Tensor completion, injectivity and state domination

Written by GPT-6.1 Sol (OpenAI), Ultra. Course exposition: CC0-1.0 to the extent of rights held. Earlier programme and human-source notices retain their own terms.

A completed tensor has two descriptions here: a Cauchy sequence for the projective norm $\pi$, and a Cauchy sequence for the symmetric norm $\sigma$. Because $\sigma\le\pi$, the algebraic identity gives a map from the first completion to the second. The random lift construction supplies a projective preimage of every self-adjoint target vector, with cost at most $(81/16)\|v\|_\sigma+\varepsilon$. Different choices of $\varepsilon$ could still give different preimages. Their difference lies in the kernel of the completion map.

We prove that this kernel is zero. A normalized functional detects a kernel vector through its real part; the bilinear exponential test makes that real part symmetric-norm continuous. Finite cosine averages and an exact binomial variation bound then turn two numerical-reality conditions into a norm bound tending to zero. A second proof supplies the full Gaussian heat estimate. Once the lift is unique, its approximate costs become an exact norm comparison. That comparison makes every bounded real bilinear form eligible for the compact affine argument, which chooses one pair of states for all inputs.

Write

The two algebraic tensor norms are

Both infima use finite decompositions. The real projective norm uses self-adjoint factors and real scalars. A bounded form acts on these completions when its value on an algebraic tensor is bounded by the corresponding norm; its dual norm is the least such bound.

Section 8.1 proves real and complex functional extension. Section 8.2 constructs the completion maps and the projective algebra. Section 8.3 constructs the states, using the scalar spectrum, continuous calculus and order proved in ITP F01–F13. The scalar foundations and continuous norm integrals are in Elementary calculus for infinite products. Section 9 proves compactness and convergent subnets for arbitrary dual balls. None of these constructions assumes separability. A unital algebra in the state assertions has a nonzero identity of norm one; the injectivity assertion also includes zero and nonunital algebras.

Can a tensor disappear in the completion?

Theorem 3.1. For any C*-algebras $A,B$, the canonical map

is injective.

The algebraic identity is contractive and therefore extends to $I$ on Cauchy representatives. Injectivity asks whether $\sigma$-convergence to zero can hide a nonzero projective limit.

Proof. If either factor is zero, both tensor spaces are zero. First take nonzero unital $A,B$, and put $C=A\widehat\otimes_\pi B$. Section 8.2 proves its completeness, submultiplicative norm, isometric involution and norm-one identity $1\otimes1$. Multiplication and involution also extend to the $\sigma$-completion by the preceding lesson's product estimate. The algebraic identities consequently pass to the completions: $I(uv)=I(u)I(v)$ and $I(u^*)=I(u)^*$. Thus $K=\ker I$ is closed, is invariant under involution, and is an ideal.

Norm-preserving extension at the identity supplies $\mathcal S_C$. For $F\in\mathcal S_C$, put $G=(F+F^*)/2$, where $F^*(u)=\overline{F(u^*)}$. Adjoint isometry gives $\|G\|\le1$, while $G(1)=1$ gives the reverse inequality. The bilinear form $V_G(x,y)=G(x\otimes y)$ therefore has norm one and value one at $(1,1)$: its norm is at most $\|G\|$, and its value on that pair is one. It is self-adjoint. Lemma 1.1 and (3) below bound its value on every tensor decomposition by its $\sigma$-cost. It extends continuously to the $\sigma$-completion. The extension composed with $I$ agrees with $G$ on the algebraic tensors and hence on $C$. Consequently $G$ vanishes on $K$. If $v=v^*\in K$, this says $\operatorname{Re}F(v)=0$ for every normalized $F$. Consequently $h=iv$ is numerical-real. Since $v^2\in K$ is self-adjoint, $ih^2=-iv^2$ is also numerical-real. Lemma 2.1 gives $v=0$. For general $u\in K$, apply this to $u+u^*$ and $i(u-u^*)$; both are self-adjoint elements of $K$. Thus $u=0$.

