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For the point interaction in a second-order differential equation $L_{\alpha }={\partial }_x^2+\alpha {\delta }_c$ on $(0,\ell )$, use homogeneous Dirichlet boundary conditions. The unperturbed Green function iswith a positive unit derivative jump at $x=\xi$. Whenever $1+\alpha G_0(c,c)\ne 0$,To derive this rank-one operator correction, move the delta coefficient to the right-hand side and apply the unperturbed inverse: $G_{\alpha }(x,\xi )=G_0(x,\xi )-\alpha G_0(x,c)G_{\alpha }(c,\xi )$. Setting $x=c$ determines $G_{\alpha }(c,\xi )$ and gives the formula. Its denominator vanishes exactly at Dirichlet resonance for a point interaction.

The kernel is symmetric. It is continuous at $x=c$ and $x=\xi$, satisfies $[{\partial }_x{G}_{\alpha }{]}_{\xi }=1$ when $\xi \ne c$, and satisfies $[{\partial }_x{G}_{\alpha }{]}_c=-\alpha G_{\alpha }(c,\xi )$ away from the source. When $\xi =c$, the single combined condition is $[{\partial }_x{G}_{\alpha }{]}_c=1-\alpha G_{\alpha }(c,c)$. These are the transmission conditions for a delta coefficient, including a source at the interaction.

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.

For the rank-one operator $Tf=⟨f,v⟩u$ on a Hilbert space with a countable Hilbertian basis $(e_k)$, the operator norm and Hilbert-Schmidt norm both equal $\Vert u\Vert \Vert v\Vert$. The Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, with equality at $f=v/\Vert v\Vert$ when $v\ne 0$. The Parseval identity for a Hilbertian basis gives ${\sum }_k\Vert T{e}_k{\Vert }^2=\Vert u{\Vert }^2{\sum }_k|⟨e_k,v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$. The zero-vector cases are immediate.

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 $H=0$.

Write $u(t)=e^t$ and $v(t)=e^{-t}$. Both lie in the Hilbert space $L^2(-\pi ,\pi )$, and the integral operator factors as the rank-one operatorThe Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, and equality is attained at $f=v/\Vert v\Vert$. Hence the operator norm is $\Vert T\Vert =\Vert u\Vert \Vert v\Vert$. Direct integration givesFor any Hilbertian basis $(h_k)$, the Parseval identity for a Hilbertian basis givesThis is the general formula for the norms of a rank-one operator. In particular $T$ is a Hilbert-Schmidt operator, andThis computation needs no explicit Fourier basis. The printed optional basis formula omits the variable $t$: the intended Hilbertian basis is $e_n(t)=(2\pi {)}^{-1/2}{e}^{int}$, $n\in \mathbb{Z}$. Taken literally, the printed expressions are constant functions and cannot form an orthonormal basis of $L^2(-\pi ,\pi )$.

If $x\in J$, part (ii) makes $1-x/\lambda$ invertible for every nonzero scalar $\lambda$. Thus $\sigma (x)\subseteq \{0\}$ and $r(x)=0$: it is a quasinilpotent element. Conversely, let $I$ be a two-sided ideal consisting of quasinilpotent elements. For $x\in I$ and $y\in A$, $yx\in I$ has spectrum $\{0\}$, so $1+yx$ is invertible. Part (ii) implies $x\in J$. This proves $J$ is the greatest ideal contained in the quasinilpotent set, the Jacobson radical as greatest quasinilpotent ideal characterization.

A semisimple Banach algebra is one with $J=\{0\}$. For $B=\mathcal{B}(X)$, its natural action on $X$ is faithful and algebraically irreducible. Indeed, given $v\ne 0$ and any $w\in X$, the Hahn-Banach theorem gives a bounded functional $g$ with $g(v)=1$. The rank-one operator $z\mapsto g(z)w$ then maps $v$ to $w$. Any invariant subspace containing $v$ consequently contains every $w$. The faithful representation has kernel zero, so zero is primitive and $B$ is semisimple.

Nevertheless, choose a nonzero bounded functional $g$ and $0\ne x_0\in \ker g$, possible when $\dim X>1$. The nonzero rank-one operator $Tz=g(z)x_0$ satisfies $T^{2}z=g(z)g(x_0)x_0=0$. Hence $\sigma (T)=\{0\}$ and $r(T)=0$, although $T\ne 0$. This proves the quasinilpotent set of $B$ is not $\{0\}$; it is not the same as the radical.

For infinite-dimensional $X$, the finite-rank operators form a nonzero proper two-sided ideal: products with bounded operators are finite-rank, and the identity has infinite rank. Thus zero is not maximal. Even if maximal ideals are required to be closed, the compact operators give a nonzero proper closed ideal containing the finite-rank operators. The identity is not compact on an infinite-dimensional Banach space, since the Riesz lemma supplies a bounded sequence with pairwise distances bounded below. Therefore zero is primitive but not maximal in infinite dimension.

For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

Write $x\otimes y$ for the rank-one operator $h\mapsto ⟨h,y⟩x$. The estimator is the kernel representation of the empirical covariance operatorSince $\mathbb{E}X=0$,Thus the estimator has the fixed bias of an estimator ${\mu }_0\otimes {\mu }_0$ for $C_X$.

The fourth-moment assumption makes $(X_i-{\mu }_0)\otimes (X_i-{\mu }_0)$ square-integrable in the Hilbert space of Hilbert-Schmidt operators. The weak law of large numbers therefore givesConsequently consistency for $C_X$ holds exactly when ${\mu }_0=0$. More precisely, if $Y=(X-{\mu }_0)\otimes (X-{\mu }_0)$, thenThe Hilbert-space central limit theorem also yieldswhere $G$ is a centered Gaussian random element in the Hilbert-Schmidt operator space with covariance determined by $Y$. Relative to $C_X$, the same fluctuation is displaced by $\sqrt{n}({\mu }_0\otimes {\mu }_0)$ and hence does not have a finite centered limit when ${\mu }_0\ne 0$.

Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.

Take an orthonormal basis $\{|j⟩\}$ and insert its completeness relation:No normalization or orthogonality of $u,v$ is required. This is the trace of a rank-one operator; in a Hilbert space the sum converges by the Cauchy-Schwarz inequality.

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.