Write $K={\sigma }_A(x)$ and $R_x(\zeta )=(\zeta 1-x{)}^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda \notin U$, the holomorphic function $1/(Z-\lambda )$ belongs to $\mathcal{O}(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda 1$. Thus existence forces $K\subset U$, even before continuity is used.

For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. DefineThe integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The Cauchy theorem shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying bounded linear functionals. We use the scalar Cauchy integral formula in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta )/(\zeta -z)$ to $2\pi if(z)$.

Linearity is immediate and, for a fixed cycle,This is a bound by one compact-uniform seminorm, proving continuity for local uniform convergence. To obtain ${\Theta }_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb{C}\setminus K$, to a large circle and integrate its Neumann series. Similarly $\zeta R_x(\zeta )=1+x{R}_x(\zeta )$ gives ${\Theta }_x(Z)=x$.

For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The resolvent identity isSubstitute it into the double integral for ${\Theta }_x(f){\Theta }_x(g)$. In the term containing $R_x(\eta )$, first integrate $f(\zeta )/(\zeta -\eta )$ around the outer cycle, obtaining $f(\eta )$. In the term containing $R_x(\zeta )$, first integrate $g(\eta )/(\zeta -\eta )$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. HenceThis proves the required continuous unital homomorphism, the holomorphic functional calculus.

For uniqueness use the following version of the allowed Runge theorem: on any open $U\subseteq \mathbb{C}$, rational functions with all finite poles outside $U$ approximate each holomorphic function locally uniformly; poles at infinity are allowed, giving polynomials. A unital homomorphism sending $Z$ to $x$ is forced on polynomials and on every inverse $(Z-\lambda {)}^{-1}$, $\lambda \notin U$, and hence on these rational functions. Continuity then forces its value on their limits. Thus ${\Theta }_x$ is unique. This argument works for disconnected $U$ as well and does not assume polynomials alone are dense. It proves continuity and uniqueness of holomorphic functional calculus.

We will use the character description of spectra, with a short justification. A character of an algebra $\phi$ is a nonzero complex multiplicative linear functional; it satisfies $\phi (1)=1$. Applying it to an inverse shows $\phi (a)\in {\sigma }_A(a)$, so $|\phi (a)|\le \Vert a\Vert$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda 1$, its principal ideal is proper because $A$ is commutative. Put it in a maximal ideal $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb{C}$ and its quotient map is a character taking $a$ to $\lambda$. This provesIt is the spectrum equals character values in a commutative Banach algebra statement and does not assume semisimplicity.

For any character, continuity permits passing it inside the contour integral. Since $\phi (R_x(\zeta ))=(\zeta -\phi (x){)}^{-1}$ and $\phi (x)\in K$, the scalar Cauchy formula givesApply the character description first to ${\Theta }_x(f)$ and then to $x$ to obtain the holomorphic spectral mapping theorem

On ${\Pi }_{+}$, use the analytic logarithm with argument in $(-\pi /2,\pi /2)$ and set $s(z)=\exp (\frac{1}{2}\mathrm{Log}z)$. It satisfies $s(z{)}^2=z$ and takes its values in ${\Pi }_{+}$. The principal square root in a commutative Banach algebra isFor uniqueness let $v$ be another root with spectrum in ${\Pi }_{+}$. For each character, $\phi (y)$ and $\phi (v)$ are scalar roots of $\phi (x)$ with positive real part, so both equal $s(\phi (x))$. Hence every character takes the nonzero value $2s(\phi (x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.

Set $a_n=\Vert x^n\Vert$ and $D={\inf }_{n\ge 1}{a}_n^{1/n}$. If some $x^m=0$, all later powers vanish, their root norms tend to zero, and $D=0$. Otherwise fix $m\ge 1$ and write $n=qm+s$, $0\le s<m$. Submultiplicativity givesSince $q/n\to 1/m$, it follows that ${\mathrm{limsup}}_n{a}_n^{1/n}\le a_m^{1/m}$. Taking the infimum over $m$ and using $a_n^{1/n}\ge D$ provesThis establishes existence of the limit and its infimum characterization before identifying it with the spectral radius $r(x)={\max }_{\lambda \in \sigma (x)}|\lambda |$.