For nonunital algebras, use forced C*-unitizations $\widetilde A,\widetilde B$. ITP F02 constructs these unitizations with isometric copies of the original algebras and a contractive scalar quotient. Thus $\|P_A(a+\lambda1)\|=\|a\|\le\|a+\lambda1\|+|\lambda|\le2\|a+\lambda1\|$; the same holds for $P_B$. On each finite decomposition, $P_A\otimes P_B$ has projective bound at most $4$, so Section 8.2 extends it to a bounded left inverse of the projective inclusion. That inclusion is therefore injective. The algebraic inclusion is $\sigma$-contractive, since all square-sum norms are preserved. The commuting square of the two canonical maps shows that an element in the kernel of (7) maps to the kernel for the unitizations. The unital result and projective injectivity of the inclusion make it zero. $\square$

Normalized functionals control the kernel

Suppose initially that $A,B$ are unital and $V:A\times B\to\mathbb C$ is bilinear, with $\|V\|=V(1,1)=1$. Each marginal has norm at most one and value one at the identity, so its norm is exactly one. Section 8.3's positivity criterion therefore makes $\varphi(x)=V(x,1)$ and $\psi(y)=V(1,y)$ states. For self-adjoint $a$, the exponential series and adjoint give $(e^{ita})^*=e^{-ita}$ and $e^{-ita}e^{ita}=1$. The C*-identity yields $\|e^{ita}\|=1$, which permits the following bilinear test.

Lemma 1.1. For self-adjoint $a,b$,

Proof. Expand the two exponential series through order two and apply $V$. If $M=\|a\|+\|b\|$ and $|t|\le1$, the terms of total degree at least three have absolute sum at most $e^M M^3|t|^3/6$; this follows from $n!\ge6(n-3)!$. The marginal values on $a,b$ are real, so the first-order term has zero real part. Thus

The left side is at most $1$, so the bracket is nonnegative. Replace $a$ by $-a$ to get $|\operatorname{Re}V(a,b)|\le[\varphi(a^2)+\psi(b^2)]/2$. Replace $a,b$ by $t a,t^{-1}b$ and minimize over $t>0$. If either quadratic value is zero, take the corresponding limit; otherwise the minimizing bound is their geometric mean. $\square$

Write $V^*(x,y)=\overline{V(x^*,y^*)}$. If also $V=V^*$, its values on self-adjoint pairs are real. For arbitrary $x=a+ic,y=b+id$, rotate $x$ by a scalar phase so that $V(x,y)$ is real. Then

The phase does not change $q(x)$. Summing (3) and applying scalar Cauchy–Schwarz proves that this normalized self-adjoint functional on $A\widehat\otimes_\pi B$ is $\sigma$-continuous, with dual norm at most $1$.

Two tests for an invisible element

For a unital complex Banach algebra $C$, assume $\|1\|=1$. Its normalized functionals are

They need not be positive on any involution. Call $h$ numerical-real if every $F\in\mathcal S_C$ takes a real value at $h$.

Lemma 2.1. If $h$ and $ih^2$ are numerical-real, then $h=0$.

Proof by finite cosine averages.

The directional norm derivative and Euler argument proved below have the following consequence: if $k$ is numerical-real, then $e^{isk}$ and $e^{-isk}$ are inverse contractions for every real $s$. Multiplication by either is an isometry. These two preliminary arguments use only norming functionals on $\operatorname{span}_{\mathbb C}\{1,k\}$, scalar series and completeness; neither uses the Gaussian argument. We apply them first to $k=h$.

Define

Both have norm at most one. Products of the exponentials add their parameters, by the absolutely convergent Cauchy product. For $m\ge0$, put

The binomial expansion, followed by multiplication by $S_s$, gives the exact finite identity

Indeed the positive exponential in $S_s$ shifts the index by one, while its negative exponential leaves the index fixed. Every exponential on the right has norm one.

