Abstract

Analytic structure, recovery, and closure

The recovery identities and integral normalization determine one σ uniquely on I = (−1/γ, ∞), where μ > 0 and γ > μ. Its explicit differential hierarchy, exact scale invariance, transform recovery, dimensional self-closure, and spherical closure follow simultaneously. The Laplace transform returns μ and γ globally, and exact Lambert inversion has two real branches that meet at the unique closure.

Dimensional self-closure is the absence of a residual inverse-square term. It selects D = 3 uniquely among integer spatial dimensions D ≥ 2. In three dimensions, the sphere x₁² + x₂² + x₃² = 1/μ − 1/γ is the complete closure set. No amplitude, finite differential data, scale prescription, spatial dimension under dimensional self-closure, or closure geometry remains independently specifiable.

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A. Albert (2026)

BibTeX 
@misc{albert2026universalinvariant,
  author       = {Albert, Alex},
  title        = {The Universal Invariant of Physical Reality},
  year         = {2026},
  publisher    = {Mathematical Research Institute of Physical Reality},
  address      = {Pécs, Hungary},
  url          = {https://www.mripr.org/en/research/the-universal-invariant-of-physical-reality/},
  note         = {Published 25 July 2026}
}
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Mathematical core

One invariant, in logical order

Every statement and proof below concerns the same σ. The numbered separation records proof dependencies and introduces no independent objects. The final theorem collects the normalization, recovery, rigidity, scale, dimensional, and spatial results proved by the sequence.

1

Definition and domain

Let

μ>0,γ>μ,\mu>0,\qquad \gamma>\mu,

and define

I=(−1γ,∞).I=\left(-\frac{1}{\gamma},\infty\right).

For r ∈ I, let

σ(r)=ln⁡ ⁣(μ2μ+γ)+ln⁡(1+γr)−μr.\sigma(r)=\ln\!\left(\frac{\mu^2}{\mu+\gamma}\right)+\ln(1+\gamma r)-\mu r.

Its exponential representation is

eσ(r)=μ2μ+γ(1+γr)e−μr.e^{\sigma(r)}=\frac{\mu^2}{\mu+\gamma}(1+\gamma r)e^{-\mu r}.

Equation (2) follows directly from (1), and (1) is recovered from (2) by the real logarithm.

μ>0\mu>0

gives convergence of the normalization integral.

γ>0\gamma>0

places [0, ∞) inside the logarithmic domain, and

γ>μ⟺r∗=1μ−1γ>0.\gamma>\mu\quad\Longleftrightarrow\quad r_*=\frac{1}{\mu}-\frac{1}{\gamma}>0.

The condition γ > μ places closure at a positive spatial radius.

2

Normalization

Proposition 1

The exponential representation is normalized on [0, ∞):

∫0∞eσ(r) dr=1.\int_0^\infty e^{\sigma(r)}\,dr=1.

The prefactor μ²/(μ + γ) is the unique normalizing factor. No independent amplitude remains.

Proof.

Since μ > 0,

∫0∞e−μr dr=1μ\int_0^\infty e^{-\mu r}\,dr=\frac{1}{\mu}

and

∫0∞re−μr dr=1μ2.\int_0^\infty r e^{-\mu r}\,dr=\frac{1}{\mu^2}.

Combining these integrals gives

∫0∞eσ(r) dr=μ2μ+γ∫0∞(1+γr)e−μr dr=μ2μ+γ(1μ+γμ2)=1.\begin{aligned} \int_0^\infty e^{\sigma(r)}\,dr &=\frac{\mu^2}{\mu+\gamma}\int_0^\infty(1+\gamma r)e^{-\mu r}\,dr\\ &=\frac{\mu^2}{\mu+\gamma}\left(\frac{1}{\mu}+\frac{\gamma}{\mu^2}\right)\\ &=1. \end{aligned}

Also,

∫0∞(1+γr)e−μr dr=μ+γμ2.\int_0^\infty(1+\gamma r)e^{-\mu r}\,dr=\frac{\mu+\gamma}{\mu^2}.

