Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 (C4.1, DMAC 2464, DMPC 2464)

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 $n$.

(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.

(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.

A character of an algebra is a nonzero multiplicative complex-linear map $\chi :A\to \mathbb{C}$. A maximal proper ideal is a proper ideal contained in no larger proper ideal. For a character, $\chi (1)=1$ and $\chi (a)\in \sigma (a)$: otherwise $a-\chi (a)1$ would be invertible although its character value is zero. The spectral bound gives $|\chi (a)|\le \Vert a\Vert$, so the character is continuous. It is onto, and its kernel is a closed maximal ideal because $A/\ker \chi \cong \mathbb{C}$.

Conversely let $M$ be maximal. Its closure is an ideal and remains proper: otherwise an element of $M$ would lie within distance one of the unit, hence would be invertible by the Neumann series, forcing $1\in M$. Maximality then gives $\overline{M}=M$. The Banach algebra $A/M$ is a field. For any $b$ in this quotient choose $\lambda \in \sigma (b)$, using nonemptiness of the complex spectrum. In a field only zero is noninvertible, so $b-\lambda 1=0$. Hence every quotient element is a scalar multiple of the unit. The quotient map therefore determines a character ${\chi }_M$, uniquely normalized by ${\chi }_M(1)=1$. These constructions are inverse:For the convolution algebra, write $e_j$ for the sequence supported at $j$ with value one. Then $e_j{e}_k=e_{j+k}$, $e_0=1$, and $e_1{e}_{-1}=1$. If $z=\chi (e_1)$, continuity and $\Vert e_{\pm 1}{\Vert }_1=1$ imply $|z|\le 1$ and $|z^{-1}|\le 1$. Thus $|z|=1$, so uniquely $z=e^{it}$ for $0\le t<2\pi$, and $\chi (e_j)=e^{ijt}$ for every integer $j$. Absolute convergence in ${\ell }^1$ now givesConversely each displayed map is bounded, nonzero, and multiplicative, since the absolutely convergent double sum can be rearranged along $k+l=j$. Thus these are all characters, and all maximal proper ideals areThey are distinct because the character is uniquely recovered from its kernel and its value on the unit.

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

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

For a bounded linear functional on a unital C*-algebra, the state criterion isEquivalently it is positive and takes the unit to one.

In the Gelfand–Naimark–Segal construction, set $N_x=\{b:x(b^{\ast }b)=0\}$. This is a left ideal. Give $A/N_x$ the inner product $⟨[b],[c]⟩=x(c^{\ast }b)$, linear in the first variable, and complete it to a Hilbert space $H_x$. Left multiplication gives the bounded *-representation ${\pi }_x(a)[b]=[ab]$. The vector $\xi =[1_A]$ is cyclic and satisfies $x(a)=⟨{\pi }_x(a)\xi ,\xi ⟩$.

The Hermitian matrix $\rho$ has eigenvalues ${\lambda }_{\pm }=1/2\pm |\alpha |$ and trace one. In an orthonormal eigenbasis $v_{+},v_{-}$,If $|\alpha |\le 1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha |>1/2$, evaluating on the positive rank-one projection onto the negative-eigenvalue eigenspace gives a negative value, so it is not a state. ThereforeTo obtain the representation dimension, cyclicity of trace givesThus $A/N_{x_{\rho }}$ identifies isometrically with the matrix space $\{b{\rho }^{1/2}:b\in M_2(\mathbb{C})\}$, with left multiplication as the representation. If $r=\mathrm{rank}\rho$, this space consists, after a unitary change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. Rank is one at $|\alpha |=1/2$ and two at $|\alpha |<1/2$, proving