The sequence $p_j$ increases up to $p_m$ and decreases thereafter. This follows directly from $p_{j+1}/p_j=(2m-j)/(j+1)$. Including its zero endpoints, its total variation is therefore $2p_m$. Taking norms in (BK1) proves

Here the variation is a finite sum of scalar absolute differences, not a measure-theoretic assertion.

There is an elementary bound sufficient for the limit:

For $m=0$ equality holds. Factorials give $c_{m+1}/c_m=(2m+1)/(2m+2)$. The induction step follows upon squaring from the exact identity

Thus (BK3) uses no asymptotic formula or Gaussian integral.

Fix $t>0$, and for $m\ge1$ set $s_m=\sqrt{t/m}$ and $H=\|h\|$. Absolute scalar series give

To justify the bound, the constant and quadratic terms of the cosine series are the two displayed terms, while its remaining terms are bounded by the exponential-series tail of degree at least four. For $r\ge0$, that tail is at most $r^4e^r/24$, since $n!\ge24(n-4)!$ for $n\ge4$.

Let $X=C_{s_m}$ and $Y=1-t h^2/(2m)$. The finite power-telescoping identity bounds their $2m$-th powers by

We used $\|X\|\le1$, $\|Y\|\le1+tH^2/(2m)$, and $1+r\le e^r$. The telescoping identity and this norm estimate are valid without a commutation assumption. The Euler formula already proved for Banach-algebra series gives $Y^{2m}\to e^{-t h^2}$. Also,

by the same exponential-tail estimate starting at degree three. Consequently

in norm. Dividing (BK2) by $s_m$ and using (BK3) now gives

Finally apply the isometric exponential argument to $k=ih^2$, which is numerical-real by the second hypothesis. The two real parameters $t,-t$ give inverse contractions $e^{-t h^2}$ and $e^{t h^2}$. Hence

Letting $t$ increase without bound proves $h=0$. This proof uses only finite averages and norm limits. The Gaussian proof below also gives the more precise intermediate estimate $1/\sqrt{\pi t}$, which is used in the exercises. $\square$

Second proof, by Gaussian averaging. We build the norm derivative, the isometric exponential group and the integral estimate in turn. The norm derivative only needs restrictions to $\operatorname{span}_{\mathbb C}\{1,a\}$. Section 8 proves the functional extension used in that step.

Directional norm derivative using only two-dimensional compactness

Let $C$ be a unital complex Banach algebra with $\|1\|=1$, and define

The norming-functional assertion at $1$ makes this set nonempty. For $a\in C$, set

The map $s\mapsto\|1+s a\|$ is convex, so $D_s\le D_t$ for $0<s<t$. Also $|D_s|\le\|a\|$. Thus $D=\lim_{s\downarrow0}D_s$ exists and is finite. Evaluating a member of $\mathcal S_C$ gives

Choose $s_j\downarrow0$, small enough that $1+s_j a\ne0$, and a norm-one $F_j\in C^*$ with $F_j(1+s_j a)=\|1+s_j a\|$. This is the norming consequence proved in Section 8.1. Restrict these functionals to $E=\operatorname{span}_{\mathbb C}\{1,a\}$. Their values at a fixed basis of $E$ are bounded, because $|F_j(x)|\le\|x\|$. A subsequence converges coordinatewise: repeatedly bisect a bounded interval containing each of their finitely many real and imaginary coordinates, retain a subinterval containing infinitely many indices, and then take a diagonal increasing subsequence. The nested lengths tend to zero, so real completeness gives convergence in every coordinate. This is the only compactness argument needed.

The limiting functional $f:E\to\mathbb C$ is linear and satisfies $|f(x)|\le\|x\|$ for every fixed $x\in E$. Moreover,

so $f(1)=1$. Apply Section 8.1 to extend $f$ to $F\in C^*$ with norm at most one; its value at $1$ forces its norm to be exactly one. Hence $F\in\mathcal S_C$. Finally,

Passing along the coordinate-convergent subsequence gives $D\le\operatorname{Re}F(a)$. Together with (B1), this proves

This proof applies even when $C$ is nonseparable. It uses finite-dimensional scalar subsequence compactness and Hahn-Banach, without the weak-* compactness of the full dual ball.

