapThe preliminary coordinate bound ensures it really takes values in ${\ell }^{\infty }$. Condition (vi) states exactly that its graph is sequentially closed: $y_r\to x$ and $\Phi (y_r)\to \beta$ in the supremum norm imply $\Phi (x)=\beta$. The product is a metric space, so sequentially closed means closed. The closed graph theorem therefore makes $\Phi$ bounded. Every coordinate projection on ${\ell }^{\infty }$ is bounded, hence so is each ${\phi }_n$.

Together with the preceding implications this proves all six conditions are equivalent. No continuity of the coordinate functionals was assumed during the preliminary bounds.

(v) implies (iv). Fix $n$ and consider any sequence $y_r\to y$ with ${\phi }_n(y_r)\to \gamma$. Apply (v) to $y_r-y$ and $\beta =\gamma -{\phi }_n(y)$. It gives $\gamma ={\phi }_n(y)$. Thus the graph of ${\phi }_n:X\to \mathbb{K}$ is closed. Both spaces are Banach, so the closed graph theorem makes ${\phi }_n$ 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 $n$.

(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.

(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 $k$.

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).

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