Its reciprocal is μ²/(μ + γ), so no other constant normalizes the expression.

Q.E.D.

3

Differential hierarchy

Proposition 2

The logarithmic representation σ is real analytic on I. Its first two derivatives with respect to r are

σ′(r)=γ1+γr−μ\sigma'(r)=\frac{\gamma}{1+\gamma r}-\mu

and

σ′′(r)=−γ2(1+γr)2.\sigma''(r)=-\frac{\gamma^2}{(1+\gamma r)^2}.

For every integer n ≥ 2 and every r ∈ I,

σ(n)(r)=(−1)n−1(n−1)!γn(1+γr)n.\sigma^{(n)}(r)=(-1)^{n-1}(n-1)!\frac{\gamma^n}{(1+\gamma r)^n}.

Proof.

Equation (4) follows by differentiating (1). Differentiating (4) gives (5). Repeated differentiation of ln(1 + γr) gives (6). The constant and linear terms vanish after the first derivative.

Q.E.D.

Corollary 1

The logarithmic representation is strictly concave with respect to r on I. The exponential representation is strictly log-concave and need not be concave.

Proof.

Equation (5) gives

σ′′(r)<0\sigma''(r)<0

for every r ∈ I. Since ln(eσ(r)) = σ(r), the exponential representation is strictly log-concave. Its second derivative is

d2dr2eσ(r)=eσ(r)[(σ′(r))2+σ′′(r)].\frac{d^2}{dr^2}e^{\sigma(r)}=e^{\sigma(r)}\left[(\sigma'(r))^2+\sigma''(r)\right].

The bracket equals μ² − 2μγ/(1 + γr). It is negative at r∗ and tends to μ² > 0 as r tends to infinity, so it has no fixed sign on I.

Q.E.D.

Proposition 3

For every integer n ≥ 3 and every r ∈ I,

(1+γr)σ(n)(r)+(n−1)γσ(n−1)(r)=0.(1+\gamma r)\sigma^{(n)}(r)+(n-1)\gamma\sigma^{(n-1)}(r)=0.

For every r ∈ I, the second-order relation is

(1+γr)σ′′(r)+γ(σ′(r)+μ)=0.(1+\gamma r)\sigma''(r)+\gamma(\sigma'(r)+\mu)=0.

Proof.

Substitute (6) into the left-hand side of (8). The two terms cancel. Equation (9) follows from

σ′(r)+μ=γ1+γr\sigma'(r)+\mu=\frac{\gamma}{1+\gamma r}

and (5).

Q.E.D.

4

Closure

Theorem 1

The logarithmic representation has one stationary point,

r∗=1μ−1γ=γ−μμγ.r_*=\frac{1}{\mu}-\frac{1}{\gamma}=\frac{\gamma-\mu}{\mu\gamma}.

It is the unique global maximum on I.

Proof.

By (4), the stationary equation is

γ1+γr=μ.\frac{\gamma}{1+\gamma r}=\mu.

Solving gives (10). Since γ > μ > 0,

r∗>0.r_*>0.

Strict concavity implies that there is at most one stationary point and that any stationary point is a strict maximum. At the two ends of I,

lim⁡r→−1/γ+σ(r)=−∞,lim⁡r→∞σ(r)=−∞.\lim_{r\to-1/\gamma^+}\sigma(r)=-\infty,\qquad \lim_{r\to\infty}\sigma(r)=-\infty.

The maximum is global.

Q.E.D.

At closure,

1+γr∗=γμ,1+\gamma r_*=\frac{\gamma}{\mu},

so

σ′′(r∗)=−μ2.\sigma''(r_*)=-\mu^2.

For n ≥ 2, equation (6) gives

σ(n)(r∗)=(−1)n−1(n−1)!μn.\sigma^{(n)}(r_*)=(-1)^{n-1}(n-1)!\mu^n.