Numerical reality gives an isometric exponential group

Call $k\in C$ numerical-real if $F(k)\in\mathbb R$ for all $F\in\mathcal S_C$. Formula (4), applied to $a=\pm ik$, gives

The exponential $e^a=\sum_{m\ge0}a^m/m!$ converges in norm, since $\|a^m\|\le\|a\|^m$ and the scalar exponential series converges. Also

To see this directly, the coefficient of $a^m$, for $m\le n$, is $\binom nm/n^m$, which tends to $1/m!$ and is at most $1/m!$. The norm of the tail above any fixed $m_0$, uniformly in $n$, is bounded by $\sum_{m>m_0}\|a\|^m/m!$, which tends to zero. The finite initial segment converges coefficientwise.

Fix $t\in\mathbb R$. With $a=itk$, (C1) and submultiplicativity give

For the last bound, fix $\varepsilon>0$. Formula (C1) gives $\|1+itk/n\|\le1+\varepsilon|t|/n$ for all sufficiently large $n$, using the appropriate sign when $t\ne0$. The scalar exponential series gives $1+r\le e^r$ for $r\ge0$, and hence

Take the limit in (C2) and then let $\varepsilon\downarrow0$. For $t=0$ the norm is already one. For any $F\in\mathcal S_C$, reality of $F(k)$ gives $\|1+itk/n\|\ge|1+itF(k)/n|\ge1$. Thus the Euler norms are also bounded below by one, proving their limit is exactly one. Absolute convergence of the exponential series permits their Cauchy product; the binomial identity then gives $e^{itk}e^{-itk}=1$. Both factors are contractions and $\|1\|=1$, so both have norm one. For every $x\in C$,

and therefore left multiplication by $e^{itk}$ is an isometry. The analogous right assertion follows by the same calculation.

The scalar Gaussian facts and the vector integral

The scalar improper integral $I=\int_{\mathbb R}e^{-u^2}\,du$ exists: outside $[-1,1]$, $e^{-u^2}\le e^{-|u|}$. Its normalization can be checked using only compact Riemann integrals. If $I_R=\int_{-R}^R e^{-u^2}\,du$, the double Riemann integral over the square equals $I_R^2$, by rectangular product sums. That square contains the disk of radius $R$ and lies inside the disk of radius $\sqrt2 R$. For a continuous radial function on a disk, upper and lower sums on concentric annuli, whose areas are $\pi(r_j^2-r_{j-1}^2)$, give the radial integral $\int_0^R2\pi r f(r)\,dr$. Uniform continuity makes the difference of the two sums tend to zero. Applying this to $f(r)=e^{-r^2}$ gives

Hence $I^2=\pi$, and positivity gives $I=\sqrt\pi$. Scalar substitution now shows that for $t>0$,

Section 8.4 constructs the disk and annulus areas from rectangular sums, and identifies the same $\pi$ in the scalar quarter-phase example of Exercise 6. For the vector integral, let $v:[a,b]\to E$ be norm continuous in a real or complex Banach space. Scalar compactness makes it uniformly continuous. If $\omega(\delta)=\sup_{|s-r|\le\delta}\|v(s)-v(r)\|$, comparison of two tagged sums through their common refinement gives

Dyadic sums are Cauchy and have a limit by completeness. The same estimate proves that every choice of tags converges to that limit as its mesh tends to zero. The scalar norm function is continuous, and the triangle inequality applied to each tagged sum yields

If a continuous scalar integrable envelope $b(s)\ge\|v(s)\|$ has vanishing tails on $\mathbb R$, (D2) makes the integrals over $[-R,R]$ Cauchy as $R\to\infty$. Their limit is the improper norm integral, and (D2) holds on the line. Bounded linear maps, including multiplication by a fixed algebra element, commute with these integrals by passage from Riemann sums. The continuous fundamental theorem proved in R04 applies to any real or complex Banach space: the integral primitive has derivative $v$, and subtracting it from a function with continuous derivative leaves a constant function. With the bilinear product estimate in R07, this gives

for scalar $u$ and Banach-valued $v$ with continuous derivatives. Thus integration by parts is a norm identity. Only continuous functions with the specified integrable envelopes enter the full-line argument.

