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Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.

Fix an arbitrary finite $T$ and use the Banach space of bounded continuous functions $C_b([0,T]\times {\mathbb{R}}^{2d})$ with the supremum norm. Set $A=\Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1$. For $f$ in this space, the velocity integralis bounded by $\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. If $(t_n,x_n)\to (t,x)$, the integrands converge pointwise and are bounded by the integrable function $\Vert f{\Vert }_{\infty }{k}_2$. The dominated convergence theorem proves continuity. Thus the rank-one gain on bounded continuous functions, $Kf=k_1{H}_f$, is continuous and bounded. The same theorem applied to the time integral shows that $\tau$ maps this space into itself, including at $t=0$.

Time ordering gives the factorial bound for a Volterra iterate:Consequently the Volterra series for the linear Boltzmann equationconverges uniformly on the whole finite time slab, so its sum is continuous and bounded and satisfies the fixed point equation. Tracking the damping factors gives the sharper pointwise bound $\Vert f(t){\Vert }_{\infty }\le e^{(A-1)t}\Vert f_0{\Vert }_{\infty }$. This proves existence for every finite $T$, without requiring $AT<1$ or uniform continuity of $f_0$. Boundedness on finite slabs does not assert a uniform bound over infinite time.

Write $\mathrm{TV}(u)=|Du|(\Omega )$ for the total variation seminorm on a domain. For a locally integrable real function its dual definition isThe bounded-variation space consists of functions $u\in L^1(\Omega )$ with finite $|Du|(\Omega )$, equipped with $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_1+|Du|(\Omega )$. In the equivalent distributional derivative description, $Du$ is a finite vector-valued Radon measure and $|Du|$ is its total variation measure.

To establish completeness of the bounded-variation space, let $(u_n)$ be Cauchy in this norm. Completeness of $L^1$ gives $u_n\to u$ in $L^1$. Given $\epsilon >0$, choose $N$ so that $\Vert u_n-u_m{\Vert }_{BV}<\epsilon$ whenever $m,n\ge N$. For fixed $n\ge N$, the given lower semicontinuity yieldsSince $\Vert u_n-u_m{\Vert }_1\to \Vert u_n-u{\Vert }_1$, we obtain $\Vert u_n-u{\Vert }_1+|D(u_n-u)|(\Omega )\le \epsilon$. In particular $u_n-u$ has finite variation; the triangle inequality then puts $u$ in the BV space. The same bound proves convergence in the full norm, so this is a Banach space.

For the disk data, put $B=B(0,R)$ and assume $\alpha >0$. The exact total variation denoising of a disk isHere $(s{)}_{+}=\max (s,0)$ is the positive part. A total variation calibration certifies global optimality, including competitors that are not radial or piecewise constant. Define the bounded vector fieldIt has $|z|\le 1$. Its normal component is continuous across the circle, so the distributional divergence has no extra boundary measure. Direct differentiation gives $q=\mathrm{div}z=(2/R){\chi }_B$. The dual definition implies $\int vq\le \mathrm{TV}(v)$: for the standard $BV\cap L^2$ domain one can cut $z$ off at radius $L$, with the error bounded by $C{L}^{-1}{\int }_{L<|x|<2L}|v|\to 0$, and then smooth the test field. In the larger homogeneous bounded-variation space, the same error is bounded by $C\Vert v{\Vert }_{L^2(L<|x|<2L)}\to 0$. Thus both usual whole-plane formulations give the same certificate.

Use the perimeter identity $\mathrm{TV}({\chi }_B)=2\pi R$ and $|B|=\pi R^2$. If $c=1-2\alpha /R>0$, then $\int c{\chi }_{B}q=2\pi Rc=\mathrm{TV}(c{\chi }_B)$ and $u_{\ast }-g+\alpha q=0$. Consequently every competitor $v$ satisfiesIf $\alpha \ge R/2$, replace $z$ by $z_{\alpha }=(R/(2\alpha ))z$. Its norm is still at most one, its divergence is $q_{\alpha }={\chi }_B/\alpha$, and equality in the calibration holds at $u_{\ast }=0$. The identical comparison proves optimality and uniqueness of zero, including the threshold $\alpha =R/2$. The only general results used are completeness of $L^1$, lower semicontinuity of variation, the indicator-perimeter identity, the distributional integration-by-parts/dual variation formula and the quadratic norm identity. For $\alpha =0$, the unique squared-error minimizer is simply $g$.

