Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 (C4.1, DMAC 2464, DMPC 2464)

The triangular kernel is nonnegative, has integral one, support in $[-\delta ,\delta ]$, and supremum $1/\delta$. Thus the convolution is defined at every point and

The kernel is globally Lipschitz with constant $1/{\delta }^2$: this holds on each of its linear pieces, and summing across the finitely many intervening breakpoints proves it across pieces as well. Consequently

Since the kernel has integral one, The Tonelli theorem and the triangle inequality giveThis proves the stated convolution approximation estimate.

For the unlabelled compactness assertion let $M$ bound the norms of $S$, and fix $\eta >0$. Choose the kernel width $\delta$ smaller than the translation radius supplied for $\eta$, so that the same bound holds for all shifts of size at most this width. The preceding estimate gives $\Vert f\ast g-f{\Vert }_1<\eta$ uniformly. The smoothed family has uniform bound $M/\delta$ and uniform Lipschitz constant $M/{\delta }^2$. Choose $K$ with all original tails outside $[-K,K]$ below $\eta$, and put $L=K+\delta$. The Tonelli theorem givesOn $[-L,L]$ the smoothed family is relatively compact in the uniform norm by the Arzelà-Ascoli theorem. Take a finite uniform net with error below $\eta /(2L)$ and extend its centers by zero outside that interval. It is a finite $L^1$ net for the smoothed-and-truncated family with error $\eta$. Each original $f$ is within another $2\eta$ of its corresponding truncated smoothing. Thus $S$ is totally bounded in $L^1$. The displayed (a),(b) are the two hypotheses used together in this argument.

For the final family, Young convolution inequality gives $\Vert f_0\ast f{\Vert }_1\le \Vert f_0{\Vert }_1$. Its translations satisfywhich tends to zero uniformly over the family. Since $f$ vanishes outside $[-1,1]$,uniformly as well. Both compactness hypotheses apply, proving $S_0$ is totally bounded.

Regard the spaces as real and set $u=\mathrm{Re}\varphi$. Its real norm equals $\Vert \varphi \Vert$, since rotating any test vector makes its functional value real and nonnegative without changing its norm. By the allowed real Hahn-Banach theorem extend $u$ to a real-linear $v:X\to \mathbb{R}$ with that norm, and define $\psi (x)=v(x)-iv(ix)$. Real linearity gives $\psi (ix)=i\psi (x)$, so $\psi$ is complex linear. On $Y$ this formula recovers $\varphi$. Also $\mathrm{Re}\psi =v$; rotating a test vector again gives $|\psi (x)|\le \Vert v\Vert \Vert x\Vert$. The reverse norm inequality follows by restriction to $Y$, proving a norm-preserving complex extension.

We use two standard consequences: a point outside a closed subspace can be separated by a bounded functional annihilating that subspace, and the norm of a vector is the supremum of its pairings with dual unit-ball functionals. If $\psi \in X^{\ast }$ annihilates $Y$, the scalar holomorphic function $\psi \circ f$ vanishes at the prescribed sequence tending to zero. Its identity theorem makes it identically zero. Separation then gives $f(z)\in Y$ for every $z\in D$.

For $\varphi \in Y^{\ast }$ extend to $X^{\ast }$ with the same norm and put $F=\varphi \circ f$. Then $F$ is holomorphic with $|F|\le \Vert \varphi \Vert$. For the understood nonzero increments $h,k$, the Cauchy integral formula on $|\zeta |=r$ givesTaking absolute values, bounding the contour length by $2\pi r$, and each denominator below by $r-|z|$, $r-|z|-|h|$, $r-|z|-|k|$ proves exactly the stated estimate.

Fix $z,r$ and restrict $|h|,|k|<(r-|z|)/2$. Taking the supremum over $\Vert \varphi \Vert \le 1$ bounds the norm of the difference of the two $Y$-valued difference quotients by a fixed constant times $|h-k|$. They are therefore Cauchy as $h\to 0$. Since $Y$ is a closed subspace of a Banach space, it is complete. Consequently the derivative exists as a norm limit in $Y$. The labelled (a)-(c) are the simultaneous hypotheses, not separate claims to prove.

A linear operator has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The closed graph theorem states that an everywhere-defined linear operator between Banach spaces is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Projection of the graph onto $X$ is a bounded linear bijection, so the open mapping theorem makes its inverse bounded. Composing that inverse with the second-coordinate projection proves $T$ is bounded.

