Global finite-energy cutoff collision evolution

The weighted measure $w_{0}dv$ is sigma-finite because $w_0$ is bounded on every bounded ball. To prove $X$ complete, let $(g_n)$ be Cauchy in its norm and choose a subsequence $(g_{n_j})$ with $\mathrm{\parallel }{g}_{n_{j+1}}-g_{n_j}{\mathrm{\parallel }}_{1,w_0}\le 2^{-j}$. By [F5], the nonnegative series ${\sum }_j{w}_0\mathrm{\mid }{g}_{n_{j+1}}-g_{n_j}\mathrm{\mid }$ is finite almost everywhere and has finite integral. Hence $g_{n_j}$ converges pointwise almost everywhere to a measurable $g$, and the integral of each tail is bounded by ${\sum }_{j\ge J}{2}^{-j}$, so $g_{n_j}\to g$ in $X$. The same estimates give $\mathrm{\parallel }g{\mathrm{\parallel }}_{1,w_0}<\mathrm{\infty }$. Since the original sequence was Cauchy, it converges to the same $g$. Thus $X$ is complete.

Put $m_1=\int v{f}_{0}dv$ and $E_2=\int \mathrm{\mid }v{\mathrm{\mid }}^2{f}_{0}dv$, and let $K$ be the subset of $X$ of nonnegative functions with mass $1$, first moment $m_1$, and second moment $E_2$. It is nonempty because $f_0\in K$. It is closed: if $g_n\to g$ in $X$, then $\int \mathrm{\mid }{g}_n-g\mathrm{\mid }\le \mathrm{\parallel }{g}_n-g{\mathrm{\parallel }}_{1,w_0}$, $\int \mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }{g}_n-g\mathrm{\mid }\le (v_0\mathrm{/}2)\mathrm{\parallel }{g}_n-g{\mathrm{\parallel }}_{1,w_0}$, and $\int \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{\mid }{g}_n-g\mathrm{\mid }\le v_0^2\mathrm{\parallel }{g}_n-g{\mathrm{\parallel }}_{1,w_0}$; also the negative part of $g$ is bounded by $\mathrm{\mid }g-g_n\mathrm{\mid }$ whenever $g_n\ge 0$. Thus all constraints pass to the limit, and $K$ is complete. The continuous path space $C([0,T],K)$ with the supremum metric is complete: a uniformly Cauchy sequence has pointwise limits in $K$, converges uniformly, and the triangle inequality with one continuous approximant proves continuity of the limit.

Let $M_0=1+E_2\mathrm{/}{v}_0^2$; every $g\in K$ has $\mathrm{\parallel }g{\mathrm{\parallel }}_{1,w_0}=M_0$. On $C([0,T],K)$ define $\mathrm{\Phi }(g{)}_t=e^{-\lambda t}{f}_0+\lambda {\int }_0^t{e}^{-\lambda (t-s)}\mathcal{G}(g_s,g_s)ds\mathrm{.}$ The integrand is norm-continuous because $s\mapsto g_s$ is continuous and [F2] bounds the bilinear map. Its Riemann sums are Cauchy in $X$ by uniform continuity on $[0,T]$ and completeness from step 1.1. The gain identities in [F2] show that $\mathrm{\Phi }(g{)}_t$ has mass $1$, first moment $m_1$, and second moment $E_2$, since $e^{-\lambda t}+\lambda {\int }_0^t{e}^{-\lambda (t-s)}ds=1$; nonnegativity is preserved as well. Therefore $\mathrm{\Phi }$ maps this path space into itself.

Bilinearity and [F2] give $\mathrm{\parallel }\mathcal{G}(g_s,g_s)-\mathcal{G}(h_s,h_s){\mathrm{\parallel }}_{1,w_0}\le 2\mathrm{\parallel }{g}_s-h_s{\mathrm{\parallel }}_{1,w_0}(\mathrm{\parallel }{g}_s{\mathrm{\parallel }}_{1,w_0}+\mathrm{\parallel }{h}_s{\mathrm{\parallel }}_{1,w_0})=4M_0\mathrm{\parallel }{g}_s-h_s{\mathrm{\parallel }}_{1,w_0}\mathrm{.}$ Hence $\mathrm{\parallel }\mathrm{\Phi }(g)-\mathrm{\Phi }(h){\mathrm{\parallel }}_{C([0,T],X)}\le 4\lambda M_{0}T\mathrm{\parallel }g-h{\mathrm{\parallel }}_{C([0,T],X)}$. Choose $T=(8\lambda M_0{)}^{-1}$, so this constant is $1\mathrm{/}2$. The nonempty complete path space from step 2.1 and [F4] yield a unique fixed point on $[0,T]$. Differentiating its norm-continuous mild integral gives $f_t^{\mathrm{\prime }}=\lambda (\mathcal{G}(f_t,f_t)-f_t)$ in $X$.