True in the specified topology. Let $D\subseteq B^{\prime }$ be the countable dense set obtained in part (ii). Scaling by a positive integer is a homeomorphism for the weak-star topology, so $nD$ is dense in $n{B}^{\prime }$. Sincethe countable set ${\bigcup }_{n}nD$ is dense in $E^{\prime }$: a nonempty open set contains a point of some $n{B}^{\prime }$, so its relative open intersection with that ball meets $nD$. This proves weak-star separability of the entire dual.

ThusThere is no assertion of norm separability here. For example, the dual of the separable Banach space ${\ell }^1$ is ${\ell }^{\infty }$, whose binary sequences form an uncountable set with pairwise norm distance one. That example is nevertheless separable in its weak-star topology.

True. Choose a countable dense sequence $(x_j)$ in the separable Banach space $E$. For $\varphi ,\psi$ in its closed dual unit ball, putThis is a metric: positivity follows because continuous functionals agreeing on a dense subset agree everywhere; symmetry is immediate; and the triangle inequality follows from subadditivity of $t/(1+t)$ for $t\ge 0$. The tail of the series is uniformly small, so convergence on these coordinates implies convergence for this metric, and conversely metric convergence controls each individual coordinate.

The uniform bound $\Vert \varphi -\psi \Vert \le 2$ upgrades coordinate convergence to pointwise convergence on all of $E$. For any $x\in E$, choose $x_j$ close to $x$ and useFor neighbourhoods involving finitely many $x$, choose finitely many such approximations. This proves equality of the induced topologies, including for arbitrary nets. Thus the weak-star metrizability of the dual ball gives

True. We prove weak-star topology on an entire infinite-dimensional Banach dual is not metrizable. Suppose instead that $E^{\prime }$ has a countable local base $(W_n)$ at zero. Choose a basic weak-star topology neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.

For any $x\in E$, the set $\{\varphi :|\varphi (x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forcesIndeed a finite-dimensional span is norm closed, and the Hahn-Banach theorem supplies a bounded linear functional vanishing on it and nonzero at any point outside it.

It follows that $E={\bigcup }_n\mathrm{span}{S}_n$. Each span is a proper finite-dimensional closed vector subspace and has empty interior in the infinite-dimensional Banach space $E$. This contradicts the Baire category theorem. HenceThe uniform norm bound that made the metric work on $B^{\prime }$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore all true.

Let $q:X^{\ast }\to X^{\ast }/Y$ be the quotient map. Since $Y$ is norm closed and has finite codimension, the quotient is a finite-dimensional normed space. Choose a basis of its continuous dual and compose its coordinate linear functionals with $q$. This gives ${\varphi }_1,\dots ,{\varphi }_n\in X^{\ast \ast }$ such thatIf $Jx\in F$, it vanishes on $Y$. Evaluation at $x$ is weak-star continuous, and $Y$ is weak-star dense, so it vanishes on all of $X^{\ast }$. The dual norm formula gives $x=0$. Hence $F\cap JX=\{0\}$.

Assume $X\ne \{0\}$; the zero space is trivially normed by any positive constant. We claim thatIf not, there would be $x_j\in S_X$ and ${\psi }_j\in F$ with $\Vert J{x}_j-{\psi }_j\Vert \to 0$. The sequence ${\psi }_j$ is bounded. Finite-dimensionality of $F$ gives a norm-convergent subsequence with limit $\psi \in F$. Then $J{x}_j\to \psi$, and $JX$ is norm closed because $X$ is Banach. Thus $\psi \in F\cap JX$ and $\Vert \psi \Vert =1$, a contradiction. This is positive distance between a unit sphere and a disjoint finite-dimensional subspace.

Fix $x\in S_X$, identifying it with $Jx$, and put $\delta =\mathrm{dist}(x,F)\ge d$. On $F+\mathbb{R}x$, define $h(\varphi +tx)=t\delta$. The distance definition givesThe Hahn-Banach theorem extends it to $\Psi \in X^{\ast \ast \ast }$ with $\Vert \Psi \Vert \le 1$, $\Psi {|}_F=0$ and $\Psi (x)=\delta$.

Apply part (ii) with the Banach space $X^{\ast }$, its finite-dimensional dual vector subspace $E=F+\mathbb{R}x\subseteq X^{\ast \ast }$, and the bidual space element $\Psi \in X^{\ast \ast \ast }$. For every $\eta >0$ it supplies $y_{\eta }\in X^{\ast }$ satisfyingThus $y_{\eta }\in Y$. Normalize by $1+\eta$ and let $\eta \downarrow 0$ to obtainHomogeneity gives, for all $x\in X$,Consequently a finite-codimensional weak-star dense dual subspace is norming, with $c=d$. Since $Y$ is a real linear vector subspace and its ball is symmetric, the same supremum is obtained if an absolute value is inserted. The argument uses near-unit interpolation and a limiting supremum, not an assertion that the supremum is attained.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

Throughout this question the Banach spaces and Hilbert spaces are complex. Separability is an additional standing hypothesis, not part of the general definition of a Hilbert space. Here and below unit ball means the closed unit ball. An open unit ball in positive finite dimension is not compact. In finite dimension, equivalence of norms identifies the closed unit ball with a closed bounded subset of Euclidean space, so the Heine-Borel theorem gives compactness in the norm topology.