For the composition assertion, suppose $x_n\to x$ and $T{x}_n\to y$. Boundedness of $S$ gives $ST{x}_n\to Sy$, and boundedness of $ST$ gives $ST{x}_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.

For the derivative bound take $Y=C^1[0,1]$ with norm $\Vert v{\Vert }_{C^1}=\Vert v{\Vert }_{\infty }+\Vert v^{\prime }{\Vert }_{\infty }$, and $Z=C[0,1]$ with its supremum norm; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\Vert Ux{\Vert }_{C^1}\le c\Vert x\Vert$. In particular $|(Ux{)}^{\prime }(t)|\le c\Vert x\Vert$ uniformly in $t$.

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

Given a bounded sequence, first pass to a subsequence on which the images under $T_1$ converge. From that subsequence extract another on which the images under $T_2$ converge. Both image sequences, and hence their sum, converge on the latter subsequence. Thus $T_1+T_2$ is compact.

Extract a subsequence whose $T$ images converge, and apply continuity of the bounded operator $S$ to those images. This proves $ST$ is compact. The reverse composition $TS$ is compact as well, since $S$ sends a bounded sequence to a bounded sequence to which compactness of $T$ applies.

Suppose $T_n$ are compact and $\Vert T_n-T\Vert \to 0$. For any $\epsilon >0$ take $n$ with $\Vert T_n-T\Vert <\epsilon$. A finite $\epsilon$-net of $T_n$ of the unit ball is then a finite $2\epsilon$-net of its $T$ image. That image is totally bounded, so $T$ is compact by the stated criterion. Hence $\mathcal{K}(X)$ is norm closed in $B(X)$.

Compact. Let $P_N$ keep the first $N$ coordinates. The operator $P_{N}T$ has finite rank and $\Vert T-P_{N}T\Vert \le 1/(N+1)$, since the remaining diagonal coefficients are at most that number. Norm-closedness of compact operators proves the claim.

Not compact. For coordinate unit vectors $e_j$ with $j\ge 2$, the images are $e_{j-1}$. Distinct such images have ${\ell }^1$ distance two, so no image subsequence converges.

Compact. This operator is $BD$, where $D$ is the compact diagonal operator in (a) and $B$ is the bounded backward shift in (b). The composition property applies. Equivalently its tails have the same vanishing-coefficient finite-rank approximation.

Compact. Its column indexed by $m$ has entries $n!/m!$ for $1\le n<m$, and zero otherwise. The sum of that column is $c_m=(1!+\cdots +(m-1)!)/m!\le 2/m$ for $m\ge 2$, by the elementary induction $1!+\cdots +k!\le 2k!$. For operators from ${\ell }^1$ the supremum of the absolute column sums is the operator norm: one inequality is the triangle inequality, and coordinate unit vectors give the reverse. Keeping just the first $N$ input columns is finite rank and its remainder norm is at most ${\sup }_{m>N}{c}_m\le 2/(N+1)$. Hence it converges to $T$ in norm, proving compactness.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

Use the source's notation for the vanishing-at-zero ideal $I=\{f\in C[0,1]:f(0)=0\}$. It is the kernel of bounded evaluation at zero, hence closed. It is stable under pointwise multiplication and complex conjugation, so inherits the C-star identity and completeness. If it had an identity $e$, multiplication against $f(t)=t$ would give $e(t)=1$ for $t>0$, contradicting continuity and $e(0)=0$. Thus $I$ is a non-unital C-star subalgebra. This explicit meaning of the notation matters because the usual notation $C_0$ for a compact space has a different convention.

Define the unitization map by $\Phi (f,\lambda )=f+\lambda 1$. Its inverse is $h\mapsto (h-h(0),h(0))$. The specified multiplication and involution are preserved by direct substitution. The pullback norm $\Vert (f,\lambda )\Vert =\Vert f+\lambda 1{\Vert }_{\infty }$ is a C-star norm, so equals the unique unitization norm supplied in the question. Therefore $\Phi$ is a unital isometric C-star isomorphism onto $C[0,1]$.

For the approximation put $h(s)=a(\sqrt{s})$ on $[0,1]$, so $h$ is continuous and $h(0)=0$. The Stone-Weierstrass theorem supplies a polynomial $p(s)$ with $\Vert h-p{\Vert }_{\infty }<\epsilon /2$. Then $b(t)=p(t^2)-p(0)$ is a polynomial in the positive even powers of $t$, with zero constant term, andThis gives exactly the required finite form.