For $|\lambda |>D$, the root test makes ${\sum }_{n\ge 0}{x}^n/{\lambda }^{n+1}$ converge absolutely. Multiplication of partial sums shows that its limit is $(\lambda 1-x{)}^{-1}$. Thus $r(x)\le D$. For the reverse inequality, take any $\rho >r(x)$ and the circle $|\lambda |=\rho$. The resolvent of an element is holomorphic off the spectrum, andTo justify this identity, first use a radius larger than $\Vert x\Vert$ and integrate the uniformly convergent Neumann series term by term. Then deform to radius $\rho$: the intervening annulus has no spectral points. The Cauchy theorem for these Banach-valued integrals follows by applying bounded linear functionals and using their point separation. Consequently, with $M_{\rho }={\max }_{|\lambda |=\rho }\Vert (\lambda 1-x{)}^{-1}\Vert$,Taking root limits gives $D\le \rho$; let $\rho \downarrow r(x)$. Hence the full spectral radius formula is

For the product comparison, if $\lambda \ne 0$ and $R=(\lambda 1-xy{)}^{-1}$, direct multiplication on both sides verifiesIndeed $(\lambda 1-yx)y=y(\lambda 1-xy)$, and the remaining terms cancel against the inverse equation. Interchanging $x,y$ proves the converse. Thus the nonzero spectra of products in opposite orders agree, soThe possible difference at zero has no effect on the maximum modulus, including when that maximum is zero.

Zero really can differ. In $A=\mathcal{B}({\ell }^2({\mathbb{N}}_0))$, let $S$ be the unilateral shift operator, $S{e}_n=e_{n+1}$, and take $x=S^{\ast }$, $y=S$. Then $xy=1$, whereas $yx=1-P_0$, with $P_0$ the projection onto $e_0$. The latter is zero on $e_0$ and identity on its orthogonal complement, and these blocks also explicitly invert it away from zero and one. Hence

Finally, for $t>r(x)$ the root norms of $(x/t{)}^n$ tend to $r(x)/t<1$, so their norms eventually decay geometrically and the powers converge to zero. Conversely, if the powers converge to zero, choose $N$ with $\Vert (x/t{)}^N\Vert <1$. Homogeneity and the infimum formula implyso $t>r(x)$. The admissible positive $t$ are exactly $(r(x),\infty )$, proving the power-decay characterization of the spectral radiusThis includes quasinilpotent elements with radius zero; no positive endpoint must be attained.

Work with a nonzero complex unital Banach algebra, so $1\ne 0$. The spectrum of an element isFor $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in norm and is a two-sided inverse, by multiplying its partial sums and letting the remainder tend to zero. Thus the spectrum lies in $|\lambda |\le \Vert x\Vert$. The assumed openness of the invertible group makes its complement closed, so the spectrum is compact.

To prove nonemptiness, suppose every $\lambda$ has an inverse $R(\lambda )=(\lambda 1-x{)}^{-1}$. Locally,so the resolvent is holomorphic. At infinity its Neumann series gives $\Vert R(\lambda )\Vert =O(|\lambda {|}^{-1})$. For any bounded linear functional $\ell \in A^{\ast }$, the scalar entire function $\ell (R(\lambda ))$ is bounded: it is bounded on compact disks by continuity, and tends to zero outside them. The Liouville theorem makes it constant, and its limit makes that constant zero. The Hahn-Banach theorem separates points of the normed space $A$, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the spectrum is nonempty and compact.

On the algebra of entire functions, setIt is finite, homogeneous and satisfies the triangle inequality and submultiplicativity. If it vanishes, the identity theorem makes $f$ identically zero, so it is an algebra norm. However, no algebra norm on the entire-function algebra can be complete. The coordinate function $Z(z)=z$ satisfies $\sigma (Z)=\mathbb{C}$ algebraically: $Z-\lambda$ vanishes at $z=\lambda$ and has no entire multiplicative inverse. A complete algebra norm would make this a Banach algebra, contradicting the proved boundedness of its spectrum. This is the entire function algebra admits no complete algebra norm obstruction; it applies to every proposed complete algebra norm, not only the displayed one.

