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A local Maxwellian can be written as $\mathcal{M}=\exp (a+b\cdot v+c|v{|}^2)$, with $c<0$ and coefficients depending on $(t,x)$. Its logarithm is a linear combination of collision invariants, so ${\mathcal{M}}^{\prime }{\mathcal{M}}_{\ast }^{\prime }=\mathcal{M}{\mathcal{M}}_{\ast }$ and $Q(\mathcal{M},\mathcal{M})=0$ at each spatial point.

Comparing powers of $v$ in $({\partial }_t+v\cdot {\nabla }_x)\log \mathcal{M}=0$ givesHere $\mathrm{sym}\nabla b$ is the symmetric part of a matrix formed by the coefficients of the quadratic polynomial. For constants $A,\alpha ,\beta >0$, choose $a=\log A-\alpha |x{|}^2$, $b=2\alpha tx$ and $c=-\beta -\alpha t^2$. All four coefficient equations hold. The resulting expanding Gaussian solution of the Boltzmann equation isDirect verification is particularly simple: $x-tv$ and $v$ are constant along free characteristic curves, so the transport derivative is zero. At fixed $(t,x)$ its logarithm is the displayed Maxwellian quadratic, so the collision term is also zero. Its spatial dependence is nonconstant, it is positive and rapidly decaying, and its Gaussian velocity integral is finite because $\beta +\alpha t^2>0$. It is not a compactly supported example, nor does this final part require one.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

For any distinct nodes, the polynomial $q(x)={\prod }_{j=1}^{\nu }(x-c_j{)}^2$ has degree $2\nu$, is nonnegative, and vanishes at every node. The quadrature rule sum for $q$ is zero, but ${\int }_0^{\pi }wqdx>0$ because the weight is positive and $q$ is positive except at finitely many points. Therefore no such rule is exact for all polynomials of degree at most $2\nu$, irrespective of the chosen weights.

For the sine weight, its first four moments are $2,\pi ,{\pi }^2-4,{\pi }^3-6\pi$. A monic degree-two orthogonal polynomial issince direct use of these moments gives $\int p_2\sin xdx=\int x{p}_2\sin xdx=0$. Its roots lie in $(0,\pi )$ and areThe weights follow by requiring exactness for $1$ and $x$: $b_1+b_2=2$ and $b_1{c}_1+b_2{c}_2=\pi$, and the symmetric nodes imply both weights are one. To prove cubic exactness, divide any cubic as $f=a{p}_2+r$, with $a,r$ of degree at most one. The integral of $a{p}_2$ is zero by orthogonality, and its nodal values vanish; the remaining linear polynomial is integrated exactly. This is the two-node Gaussian quadrature with sine weight; the argument proves its Gaussian quadrature property without assuming a general quadrature theorem.

The characteristic equations for a transport equation are $\dot{X}=V$ and $\dot{V}=X$, so $\ddot{X}=X$. For an initial point $(x_0,v_0)$ their solution isThis is the hyperbolic characteristic flow for an inverted oscillator. The addition formulas give $A_s{A}_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward characteristic flow map from the point $(x,v)$ at time $t$ to time $s$ isAlong this characteristic curve, the chain rule changes the transport equation into $d[f(s,S_{s,t}(x,v))]/ds=h(s,S_{s,t}(x,v))$. Integrating from zero to $t$ givesThe assumed $C^1$ regularity makes this a classical solution: on every compact set, the integrand and its needed derivatives are continuous, so differentiation under the finite-time integral is justified. At $t=0$ it has the required initial value, and the characteristic calculation verifies the equation. Conversely every classical solution must satisfy the same integrated identity, proving uniqueness. This is the Duhamel formula for Hamiltonian transport, with Hamiltonian $(v^2-x^2)/2$.

The derivative of the backward characteristic flow map isThus the change of variables formula preserves phase-space Lebesgue measure. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,Taking the $p$th root proves $\Vert f_t{\Vert }_p=\Vert f_0{\Vert }_p$. For $0<p<1$ this is a quasi-norm, and the argument still works because it uses only a change of variables, not the triangle inequality. The identity also holds in the extended sense when an integral is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the essential supremum, so the same conclusion holds for $p=\infty$.

Taking the essential supremum in the characteristic solution gives the sharper estimateIndeed, the bijective characteristic flow map preserves each spatial-velocity essential supremum. If $H_t={\mathrm{esssup}}_{0\le s\le t,x,v\in \mathbb{R}}|h(s,x,v)|$, this proves $\Vert f_t{\Vert }_{\infty }\le \Vert f_0{\Vert }_{\infty }+t{H}_t$. A time-independent source has $H_t=\Vert h{\Vert }_{L^{\infty }({\mathbb{R}}^2)}$, which is the displayed form. For a time-dependent source, the same symbol must mean a bound uniform over the elapsed time interval; its norm at the final time alone need not bound the accumulated forcing.

The bound is sharp. Take $f_0=1$ and $h=1$, giving $f(t,x,v)=1+t$ and equality for every $t\ge 0$. Both functions are smooth and bounded, although their finite-$p$ integrals over the whole plane are infinite.