The value at closure is

σ(r∗)=ln⁡ ⁣(μγμ+γ)−1+μγ,\sigma(r_*)=\ln\!\left(\frac{\mu\gamma}{\mu+\gamma}\right)-1+\frac{\mu}{\gamma},

and therefore

eσ(r∗)=μγμ+γe−1+μ/γ.e^{\sigma(r_*)}=\frac{\mu\gamma}{\mu+\gamma}e^{-1+\mu/\gamma}.

Corollary 2

The exponential representation has the same unique maximizing value r∗.

Proof.

Since eσ(r) > 0,

ddreσ(r)=eσ(r)σ′(r).\frac{d}{dr}e^{\sigma(r)}=e^{\sigma(r)}\sigma'(r).

Its stationary points are the stationary points of σ. The exponential function preserves the ordering of real numbers.

Q.E.D.

5

Recovery

Theorem 2

For every r ∈ I,

γ=−σ′′(r)1−r−σ′′(r)\gamma=\frac{\sqrt{-\sigma''(r)}}{1-r\sqrt{-\sigma''(r)}}

and

μ=−σ′(r)+−σ′′(r).\mu=-\sigma'(r)+\sqrt{-\sigma''(r)}.

The identities recover μ and γ from σ itself at every r ∈ I.

Proof.

Equation (5), together with γ > 0 and 1 + γr > 0, gives

−σ′′(r)=γ1+γr.\sqrt{-\sigma''(r)}=\frac{\gamma}{1+\gamma r}.

It follows that

1−r−σ′′(r)=1−γr1+γr=11+γr>0.1-r\sqrt{-\sigma''(r)}=1-\frac{\gamma r}{1+\gamma r}=\frac{1}{1+\gamma r}>0.

Substitution into (16) returns γ. Combining (18) with (4) gives (17).

Q.E.D.

Proposition 4

The same γ is recovered by

∫0∞−σ′′(r) dr=γ.\int_0^\infty-\sigma''(r)\,dr=\gamma.

Proof.

Using (5),

∫0∞−σ′′(r) dr=∫0∞γ2(1+γr)2 dr=[−γ1+γr]0∞=γ.\begin{aligned} \int_0^\infty-\sigma''(r)\,dr &=\int_0^\infty\frac{\gamma^2}{(1+\gamma r)^2}\,dr\\ &=\left[-\frac{\gamma}{1+\gamma r}\right]_0^\infty\\ &=\gamma. \end{aligned}

Equivalently,

∫0∞−σ′′(r) dr=σ′(0)−lim⁡r→∞σ′(r)=(γ−μ)−(−μ).\int_0^\infty-\sigma''(r)\,dr=\sigma'(0)-\lim_{r\to\infty}\sigma'(r)=(\gamma-\mu)-(-\mu).

Q.E.D.

6

Rigidity

Theorem 3

Let u ∈ C²(I). Assume

u′′(r)<0u''(r)<0

and

1−r−u′′(r)≠01-r\sqrt{-u''(r)}\ne0

on I. Suppose there are constants μ > 0 and γ > μ such that

−u′′(r)1−r−u′′(r)=γ\frac{\sqrt{-u''(r)}}{1-r\sqrt{-u''(r)}}=\gamma

and

−u′(r)+−u′′(r)=μ-u'(r)+\sqrt{-u''(r)}=\mu

for every r ∈ I, and suppose

∫0∞eu(r) dr=1.\int_0^\infty e^{u(r)}\,dr=1.

then

u(r)=ln⁡ ⁣(μ2μ+γ)+ln⁡(1+γr)−μru(r)=\ln\!\left(\frac{\mu^2}{\mu+\gamma}\right)+\ln(1+\gamma r)-\mu r

for every r ∈ I.

Proof.

Equation (20) gives

−u′′(r)=γ(1−r−u′′(r)).\sqrt{-u''(r)}=\gamma\left(1-r\sqrt{-u''(r)}\right).