For all continuous functions on $\mathbb{C}$, choose continuous cutoffsTheir supports are disjoint and escape every compact set. Thus $f(z)={\sum }_{n\ge 1}n{h}_n(z)$ is a locally finite sum and is continuous. Let $g_n(z)=\max (0,1-4|z-3n|)$. This nonzero continuous function is supported where $h_n=1$ and all the other cutoffs vanish. Hence $f{g}_n=n{g}_n$. Any algebra norm would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Therefore the algebra of all continuous functions on $\mathbb{C}$ admits no algebra norm, even an incomplete one. This proves the continuous functions on the complex plane admit no algebra norm obstruction directly.

A point $x$ of a convex set $K$ is an extreme point if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.

Here is a proof of the Krein-Milman theorem. The empty-set case is immediate, so take $K\ne \varnothing$. A face of a convex set is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the finite intersection property. The intersection remains a compact face. The Zorn lemma, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.

If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real linear functional $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same minimal compact face argument inside any nonempty compact face shows that each such face contains an extreme point of $K$.

Let $H=\overline{\mathrm{conv}}(\mathrm{ext}K)$. Since $K$ is compact, hence closed in the Hausdorff space, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous linear functional withThe maximizer face of $\ell$ on $K$ contains an extreme point $e$ by the preceding argument. Then $e\in H$ but $\ell (e)={\max }_K\ell \ge \ell (p)>{\sup }_H\ell$, a contradiction. ThereforeAll closures refer to the given locally convex topology, and compactness is what justified both maxima and chain intersections.

For the real l-infinity sequence space, if $|x_j|<1$ for any coordinate, choose $0<ϵ\le 1-|x_j|$. The two distinct sequences $x\pm ϵe_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. ThusFor the space of sequences converging to zero, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, givingThese are extreme points of real sequence-space unit balls. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not compact in the norm topology, as the coordinate vectors have mutual distance one.

First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$. This gives $v\in E$ withFor zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-T{v}_j$. Inductively,The Banach space $E$ therefore contains the sum $x={\sum }_{j\ge 0}{v}_j$, withThe partial sums satisfy $T{\sum }_{j=0}^{n-1}{v}_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to 0$ gives $Tx=y$. Thus $T$ is surjective with the stated lifting bound. This geometric correction for approximate surjectivity used completeness of $E$, not any unproved completeness of $F$.

Put $C=R/(1-k)$. Given a Cauchy sequence $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$ for $j\ge 1$. Lift each difference to $u_j\in E$ with $T{u}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert u_j\Vert \le C{2}^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+{\sum }_{j\ge 1}{u}_j$ exists in $E$, and its image is the limit of the subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit: use the triangle inequality between an arbitrary late term, a later subsequence term, and the limit. This proves $F$ is complete, the completeness forced by uniformly bounded lifting principle.

The open mapping theorem states that a bounded surjective linear operator between Banach spaces maps open sets to open sets. In particular a bounded linear bijection between Banach spaces has a bounded inverse. To deduce the closed graph theorem, let $S:E_1\to F_1$ be an everywhere-defined linear map between Banach spaces with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$, so it is itself Banach. The projectionis a bounded linear bijection. The open mapping theorem makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.

Finally consider the identity map $I:(C(X),\Vert \cdot \Vert )\to (C(X),\Vert \cdot {\Vert }_{\infty })$. If $f_n\to f$ in the new norm and $f_n\to g$ uniformly, each assumed continuous point evaluation functional gives $f_n(x)\to f(x)$, while uniform convergence gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both norms are complete, so the closed graph theorem makes $I$ bounded. Its inverse is bounded by the open mapping theorem. Consequently there are constants $c,C^{\prime }>0$ such thatThis proves Banach norm rigidity from continuous point evaluations. Pointwise continuity alone would not give a common bound on all evaluations; completeness supplies that through the closed graph argument.

For each hidden variable $\lambda$, a deterministic local hidden-variable model assigns values $a_i(\lambda ),b_j(\lambda )\in \{0,\dots ,d-1\}$ to all settings. The responses depend only on the local setting; their averaging distribution is setting independent, as required by measurement independence.

Let $T(\lambda )$ be the sum of the nonnegative modular representatives in the chained modular Bell inequality. Before reduction modulo $d$, the chain telescopes:Reduction changes this integer only by multiples of $d$. Since all reduced terms are nonnegative,The smallest possible nonnegative value in this residue class is $d-1$. Averaging over the hidden variable provesAdding local random seeds to $\lambda$ gives the same result for stochastic local hidden-variable theories.