The continuous functional calculus for a Hermitian element of a C-star algebra identifies its generated unital C-star algebra with continuous functions on its real spectrum, preserves the involution and norm, and sends functions to the corresponding spectral values by spectral mapping. Consequently an element is a square of a Hermitian element of a C-star algebra exactly when it is a Hermitian element of a C-star algebra with nonnegative spectrum: the reverse direction uses the continuous square root on that spectrum.

Nonnegative scalar multiples preserve this property directly. For positive $a,b$ put $C=\Vert a\Vert +\Vert b\Vert$. Functional calculus gives $\Vert \Vert a\Vert 1-a\Vert \le \Vert a\Vert$ and similarly for $b$, so $\Vert C1-(a+b)\Vert \le C$. The sum is a Hermitian element of a C-star algebra, and spectral mapping now places its spectrum in $[0,2C]$. It therefore has a positive square root in a C-star algebra. If $C=0$ both elements are zero. Thus $A_{+}$ is a cone. It is pointed because an element positive with its negative has spectrum contained in $\{0\}$ and hence is zero by the self-adjoint norm formula. It is closed as well: for a converging sequence of Positive elements of a C-star algebra use a common norm bound $C$ and pass to the limit in $\Vert C1-a_n\Vert \le C$.

For the two-by-two matrix the exact criterion isA positive matrix is Hermitian with nonnegative eigenvalues; evaluating its quadratic form at the coordinate vectors gives the diagonal conditions, and its determinant, the eigenvalue product, gives the final inequality. Conversely if $a_{11}>0$, completing the square writes the quadratic form as $a_{11}|z_1+(a_{12}/a_{11})z_2{|}^2+(a_{22}-|a_{12}{|}^2/a_{11})|z_2{|}^2$, which is nonnegative. If $a_{11}=0$, the determinant inequality forces $a_{12}=0$ and the conclusion is immediate. The spectral theorem then gives nonnegative eigenvalues and their Hermitian square root, proving positivity in the question's sense.

A state on a C-star algebra is a positive linear functional of norm one. A cyclic representation of a C-star algebra is a representation of a C-star algebra $\pi :A\to B(H)$ with a vector $\xi$ whose orbit $\pi (A)\xi$ has dense linear span in $H$. On $c$, pointwise multiplication and coordinatewise complex conjugation preserve convergence; the constant sequence one is the unit. The supremum norm is submultiplicative and obeys $\Vert {\alpha }^{\ast }\alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }^2$. Completeness is given, so $c$ is a commutative unital C-star algebra.

If $F=\varphi (\beta )$ is a state, testing its positivity on the coordinate projections gives ${\beta }_{j+1}\ge 0$. To recover the independent coefficient at infinity, let $q_N$ have its first $N$ coordinates zero and all later ones one. This is a positive element of $c$, and $F(q_N)={\beta }_1+{\sum }_{j>N}{\beta }_{j+1}\ge 0$. Passing to the limit gives ${\beta }_1\ge 0$. By the given isometry, $\Vert F\Vert =\Vert \beta {\Vert }_1={\sum }_j{\beta }_j$, which is one. Conversely these nonnegative unit-sum coefficients make $F({\alpha }^{\ast }\alpha )$ a sum of nonnegative weighted squares, and the isometry gives norm one. Thus the stated coefficient conditions are necessary and sufficient.

For the specified state, all other coefficients are zero because they are nonnegative and the first three already sum to one. Its GNS null ideal isEvaluation at infinity and the first two coordinates identifies $c/N_x$ with ${\mathbb{C}}^3$, with inner product $\frac{1}{2}{u}_{\infty }\overline{v_{\infty }}+\frac{1}{4}{u}_1\overline{v_1}+\frac{1}{4}{u}_2\overline{v_2}$. Left multiplication induces the diagonal representation; the class of the unit is $(1,1,1)$. In the equivalent standard inner product the resulting GNS triple isIts coefficient is exactly $\frac{1}{2}{\alpha }_{\infty }+\frac{1}{4}{\alpha }_1+\frac{1}{4}{\alpha }_2=F(\alpha )$. The three evaluations can be assigned arbitrary independent values by a convergent sequence, and all three entries of ${\xi }_x$ are nonzero. Therefore its orbit spans ${\mathbb{C}}^3$, verifying cyclicity and the required state identity directly.