Collecting the terms containing the square root gives

−u′′(r)=γ1+γr.\sqrt{-u''(r)}=\frac{\gamma}{1+\gamma r}.

The denominator is positive because r ∈ I. Substituting (24) into (21) yields

u′(r)=γ1+γr−μ.u'(r)=\frac{\gamma}{1+\gamma r}-\mu.

Integration gives

u(r)=u(0)+ln⁡(1+γr)−μr.u(r)=u(0)+\ln(1+\gamma r)-\mu r.

Normalization now gives

1=eu(0)∫0∞(1+γr)e−μr dr=eu(0)μ+γμ2.\begin{aligned} 1 &=e^{u(0)}\int_0^\infty(1+\gamma r)e^{-\mu r}\,dr\\ &=e^{u(0)}\frac{\mu+\gamma}{\mu^2}. \end{aligned}

Normalization fixes the value

u(0)=ln⁡ ⁣(μ2μ+γ),u(0)=\ln\!\left(\frac{\mu^2}{\mu+\gamma}\right),

which gives u = σ on I.

Q.E.D.

Corollary 3

Let u satisfy the differential hypotheses of Theorem 3. The recovery identities (20) and (21), together with normalization (22), are an exact characterization:

u=σ  on I⟺(20), (21), and (22) hold on their stated domains.u=\sigma\ \text{ on }I\quad\Longleftrightarrow\quad\text{(20), (21), and (22) hold on their stated domains}.

Proof.

If u = σ on I, Theorem 2 gives (20) and (21), while Proposition 1 gives (22). Conversely, Theorem 3 gives u = σ on I.

Q.E.D.

7

Parameter uniqueness

Theorem 4

Let (μ1, γ1) and (μ2, γ2) satisfy

μi>0,γi>μi,i∈{1,2}.\mu_i>0,\qquad \gamma_i>\mu_i,\qquad i\in\{1,2\}.

Suppose their canonical exponential representations agree on a nonempty open interval contained in

(−1γ1,∞)∩(−1γ2,∞).\left(-\frac{1}{\gamma_1},\infty\right)\cap\left(-\frac{1}{\gamma_2},\infty\right).

Then

μ1=μ2,γ1=γ2.\mu_1=\mu_2,\qquad \gamma_1=\gamma_2.

Proof.

Injectivity of the exponential function gives equality of the logarithmic representations. Equality of their second derivatives gives

γ12(1+γ1r)2=γ22(1+γ2r)2.\frac{\gamma_1^2}{(1+\gamma_1r)^2}=\frac{\gamma_2^2}{(1+\gamma_2r)^2}.

All quantities in the denominators are positive on the common interval. Taking positive square roots gives

γ11+γ1r=γ21+γ2r.\frac{\gamma_1}{1+\gamma_1r}=\frac{\gamma_2}{1+\gamma_2r}.

Cross-multiplication yields

γ1+γ1γ2r=γ2+γ1γ2r,\gamma_1+\gamma_1\gamma_2r=\gamma_2+\gamma_1\gamma_2r,

so

γ1=γ2.\gamma_1=\gamma_2.

Equality of the first derivatives then gives

μ1=μ2.\mu_1=\mu_2.

Q.E.D.

8

Finite differential exhaustion

For a fixed integer k ≥ 0, let O be a fixed smooth function of the arguments displayed below. A finite-order local differential expression has the form

O ⁣(r;u(r),u′(r),…,u(k)(r)),O\!\left(r;u(r),u'(r),\ldots,u^{(k)}(r)\right),

Theorem 5

Let u satisfy the hypotheses of Theorem 3. For every fixed integer k ≥ 0, every finite-order local differential expression

O ⁣(r;u(r),u′(r),…,u(k)(r))O\!\left(r;u(r),u'(r),\ldots,u^{(k)}(r)\right)

is determined entirely by r, μ, and γ. No additional finite local differential data remain.

Proof.