In an infinite-dimensional Hilbert space, the Gram-Schmidt process constructs an infinite orthonormal sequence $(e_n)$. Put $f_n=e_n/\sqrt{2}$. Then $\Vert f_n\Vert =1/\sqrt{2}\le 1$ and, for $m\ne n$, $\Vert f_n-f_m{\Vert }^2=(1+1)/2=1$. No subsequence is a Cauchy sequence, contradicting sequential compactness. Thus the closed unit ball is norm compact exactly in finite dimension.

The finite-dimensional implication again follows from equivalence of norms and the Heine-Borel theorem. In an infinite-dimensional Banach space, inductively apply Riesz lemma to the proper closed finite-dimensional linear span $Y_n=\mathrm{span}(f_1,\dots ,f_n)$ to obtain a unit vector $f_{n+1}$ with $\mathrm{dist}(f_{n+1},Y_n)>1/2$. Therefore $\Vert f_n-f_m\Vert >1/2$ for distinct indices, precluding a Cauchy subsequence.

For completeness, Riesz lemma here follows by taking $z\notin Y$, setting $d=\mathrm{dist}(z,Y)>0$, choosing $y\in Y$ with $\Vert z-y\Vert <2d$, and putting $f=(z-y)/\Vert z-y\Vert$. Its distance from $Y$ exceeds $1/2$. This proves the closed unit ball is norm compact if and only if the Banach space is finite-dimensional. The printed hint's equality of all pairwise distances is unnecessary; the separation inequality is what the argument supplies.

For the subsequent essential spectrum arguments, the discrete-spectrum definition must use $\mathrm{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin {\Sigma }_{\mathrm{e}}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\mathrm{ran}L$ is closed; this includes the case $0$ is in the resolvent set.

A closed-range bound on the kernel complement supplies the useful equivalenceFor the forward implication, $L:N^{\perp }\to \mathrm{ran}L$ is a bounded bijection between Banach spaces, so the bounded inverse theorem applies. For the reverse implication, any Cauchy sequence of image points has a Cauchy sequence of preimages in $N^{\perp }$, and completeness gives a preimage of its limit.

Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^{\perp }$. If $L{f}_n$ converges, the lower bound makes $(v_n)$ a Cauchy sequence. The finite-dimensional vector space $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.

Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an orthonormal sequence in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^{\perp }$ with $L{v}_n\to 0$. Any norm limit would lie in both $N$ and $N^{\perp }$, hence be zero, contradicting its unit norm. This proves the required sequential properness for a self-adjoint operator equivalence.

The spectral-shift repair is essential for later parts. On ${\ell }^2$, take $L=\mathrm{diag}(2,1,1/2,1/3,\dots )$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the essential spectrum, although no singular Weyl sequence exists there: on the complement of that eigenspace, $\Vert (L-2I)v\Vert \ge \Vert v\Vert$.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

Work in the Banach space ${\mathcal{B}}_T$ of bounded strongly measurable maps $[0,T]\to L_{x,v}^2$, with norm $\Vert g{\Vert }_{{\mathcal{B}}_T}={\sup }_{0\le t\le T}\Vert g(t){\Vert }_2$. Keeping actual representatives at every time matches the pointwise-in-time mild formulation. The Boltzmann Volterra operator is bounded on this space, and the factorial bound for a Volterra iterate givesThus the Volterra series for the linear Boltzmann equation converges in operator norm for every finite $T$, even when $CT\ge 1$. SetFor its partial sums, $(I-\tau ){\sum }_{n=0}^N{\tau }^{n}F=F-{\tau }^{N+1}F$. The remainder tends to zero by the factorial estimate. Hence $(I-\tau )f=F$, precisely the required characteristic integral equation.

The norm bound in (b) gives an explicit choice of the existence constant:Also $f(0)=f_0$, since every term with $n\ge 1$ vanishes at zero and the damping interval has length zero. This proves existence in the paper's weak, characteristic-integral sense. With merely measurable nonnegative $a$, that sense does not itself require a continuous initial trace; that trace follows under the additional local characteristic-integrability condition described in (b).