The heat identity, with all exchanges justified

Let $h\in C$. The series for $e^{ish}$ converges uniformly on every compact $s$-interval, so it may be integrated there termwise. For the full-line integral, its absolute series has scalar envelope

This is integrable, since with $A=\|h\|$,

Fix a large compact interval; uniform convergence gives termwise integration there. Outside the interval, the same integrable envelope controls every partial sum and the full sum. Its tail tends to zero. These two observations justify interchange of the series and the improper norm integral, without appealing to an unproved general Bochner dominated-convergence theorem.

Odd Gaussian moments vanish by symmetry. Integration by parts for the even moments uses $g_t'=-s g_t/(2t)$; polynomial times $g_t$ tends to zero at both endpoints. Indeed $e^u\ge u^N/N!$ for $u\ge0$; choosing $2N$ larger than a given polynomial degree shows that $|s|^m e^{-s^2/(8t)}\to0$. Keeping the other factor $e^{-s^2/(8t)}$ bounds each polynomial Gaussian tail by an integrable Gaussian, and eventually by a constant times $e^{-|s|}$. For $m\ge1$,

The last equality follows inductively from (D1). Consequently,

Derivative estimate and the kernel conclusion

Now assume $h$ and $ih^2$ are numerical-real. The isometric exponential argument gives $\|e^{ish}\|=1$. Uniform differentiation of the exponential series on compact intervals gives $(e^{ish})'=ih e^{ish}$. Integration by parts on $[-R,R]$, and then passage to the limit, gives

The endpoint terms vanish because $g_t(\pm R)\to0$ and $\|e^{\pm iRh}\|=1$; both integrals have scalar integrable envelopes. Thus (5) yields

The Gaussian is even and decreasing on the positive half-line, so scalar integration gives

Numerical reality of $ih^2$ and the isometric exponential argument make both $e^{t h^2}$ and $e^{-t h^2}$ contractions: these are $e^{ir(ih^2)}$ for the two real values $r=-t,t$. Hence right multiplication by $e^{-t h^2}$ is an isometry. Equations (6) and (F2) give

Letting $t\to\infty$ gives $h=0$, as required. The first hypothesis bounds the Gaussian integrand; the second allows the inverse damping operator without increasing the norm. These operations give $\|h\|\le1/\sqrt{\pi t}$ for every $t>0$, and therefore prove the lemma. $\square$

From small lifts to the real tensor bound

The kernel theorem makes a small preimage unique. Before transferring its cost, we identify the real completion inside the complex one. The real projective space embeds in the complex projective space. To see this without assuming a tensor-inclusion theorem, expand any complex decomposition of a self-adjoint tensor as in (6) of the preceding lesson. It costs at most twice the original projective cost, since $\|\operatorname{Re}x\|^2+\|\operatorname{Im}x\|^2\le2\|x\|^2$. Thus

on the algebraic real tensor space, and the inclusion of its completion is injective. Its image is the self-adjoint part of the complex completion: symmetrized algebraic approximants are Cauchy by (8).

Theorem 4.1. On the completed real self-adjoint tensor space,

Every bounded real bilinear form $W$ on $A_{\rm sa}\times B_{\rm sa}$ therefore has $\sigma$-dual norm at most $(81/16)\|W\|$.