Rigidity gives u = σ on I. Equations (1), (4), and the explicit derivative formula (6) determine every displayed argument from r, μ, and γ. Substitution into O proves the statement.

Q.E.D.

9

Exact scale invariance

In this section only, display the parameter dependence as

σ(r;μ,γ).\sigma(r;\mu,\gamma).

Theorem 6

For every λ > 0 and r > −1/γ,

σ ⁣(rλ;λμ,λγ)=σ(r;μ,γ)+ln⁡λ.\sigma\!\left(\frac r\lambda;\lambda\mu,\lambda\gamma\right)=\sigma(r;\mu,\gamma)+\ln\lambda.

For r > −1/(λγ), equivalently,

σ(r;λμ,λγ)=σ(λr;μ,γ)+ln⁡λ.\sigma(r;\lambda\mu,\lambda\gamma)=\sigma(\lambda r;\mu,\gamma)+\ln\lambda.

Proof.

Using (1),

σ(r;λμ,λγ)=ln⁡ ⁣((λμ)2λμ+λγ)+ln⁡(1+λγr)−λμr=ln⁡λ+ln⁡ ⁣(μ2μ+γ)+ln⁡(1+γ(λr))−μ(λr)=σ(λr;μ,γ)+ln⁡λ.\begin{aligned} \sigma(r;\lambda\mu,\lambda\gamma) &=\ln\!\left(\frac{(\lambda\mu)^2}{\lambda\mu+\lambda\gamma}\right)+\ln(1+\lambda\gamma r)-\lambda\mu r\\ &=\ln\lambda+\ln\!\left(\frac{\mu^2}{\mu+\gamma}\right)+\ln(1+\gamma(\lambda r))-\mu(\lambda r)\\ &=\sigma(\lambda r;\mu,\gamma)+\ln\lambda. \end{aligned}

Replacing r by r/λ gives (25).

Q.E.D.

Corollary 4

For r > −1/(λγ), the exponential representation satisfies

eσ(r;λμ,λγ)=λeσ(λr;μ,γ).e^{\sigma(r;\lambda\mu,\lambda\gamma)}=\lambda e^{\sigma(\lambda r;\mu,\gamma)}.

For r > −1/γ, the normalized differential identity is

eσ(r/λ;λμ,λγ) d ⁣(rλ)=eσ(r;μ,γ) dr.e^{\sigma(r/\lambda;\lambda\mu,\lambda\gamma)}\,d\!\left(\frac r\lambda\right)=e^{\sigma(r;\mu,\gamma)}\,dr.

Closure scales according to

r∗(λμ,λγ)=r∗(μ,γ)λ.r_*(\lambda\mu,\lambda\gamma)=\frac{r_*(\mu,\gamma)}{\lambda}.

Successive positive scale transformations by λ1 and λ2 equal the direct transformation by λ1λ2.

Proof.

Equation (27) is the exponential of (26). Exponentiating (25) and multiplying by d(r/λ) = dr/λ gives (28). Equation (10) evaluated at λμ and λγ gives (29). For λ1, λ2 > 0, two applications of (26) give

σ(r;λ1λ2μ,λ1λ2γ)=σ(λ2r;λ1μ,λ1γ)+ln⁡λ2=σ(λ1λ2r;μ,γ)+ln⁡λ1+ln⁡λ2=σ(λ1λ2r;μ,γ)+ln⁡(λ1λ2).\begin{aligned} \sigma(r;\lambda_1\lambda_2\mu,\lambda_1\lambda_2\gamma) &=\sigma(\lambda_2r;\lambda_1\mu,\lambda_1\gamma)+\ln\lambda_2\\ &=\sigma(\lambda_1\lambda_2r;\mu,\gamma)+\ln\lambda_1+\ln\lambda_2\\ &=\sigma(\lambda_1\lambda_2r;\mu,\gamma)+\ln(\lambda_1\lambda_2). \end{aligned}

This is the direct transformation by λ1λ2.

Q.E.D.