For the coplanar qubit realization of the chained Bell inequality, shareAt angle $\theta$ in the $xz$ plane, use the projective measurement of the observableHere ${\sigma }_x,{\sigma }_z$ are Pauli matrices, and the classical bit $a$ records eigenvalue $(-1{)}^a$. Since $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and both crossed expected values vanish in this Bell state,The one-party expected values vanish, so the complete joint probabilities areFor binary outcomes, $[a-b]$ is one exactly when $a\ne b$, while $[b-a-1]$ is one exactly when $a=b$. Consequently

For $N\ge 2$, set $\delta =\pi /(2N)$ and choose Alice's angle ${\theta }_i=2(i-1)\delta$ and Bob's angle ${\varphi }_i=(2i-1)\delta$. Every adjacent pair $A_i,B_i$ or $B_i,A_{i+1}$ differs by $\delta$, giving expectation ${\sin }^2(\delta /2)$. The closing pair has angle difference $(2N-1)\delta =\pi -\delta$; its agreement probability is also ${\sin }^2(\delta /2)$. Thus every one of the $2N$ terms contributes equally:Indeed $\sin x<x$ for $x>0$, so this value is below ${\pi }^2/(8N)\le {\pi }^2/16<1$. For an explicit two-setting instance, Alice measures ${\sigma }_z,{\sigma }_x$ and Bob measures $({\sigma }_z+{\sigma }_x)/\sqrt{2},(-{\sigma }_z+{\sigma }_x)/\sqrt{2}$. The four relevant expected values sum toviolating the local bound one. At $N=1$ there is no violation; the two terms always sum to one. As $N\to \infty$, the displayed quantum value is asymptotic to ${\pi }^2/(8N)$ and approaches the algebraic lower bound zero.

The violation rules out a local hidden-variable theory with setting-independent preassigned local responses reproducing these quantum correlations. It does not permit faster-than-light communication: the joint formula gives each local outcome probability $1/2$, independent of the remote setting, as required by the no-signalling theorem. The contradiction concerns the joint correlations and the Bell assumptions, not the ability to send a signal.

For the Kraus operators of the measurement in quantum mechanics, the Born rule givesThe density matrix $A_i\rho A_i^{†}$ is the unnormalized state of that outcome. It is positive, its trace is $p_i$, and completeness gives ${\sum }_i{p}_i=\mathrm{Tr}\rho =1$. There is no conditional state to assign to an event with $p_i=0$.

For a projective measurement, each $P_i$ is an orthogonal projection, satisfying $P_i=P_i^{†}=P_i^2$. Completeness becomes ${\sum }_i{P}_i=I$. Fix $v$ in the range of $P_j$. ThenSince $P_{j}v=v$, the $j$th summand already equals $\Vert v{\Vert }^2$. All other summands are nonnegative, so $P_{i}v=0$ for $i\ne j$. Consequently $P_i{P}_j=0$, and taking adjoints also gives $P_j{P}_i=0$. This proves the stronger completeness forces orthogonality of measurement projections:

For the unitary dilation of a measurement instrument, take a quantum ancilla with orthonormal states $|0⟩,|1⟩,\dots ,|n⟩$, initially in $|0⟩$. DefineThe completeness equation impliesso $V$ is an isometry. Its action on the input subspace can be extended to a unitary operator $U$ on the enlarged system:Choose orthonormal bases in the two complementary subspaces and map one to the other. Their dimensions agree; the extra unused quantum ancilla level also ensures this extension is possible for an infinite-dimensional system. After the coupling, the joint density matrix is

Now perform a projective measurement on the larger system, withThese are pairwise orthogonal and sum to the full identity. The unused $|0⟩$ level has zero weight in $\Omega$, so merging it into outcome one adds no probability. For every $i$,Its trace is $p_i$, and taking the partial trace over the quantum ancilla after normalization gives exactly ${\rho }_i$. Hence unitary coupling to an quantum ancilla, a projective measurement, and discarding the quantum ancilla reproduce both the probabilities and the conditional states of the general measurement. This is the instrument version of a Stinespring dilation, rather than a construction reproducing only the outcome probabilities.