Proof. Let $v$ be in the completed real $\sigma$-space. For every $\varepsilon>0$, the preceding random-lift lesson supplies a real projective vector $u_\varepsilon$ with

Theorem 3.1 above and (8) show that the real canonical map is injective. All these $u_\varepsilon$ are therefore one vector $u$. Taking the infimum of its bounds gives $\|u\|_{\pi_{\mathbb R}}\le(81/16)\|v\|_\sigma$. The reverse inequality follows from the contractive canonical map. This identifies the two completed real spaces and proves (9). Evaluation of a bounded real bilinear form on projective decompositions gives $|W(u)|\le\|W\|\|u\|_{\pi_{\mathbb R}}$; (9) gives the dual norm bound and extension to the completion. $\square$

One pair of states for all tests

A finite list of affine inequalities can be encoded by a subset of Euclidean space. Let $Q$ be a nonempty compact convex set and let $F_1,\ldots,F_m$ be continuous real affine functions. Their joint image $K\subset\mathbb R^m$ is compact and convex. Write $y_+$ for the coordinatewise positive part and $y_-=(-y)_+$. If $K$ misses the positive orthant, the continuous function $\|y_-\|^2$ has a strictly positive minimum at $y_0\in K$. Put $z_0=(y_0)_+$ and $w=z_0-y_0=(y_0)_-$. For every $y\in K$, $z_0$ is a possible orthant point, so

The right derivative at zero gives $(y_0-z_0)\cdot(y-y_0)\ge0$. Hence

The nonnegative coefficients $w_j$ therefore give a strictly negative maximum of $\sum_jw_jF_j$. If every finite nonnegative combination of functions in a family $\mathcal F$ has nonnegative maximum, this contradiction proves that each finite list of its closed nonnegative sets has a common point. Compactness of $Q$ gives one point satisfying the entire family. The finite-dimensional minimum uses scalar compactness; the last step uses arbitrary compactness, without a countability or metrizability assumption.

Here $K=\{(-2+t,1-2t):0\le t\le1\}$. Its nearest point to the positive orthant is $y_0=(-6/5,-3/5)$, with $z_0=(0,0)$ and $w=(6/5,3/5)$. The line $w\cdot y=-9/5$ contains the entire segment, while $w\cdot z\ge0$ on the orthant. These are the signs and strict gap in (S1). The picture illustrates one finite test family; it does not represent an infinite-dimensional state space.

Section 8.3 constructs states norming every positive element, including the zero element, on a nonzero unital algebra. It also proves that the state space is a closed subset of the weak-* compact dual ball from Section 9. The product $S(A)\times S(B)$ is compact by the same disk-product construction with the coordinates of both algebras together. The two states norming a finite test's two square sums may depend on that test. The orthant argument will replace those separate choices by one pair that passes every test.

Lemma 5.1. For a real $\sigma$-continuous bilinear form $W$ on self-adjoint pairs in unital $A,B$, there are states $\varphi,\psi$ with

Proof. If $c=0$, the form is zero and any pair of states works. Suppose $c>0$. On $Q=S(A)\times S(B)$, test every self-adjoint pair with

For a finite sum $\sum_j\lambda_jF_{a_j,b_j}$, $\lambda_j\ge0$, replace $a_j,b_j$ by $\sqrt{\lambda_j}a_j,\sqrt{\lambda_j}b_j$. Set $S_A=\|\sum_j a_j^2\|$, $S_B=\|\sum_j b_j^2\|$, and choose states separately norming these two positive sums. The definition of the $\sigma$-dual norm gives $\sum_jW(a_j,b_j)\le c\sqrt{S_A S_B}$. The combined test at those states has value at least $c(S_A+S_B)/2-c\sqrt{S_A S_B}\ge0$.

The finite-orthant minimum and compactness argument therefore give one pair $(\varphi,\psi)$ passing every test. Testing $(a,b)$ and $(-a,b)$ gives

Apply this to $(ta,t^{-1}b)$. If both quadratic values are positive, take $t^4=\psi(b^2)/\varphi(a^2)$. If either value is zero, send $t$ to the corresponding endpoint. This proves (10), with the same states for all pairs. $\square$

The complex form and its two common states

Theorem 6.1. For unital C*-algebras and any bounded complex bilinear form $V$, there are states $\varphi,\psi$ such that