10

Transform recovery

Theorem 7

For a complex variable s with Re(s) > −μ,

∫0∞e−sreσ(r) dr=μ2μ+γs+μ+γ(s+μ)2.\int_0^\infty e^{-sr}e^{\sigma(r)}\,dr=\frac{\mu^2}{\mu+\gamma}\frac{s+\mu+\gamma}{(s+\mu)^2}.

Its meromorphic continuation has a double pole at s = −μ and a zero at s = −(μ + γ). The pole recovers μ, and the pole-zero separation recovers γ.

Proof.

Using (2),

∫0∞e−sreσ(r) dr=μ2μ+γ∫0∞(1+γr)e−(s+μ)r dr=μ2μ+γ(1s+μ+γ(s+μ)2)=μ2μ+γs+μ+γ(s+μ)2.\begin{aligned} \int_0^\infty e^{-sr}e^{\sigma(r)}\,dr &=\frac{\mu^2}{\mu+\gamma}\int_0^\infty(1+\gamma r)e^{-(s+\mu)r}\,dr\\ &=\frac{\mu^2}{\mu+\gamma}\left(\frac{1}{s+\mu}+\frac{\gamma}{(s+\mu)^2}\right)\\ &=\frac{\mu^2}{\mu+\gamma}\frac{s+\mu+\gamma}{(s+\mu)^2}. \end{aligned}

At s = −μ the numerator equals γ > 0, so the pole is double. The numerator vanishes at s = −(μ + γ). Their separation is γ.

Q.E.D.

On [−1/e, 0), W0 is the real Lambert branch with values in [−1, 0), and W−1 is the real Lambert branch with values in (−∞, −1].

Theorem 8

Let

0<y≤eσ(r∗).0<y\le e^{\sigma(r_*)}.

The real solutions of

eσ(r)=ye^{\sigma(r)}=y

on I are

r=−1μWk ⁣(−μγyeσ(0)e−μ/γ)−1γ,k∈{0,−1}.r=-\frac{1}{\mu}W_k\!\left(-\frac{\mu}{\gamma}\frac{y}{e^{\sigma(0)}}e^{-\mu/\gamma}\right)-\frac{1}{\gamma},\qquad k\in\{0,-1\}.

For 0 < y < eσ(r∗), W0 gives the solution below r∗, and W−1 gives the solution above r∗. At y = eσ(r∗), the branches meet at r∗.

Proof.

Equation (31) gives

yeσ(0)=(1+γr)e−μr.\frac{y}{e^{\sigma(0)}}=(1+\gamma r)e^{-\mu r}.

Equivalently,

−μγyeσ(0)e−μ/γ=−μγ(1+γr)exp⁡ ⁣[−μγ(1+γr)].-\frac{\mu}{\gamma}\frac{y}{e^{\sigma(0)}}e^{-\mu/\gamma}=-\frac{\mu}{\gamma}(1+\gamma r)\exp\!\left[-\frac{\mu}{\gamma}(1+\gamma r)\right].

Applying Wk and solving for r gives (32). At closure, (15) makes the Lambert argument −1/e. For arguments in (−1/e, 0), W0 lies in (−1, 0) and W−1 lies below −1, which gives the stated positions relative to r∗.

Q.E.D.

11

Dimensional identity and self-closure

Let x > 0, set

r=x2,r=x^2,

and retain

f(x)=eσ(x2)/2.f(x)=e^{\sigma(x^2)/2}.

Theorem 9 — Dimensional self-closure

For every positive integer D,

f′′(x)f(x)+D−1xf′(x)f(x)=r(σ′(r))2+2rσ′′(r)+Dσ′(r).\frac{f''(x)}{f(x)}+\frac{D-1}{x}\frac{f'(x)}{f(x)}=r\bigl(\sigma'(r)\bigr)^2+2r\sigma''(r)+D\sigma'(r).