Suppose $f_0\in {\mathcal{M}}_{u,a,b}$. The class is closed and convex by definition. If a path takes values in this class, [F3] puts every gain value in the class. For a partition $0=s_0<\cdots <s_N=t$, assign each subinterval $I_j=[s_{j-1},s_j]$ the exact nonnegative weight $\lambda {\int }_{I_j}{e}^{-\lambda (t-s)}ds$ and evaluate the gain at one point of that interval; these weights sum to $1-e^{-\lambda t}$. Adding the weight $e^{-\lambda t}$ on $f_0$ gives convex combinations in the class. As the mesh tends to zero, the Riemann sums converge in $X$ to $\mathrm{\Phi }(g{)}_t$, so closedness puts $\mathrm{\Phi }(g{)}_t$ in the class. Thus Picard iteration of $\mathrm{\Phi }$ starting at the constant path $f_0$ remains in ${\mathcal{M}}_{u,a,b}$, and [F4] makes its uniform limit the fixed point. Therefore every $f_t$ lies in this class and has the smooth positive representative and Gaussian envelopes from [F3].

At $t=T$ the solution is again in $K$. Restart the same construction with initial value $f_T$; the same $M_0$ gives the same positive interval length. Iterating constructs a solution on every $[0,kT]$, and uniqueness on overlaps makes the concatenation a unique global mild solution. Substituting the restarted mild formula into the preceding interval gives the displayed global formula. Since the gain is nonnegative, the formula also gives $f_t\ge e^{-\lambda t}{f}_0$ almost everywhere.

On any compact time interval, the uniform spatial derivative bounds from [F3] make the family of representatives locally relatively compact in $C^{\mathrm{\infty }}$. If $t_j\to t$, continuity in $X$ gives $f_{t_j}\to f_t$ in weighted $L^1$. Every subsequence has a further locally uniformly convergent subsequence by [F3]; its limit equals almost everywhere the continuous representative of $f_t$, hence equals it everywhere. A subsequence on which local uniform convergence failed would contradict this property, so $t\mapsto f_t$ is locally uniformly continuous in velocity. For $v$ in a compact velocity set, the gain integrand $f_s(v^{\mathrm{\prime }})f_s(w^{\mathrm{\prime }})$ is continuous in $(s,v)$, and pair-energy preservation gives the integrable majorant $C{e}^{-\mathrm{\mid }w-u{\mathrm{\mid }}^2\mathrm{/}(2b)}$, uniform on that compact set and compact time intervals. Dominated convergence [F6] therefore makes the pointwise gain integral finite and jointly continuous in $(s,v)$; it is a continuous representative of the $X$-class gain. In particular, for fixed $v$ this supplies the integrable majorant $C(v)e^{-\mathrm{\mid }w-u{\mathrm{\mid }}^2\mathrm{/}(2b)}$ uniformly on compact time intervals.

For each fixed $t$, the Riemann sums defining the mild integral converge in $X$; their pointwise limits are the ordinary time integral because the gain is continuous in time by step 5.1. Choose a subsequence whose successive $X$-norm differences are summable; the Tonelli argument in step 1.1, using [F5], shows it converges almost everywhere to the $X$-class limit. Hence the pointwise time integral represents the mild right-hand side. It is continuous in velocity by the same compact-velocity Gaussian majorant in step 5.1. The left side has the smooth representative from step 3.2. Since these continuous representatives agree as $X$ classes, they agree everywhere; a continuous function vanishing almost everywhere vanishes everywhere. The ordinary fundamental theorem of calculus applied to the pointwise mild integral gives ${\mathrm{\partial }}_t{f}_t(v)=\lambda (\mathcal{G}(f_t,f_t)(v)-f_t(v))$. This proves the stated pointwise $C^1$ time regularity.