Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 3

Use the weighted Hilbert structure of velocity-reset relaxation, with inner product $⟨f,g⟩=\int f\overline{g}/M$ and norm $\Vert f{\Vert }_H$. The Cauchy-Schwarz inequality givesThus $\rho (f)$ is an absolutely convergent integral and a bounded linear functional. Since $\Vert M{\Vert }_H^2=\int M=1$, the normalized velocity-reset collision operator obeysThis establishes boundedness. The subsequent orthogonal decomposition improves the operator norm to exactly one.

For the normalized velocity-reset collision operator,This bounded everywhere-defined operator is self-adjoint. Put $h=f/M$. Expanding the square with the probability measure $M(v)dv$ givesIt follows thatReplacing the difference by $(f(v_{\ast })M(v)-f(v)M(v_{\ast }))/(M(v)M(v_{\ast }))$ gives the other printed integral expression. For real functions the modulus squares are ordinary squares; the modulus version also proves the complex-space statement.

The weighted inner product satisfies $⟨f,M⟩=\int f=\rho (f)$. ThereforeThe identity operator and $L$ are linear, bounded and self-adjoint, hence so is $\Pi$. Directly, $\Vert \Pi f{\Vert }_H=|\rho (f)|\le \Vert f{\Vert }_H$; equality holds at $f=M$. Thus $\Vert \Pi \Vert =1$.

Since $\rho (M)=1$, ${\Pi }^{2}f=\rho (f)\rho (M)M=\Pi f$. Also $Lf=0$ exactly when $f=\rho (f)M$, which is exactly the range of $\Pi$. ThusA bounded self-adjoint idempotent is an orthogonal projection: if $u=\Pi u$ and $w=(I-\Pi )w$, then $⟨u,w⟩=⟨u,\Pi w⟩=0$. Here the orthogonal complement is precisely the zero-mass subspace $\rho (f)=0$.

Using $L=\Pi -I$ and the orthogonal projection identity ${\Pi }^{\ast }\Pi =\Pi$, we have $⟨\Pi f,f⟩=\Vert \Pi f{\Vert }_H^2$. The orthogonal decomposition $f=\Pi f+(I-\Pi )f$ then givesThe sign is negative, as in the PDF; the converted TeX loses it at the last equality. This establishes the exact spectral gap of normalized velocity relaxation: on the zero-mass subspace $L=-I$, with gap one rather than the weaker half-gap estimate.

The orthogonal projection satisfies $\Pi L=\Pi (\Pi -I)=0$, hence $d\Pi f_t/dt=0$ and $\Pi f_t=\Pi f_0=\rho (f_0)M$. Set $h_t=f_t-\Pi f_0$. Then $\Pi h_t=0$ and ${\dot{h}}_t=L{h}_t$. Crucially, applying the energy identity of part (e) to $h_t$ givesSolving this scalar equation and taking square roots provesThis argument uses (e) explicitly, rather than bypassing the requested energy method with an explicit solution formula.