Also,

(x(D−1)/2f(x))′′x(D−1)/2f(x)−[f′′(x)f(x)+D−1xf′(x)f(x)]=(D−1)(D−3)4x2.\frac{(x^{(D-1)/2}f(x))''}{x^{(D-1)/2}f(x)}-\left[\frac{f''(x)}{f(x)}+\frac{D-1}{x}\frac{f'(x)}{f(x)}\right]=\frac{(D-1)(D-3)}{4x^2}.

Exact dimensional self-closure is the absence of a residual inverse-square term:

(x(D−1)/2f(x))′′x(D−1)/2f(x)=f′′(x)f(x)+D−1xf′(x)f(x)(x>0).\frac{(x^{(D-1)/2}f(x))''}{x^{(D-1)/2}f(x)}=\frac{f''(x)}{f(x)}+\frac{D-1}{x}\frac{f'(x)}{f(x)}\qquad(x>0).

The dimensional identity then gives

(D−1)(D−3)=0.(D-1)(D-3)=0.

The solutions are

D=1orD=3.D=1\qquad\text{or}\qquad D=3.

Among integer spatial dimensions D ≥ 2, the unique solution is

D=3.D=3.

Proof.

Equation (33) gives

f′(x)f(x)=xσ′(r).\frac{f'(x)}{f(x)}=x\sigma'(r).

Differentiating and using r = x² gives

f′′(x)f(x)=σ′(r)+2rσ′′(r)+r(σ′(r))2.\frac{f''(x)}{f(x)}=\sigma'(r)+2r\sigma''(r)+r\bigl(\sigma'(r)\bigr)^2.

Adding

D−1xf′(x)f(x)=(D−1)σ′(r)\frac{D-1}{x}\frac{f'(x)}{f(x)}=(D-1)\sigma'(r)

gives (34). Direct differentiation gives

(x(D−1)/2f(x))′′x(D−1)/2f(x)=f′′(x)f(x)+D−1xf′(x)f(x)+(D−1)(D−3)4x2,\frac{(x^{(D-1)/2}f(x))''}{x^{(D-1)/2}f(x)}=\frac{f''(x)}{f(x)}+\frac{D-1}{x}\frac{f'(x)}{f(x)}+\frac{(D-1)(D-3)}{4x^2},

which gives (35). Exact dimensional self-closure holds precisely when (D − 1)(D − 3) = 0. Its integer solutions are (36), and restriction to D ≥ 2 gives (37).

Q.E.D.

12

Three-dimensional spherical closure

For D = 3, let

r=x12+x22+x32.r=x_1^2+x_2^2+x_3^2.

Theorem 10

The spatial evaluation

σ(x12+x22+x32)\sigma(x_1^2+x_2^2+x_3^2)

has the complete global maximizing set

x12+x22+x32=r∗=1μ−1γ.x_1^2+x_2^2+x_3^2=r_*=\frac{1}{\mu}-\frac{1}{\gamma}.

The radius is

1μ−1γ.\sqrt{\frac{1}{\mu}-\frac{1}{\gamma}}.

Write

x=(x1,x2,x3)T,∥x∥2=r.\mathbf{x}=(x_1,x_2,x_3)^{\mathsf T},\qquad \lVert\mathbf{x}\rVert^2=r.

The spatial gradient is

∇σ(∥x∥2)=2x σ′(r).\nabla\sigma(\lVert\mathbf{x}\rVert^2)=2\mathbf{x}\,\sigma'(r).

With I3 the 3 × 3 identity matrix, the Hessian is

∇2σ(∥x∥2)=2σ′(r)I3+4σ′′(r)xxT,r=∥x∥2.\nabla^2\sigma(\lVert\mathbf{x}\rVert^2)=2\sigma'(r)I_3+4\sigma''(r)\mathbf{x}\mathbf{x}^{\mathsf T},\qquad r=\lVert\mathbf{x}\rVert^2.