The target states here are qubit states, as appropriate to sharing singlet states of two qubits. Write $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ and choose $|{\psi }^{\perp }⟩=-{\beta }^{\ast }|0⟩+{\alpha }^{\ast }|1⟩$. These form an orthonormal basis, and the shared Bell state can be expressed asAlice knows the classical description of $|\psi ⟩$, so she can perform the projective measurement in this basis. The outcome $|{\psi }^{\perp }⟩$ has probability $1/2$ and leaves Bob in $|\psi ⟩$, up to an irrelevant overall phase. Alice sends a success bit: one for this outcome and zero for the other. Bob keeps his qubit only on success. This proves heralded remote preparation of an arbitrary qubit:The other outcome leaves $|{\psi }^{\perp }⟩$ at Bob's side; the protocol makes no claim of success in that branch.

For a block of $n$ targets, share $n$ independent singlet states. Alice measures her $j$th qubit in the basis $|{\psi }_j⟩,|{\psi }_j^{\perp }⟩$. She sends one bit indicating whether all $n$ outcomes were of the successful kind. Because the pairs are independent, all-success has probability $2^{-n}$; conditioned on it, Bob's joint state is precisely ${⨂}_{j=1}^n|{\psi }_j⟩$. Thus one-bit heralding of a remote state preparation block givesOn failure Bob discards the block, so he does not need a list of which component states failed.

For asymptotic remote state preparation by block indexing, prelabel a supply of independent candidate blocks, each containing $n$ singlet states. Alice measures blocks until she finds the first all-success block, of index $J$. She does not send a flag after each trial. Since $q=2^{-n}$ is the success probability of each block,This geometric distribution is independent of the target states. The probability of never succeeding is ${\lim }_{k\to \infty }(1-q{)}^k=0$. Alice needs to communicate only $J$; Bob retrieves the labelled qubits in that block, and thereby knows that every requested state was prepared exactly.

Here is an explicit prefix code for a rare geometric success. Set $M=2^n$ and divide $J-1=MQ+R$, where $0\le R<M$. Encode $Q$ as $Q$ ones followed by zero, then encode $R$ in exactly $n$ bits. Bob can decode this without knowing any target-state description. The message length is $L=n+Q+1$. With $r=(1-q{)}^M$,Since $(1-1/M{)}^M\le e^{-1}$,This version succeeds exactly with probability one and has the stated expected classical communication cost. It consumes $n\mathbb{E}J=n{2}^n$ singlet states on average, compatible with the unlimited entanglement assumption.

There is also a fixed-message-length version. Try at most $K=n{2}^n$ candidate blocks and send the first successful index, or a zero codeword reporting that none succeeded. Encoding the $K+1$ possibilities uses $\lceil {\log }_2(K+1)\rceil =n+{\log }_{2}n+O(1)$ bits. Its failure probability isThus fixed-length communication approaches one bit per state with success probability tending to one; variable-length communication achieves exact eventual success with expected cost approaching one bit per state. Before the index arrives, Bob's unconditional state remains independent of the target descriptions, in agreement with the no-signalling theorem.

Use $\oplus$ for exclusive or and regard all bits as elements of ${\mathbb{F}}_2$. For every public label $z\in \{0,1{\}}^m$, Alice computes the indicator functionwhile Bob computes $v_z(y)=f(z,y)$ from the known truth table. Exactly one of the $u_z(x)$ is one, so the separated Boolean decomposition is

For each $z$, they invoke an independent PR box with respective input bits $u_z(x),v_z(y)$ and receive outputs $a_z,b_z$. These satisfy $a_z\oplus b_z=u_z(x)v_z(y)$. Alice forms $A={⨁}_z{a}_z$, and Bob forms $B={⨁}_z{b}_z$. Taking the exclusive or of all box constraints provesAlice sends the single bit $A$. Bob combines it with his locally known $B$, obtainingThis is one-bit computation using Popescu–Rohrlich boxes. It uses $2^m$ boxes; decomposing by Bob's input instead gives $2^n$, so the smaller truth-table factorization uses at most $2^{\min (m,n)}$. No efficiency bound on local calculation or box use is needed for the stated unlimited resource.

Each PR box has uniform local output distributions regardless of the other party's input. Independent invocations therefore give Bob a uniform output string before the message, carrying no information about $x$. The useful correlation becomes accessible through Alice's final bit; the construction respects the no-signalling property of the boxes.