Use the unit-period circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, with total measure one. The free characteristic flow map givesTranslations in $x$ preserve its periodic measure. The Tonelli theorem and the integral triangle inequality yieldThe Fubini's theorem therefore applies also to signed data, and the same translation givesThis holds for positive or negative time. The printed $L^1(\mathbb{R})$ in this subpart is a domain typo: the spatial variable is periodic, so the correct space is $L^1(\mathbb{T})$. For example, the smooth initial value $f_0(x,v)=e^{-v^2}$ gives the constant density $\sqrt{\pi }$, whose periodic extension is not integrable on the real line.

The Fubini's theorem and integral triangle inequality giveDefine the normalized velocity-reset collision operator using the normalized velocity-reset projection $Pg(x,v)=\rho (g)(x)M(v)$ and $B=P-I$. Since $M\ge 0$ and $\int M=1$,Also $P^2=P$; the collision gain replaces the velocity distribution by $M$ while preserving the spatial mass. In Bochner integral notation the printed, undamped source operator isUsing the transport isometry and the integral triangle inequality provesThese estimates hold for measurable, locally time-bounded $E$-valued functions. If the displayed supremum is infinite, the numerical bound is interpreted in the extended sense; the construction below works in a space where it is finite.

Fix $T$ and set $A_T={\sup }_{0\le s\le T}\Vert f(s){\Vert }_1$. The preceding integral estimate gives the result for one iterate. If for some $n\ge 0$ the bound $\Vert {\tau }^{n}f(s){\Vert }_1\le A_T{2}^n{s}^n/n!$ holds for $s\le T$, thenTaking $T=t$ yields exactlyThe case $n=0$ uses the identity operator. This factorial bound for a Volterra iterate comes from time ordering, so no commutation between free transport and the collision projection is assumed.

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.

For each rotation, unitarity givesAveraging and using self-adjointness of $Q$ yields the Dirichlet form of the Kac collision operatorThe velocity integral is necessary: its omission from the printed right-hand side would leave a function of $\mathbf{v}$ rather than a scalar. This identity applies to every $L^2$ function, with complex modulus when necessary, and is nonnegative.

If $(I-Q)F=0$, every nonnegative angular integral is zero, so $\Vert U_{ij,\theta }F-F{\Vert }_2=0$ for almost every angle. Strong continuity in angle extends equality to every angle. The coordinate-plane Givens rotations generate the special orthogonal group $SO(N)$, hence $F$ is invariant in $L^2$ under every element of this group. To identify its shape rigorously despite almost-everywhere representatives, average over the normalized Haar measure of $SO(N)$. This averaging leaves $F$ unchanged, while transitivity of the rotation group on each sphere makes the average a radial function. Thus $F(\mathbf{v})=\varphi (|\mathbf{v}|)$ almost everywhere.

Conversely, every radial function is fixed by every coordinate-plane rotation, and so by $Q$. ThereforeThis is the radial kernel of the Kac collision operator. The rotation correction is essential to this conclusion: with the literal printed map, even $F(\mathbf{v})=e^{-|\mathbf{v}{|}^2}$ in dimension two is not fixed. At $\mathbf{v}=(0,1)$ its printed-map angular average is the average of $e^{-{\sin }^2\theta }$, strictly greater than its value $e^{-1}$.

Write $F=F_N$ and integrate the Kac master equation over $v_2,\dots ,v_N$. For a pair $i,j\ge 2$, the rotation acts only on integrated variables. Its unit Jacobian determinant makes the integrated gain identical to the integrated loss, so all those pairs cancel.

The only remaining pairs are $(1,j)$, $j=2,\dots ,N$. For such a pair, first integrate over every variable except $v_1$ and $v_j$. This yields the corresponding two-coordinate marginal distribution evaluated at the rotated pair. Permutation symmetry of $F_N$ makes all $N-1$ resulting integrals identical to the one for $(1,2)$. The coefficient isConsequently the Kac marginal evolution equation isThe time argument has been suppressed on the right. The loss is consistent with normalization, since $\int {\Pi }_2(v_1,v_2)d{v}_2={\Pi }_1(v_1)$. Under the printed definition $k<N$, this use of ${\Pi }_2$ requires $N\ge 3$. For $N=2$ the same formula holds with the natural extension ${\Pi }_N(F_N)=F_N$.

This identity is exact and generally unclosed. Replacing the two-coordinate marginal distribution by the product of one-coordinate marginals would produce the quadratic collision equation associated with Kac chaos. Permutation symmetry alone does not imply that product approximation.

Put $u=F_N/{\gamma }_N$, where the Gaussian density is strictly positive. Extend $u\log u$ continuously at zero by $0\log 0=0$. Both densities have integral one, so the relative entropy can be written asThe bracket is nonnegative and vanishes only at $u=1$: its derivative for $u>0$ is $\log u$, with a unique minimum at one. HenceThe inequality holds also for infinite entropy. Its negative integrand part is integrable, since $u\log u\ge -1/e$ and ${\gamma }_N$ has integral one, so the extended-value integral is well defined. This is relative entropy in Kac's model; no differentiation is needed for nonnegativity.