At the origin,

∇2σ(∥x∥2)∣x=0=2(γ−μ)I3>0.\left.\nabla^2\sigma(\lVert\mathbf{x}\rVert^2)\right|_{\mathbf{x}=0}=2(\gamma-\mu)I_3>0.

The origin is a strict local minimum and is not closure. On the closure sphere,

σ′(r∗)=0.\sigma'(r_*)=0.

The Hessian has two zero tangential eigenvalues. Its radial eigenvalue is

4r∗σ′′(r∗)=−4μ2r∗<0.4r_*\sigma''(r_*)=-4\mu^2r_*<0.

The sphere is the complete global maximizing set. The spatial evaluation is constant in the tangential directions and strictly maximal in the radial direction.

Proof.

The map

(x1,x2,x3)⟼x12+x22+x32(x_1,x_2,x_3)\longmapsto x_1^2+x_2^2+x_3^2

has image [0, ∞). Theorem 1 gives the unique maximum at r = r∗, so its complete spatial preimage is (39), with radius (40). The chain rule gives (41), and a second differentiation gives (42). At the origin, σ′(0) = γ − μ, which gives (43). On the closure sphere σ′(r∗) = 0, so the two tangential eigenvalues vanish. Equation (12) gives the radial eigenvalue (44), which is negative because r∗ > 0.

Q.E.D.

13

Theorem — Complete closure

Theorem 11

Let

μ>0,γ>μ,I=(−1γ,∞),\mu>0,\qquad\gamma>\mu,\qquad I=\left(-\frac1\gamma,\infty\right),

and let

σ(r)=ln⁡ ⁣(μ2μ+γ)+ln⁡(1+γr)−μr.\sigma(r)=\ln\!\left(\frac{\mu^2}{\mu+\gamma}\right)+\ln(1+\gamma r)-\mu r.

The same uniquely determined σ simultaneously satisfies all eleven conclusions below.

  1. Its normalization is unique, and no independent amplitude remains.

  2. Its derivative is explicit at every finite order.

  3. Its parameters μ and γ are recovered pointwise on I.

  4. Recovery and normalization characterize σ exactly by rigidity.

  5. Its canonical parameters are unique.

  6. Every finite local differential expression is exhausted by r, μ, and γ.

  7. It has exact scale invariance, and successive positive scale transformations close under multiplication.

  8. Its Laplace transform recovers μ and γ globally.

  9. Its exact Lambert inversion has two real branches that meet at closure.

  10. Under dimensional self-closure, D = 3 is the unique integer spatial dimension D ≥ 2.

  11. In three dimensions, its complete closure set is the sphere (39).

Proof.

Proposition 1 and equation (3) establish clause 1. Propositions 2 and 3, with equations (4)–(9), establish clause 2. Theorem 2 and Proposition 4, with equations (16)–(19), establish clause 3. Theorem 3 and Corollary 3, with equations (20)–(24), establish clause 4. Theorem 4 establishes clause 5. Theorem 5 establishes clause 6. Theorem 6 and Corollary 4, with equations (25)–(29), establish clause 7. Theorem 7 and equation (30) establish clause 8. Theorem 8 and equations (31)–(32) establish clause 9. Theorem 9 and equations (33)–(37) establish clause 10. Theorem 10 and equations (38)–(44) establish clause 11.

Q.E.D.

14

Exhaustion of independent data

Corollary 5

  • Normalization fixes the amplitude.
  • The local second-order structure recovers μ and γ.
  • Rigidity fixes the complete analytic expression.
  • The analytic expression fixes every finite derivative.
  • The scale law fixes every positive rescaling.
  • The transform returns the same parameters globally.
  • Dimensional self-closure fixes D = 3 among integer spatial dimensions D ≥ 2.
  • Three-dimensional spatial closure fixes the sphere.

No amplitude, parameter information, finite local differential structure, scale prescription, spatial dimension, or closure geometry remains independently specifiable.

Proof.

These are clauses 1, 3, 4, 2, 7, 8, 10, and 11 of Theorem 11, respectively.

Q.E.D.