Proof. Put $V_1=(V+V^*)/2$, $V_2=(V-V^*)/(2i)$. Each is self-adjoint and has norm at most $\|V\|$. Its real restriction is $\sigma$-continuous by Theorem 4.1. Lemma 5.1 gives states $\varphi_i,\psi_i$ with constant $K\|V\|$, where $K=81/16$. Expanding $x=a+ic,y=b+id$ and taking real parts gives the same two-term estimate as (3). Rotating $x$ then gives

Sum the bounds for $V_1,V_2$, apply Cauchy–Schwarz to this two-term sum, and set $\varphi=(\varphi_1+\varphi_2)/2$, $\psi=(\psi_1+\psi_2)/2$. The result is $2K\|V\|\sqrt{\varphi(q(x))\psi(q(y))}$, which equals the right side of (12). $\square$

This finite universal constant is the input to the sharp inequality. Its existence, rather than its preliminary numerical size, is what the improvement argument needs.

7. Exercises with complete solutions

The first three checks build the functional that tests a kernel vector. Exercises 4–6 examine the compactness and Gaussian estimates that make the test decisive. Exercises 7–8 transfer a completed lift's cost. Exercises 9–10 turn those norm estimates into common-state bounds.

Exercise 1. Let $V$ be the normalized bilinear form used in Lemma 1.1, with marginal states $\varphi,\psi$, and let $a,b$ be self-adjoint. Test $V$ on $(e^{ita},e^{itb})$. Determine the sign of the second-order bracket in (2), and determine what changes when the first input is replaced by its negative.

Solution. Put $M=\|a\|+\|b\|$. Multiplying the two norm-convergent exponential series through $V$ gives

For $|t|\le1$, group the absolute terms by their total degree $n$. The binomial identity bounds them by $|t|^nM^n/n!$. Since $n!\ge6(n-3)!$ for $n\ge3$,

The first-order coefficient is real before multiplication by $i$, because the marginals are states. For $B=[\varphi(a^2)+\psi(b^2)]/2+\operatorname{Re}V(a,b)$ we therefore have $\operatorname{Re}V(e^{ita},e^{itb})=1-t^2B+\operatorname{Re}R(t)$. Both exponentials have norm one, so their bilinear value has real part at most one. Subtract one and divide by $t^2>0$:

The last quotient tends to zero, proving $B\ge0$. Thus $\operatorname{Re}V(a,b)\ge-[\varphi(a^2)+\psi(b^2)]/2$. For $(-a,b)$, the square terms remain the same and the cross term changes sign. The resulting bracket inequality gives the opposite bound, and hence the absolute-value estimate that reciprocal scaling will sharpen.

Exercise 2. The bilinear test gives an additive quadratic bound, while the desired state bound is multiplicative. For $\alpha,\beta\ge0$, compute the infimum over $t>0$ of

Specify its minimizer when both values are positive and handle every case $\alpha\beta=0$, including whether the infimum is attained.

Solution. The identity

proves $H(t)\ge\sqrt{\alpha\beta}$ for all $t>0$. When $\alpha,\beta>0$, equality occurs precisely at $t^4=\beta/\alpha$, so the positive minimizer is $t=(\beta/\alpha)^{1/4}$ and the minimum is $\sqrt{\alpha\beta}$.

When $\alpha=0<\beta$, $H(t)=\beta/(2t^2)>0$ at each finite positive $t$, but it tends to zero as $t\to\infty$. When $\beta=0<\alpha$, $H(t)=\alpha t^2/2>0$, tending to zero as $t\downarrow0$. Both infima are zero and neither is attained. If $\alpha=\beta=0$, every value of $H$ is zero. Applying these cases to $\alpha=\varphi(a^2)$, $\beta=\psi(b^2)$ proves that a zero quadratic value forces the tested bilinear value to vanish; no division by a zero value is needed.

Exercise 3. A kernel vector is tested by the self-adjoint part of a normalized functional. Let $C$ be an involutive complex Banach algebra whose involution is isometric and whose identity has norm one. If $F\in\mathcal S_C$, verify that

is normalized and self-adjoint.

Solution. The adjoint is complex linear, since