Notes for "Posterior consistency and convergence rates for Bayesian inversion with hypoelliptic o...

H. Kekkonen, M. Lassas, S. Siltanen, Posterior consistency and convergence rates for Bayesian inversion with hypoelliptic operators, Inverse Problems, 32, 2016, 085005

Page 10, line 6, Section 3 Generalised random variables

The connection between $B_V$ and $C_V$ is $B_V = C_V (I-\Delta)^s : H^{s}\rightarrow H^{s}$

Notes:
Consider $\phi, \psi\in H^s(N)$ , then we have
$\langle C_V\phi, \psi \rangle_{H^{s}\times H^{-s}} = \langle \psi, C_V^* \psi\rangle_{H^{s}\times H^{-s}},$
where $C_V^* : H^{s}\rightarrow H^{-s}$ . Since $(B_{V}\phi, \psi)_{H^s} = (\phi, B_V \psi)_{H^s}$ and $(B_{V}\phi, \psi)_{H^s} = \langle C_V\phi, \psi \rangle_{H^{s}\times H^{-s}}$ , we obtain
$\int_{N}(I-\Delta)^{s/2}\phi(x) (I-\Delta)^{s/2}B_V\psi(x) ds = \int_{N}\phi(x)C_V^{*} \psi(x)dx.$
Hence, we find that
$(I-\Delta)^{s}B_V = C_V^{*} : H^{s}\rightarrow H^{-s},$
which implies
$C_V = B_V (I-\Delta)^{s}.$

Page 11, line (from end) 7, Section 3.2

Condition $B_U \in \mathfrak{C}^{1}(H^{\tau})$ guarantees that $\mathbb{E}(U, U)_{H^{\tau}} < \infty$ .

Notes:
Taking $(\lambda_j, \{\varphi_{j}\})_{j=1}^{\infty}$ be eigensystem of $B_U$ on $H^{\tau}$ , we have
$\begin{align} \mathbb{E}(U,U)_{H^{\tau}} & = \mathbb{E}\sum_{j=1}^{\infty}(U,\varphi_{j})(U,\varphi_{j})=\sum_{j=1}^{\infty}\mathbb{E}{((U,\varphi_{j})_{H^{\tau}}(U,\varphi_{j})_{H^{\tau}})} \\ & = \sum_{j=1}^{\infty}(B_U \varphi_{j}, \varphi_{j})_{H^{\tau}} = \sum_{j=1}^{\infty}\lambda_{j} < \infty, \end{align}$
which is just the required estimation.
Here, in this part, we may see $C_U : H^{-\tau} \rightarrow H^{\tau}$ ( $\tau\leq r$ ) and $B_U : H^{\tau} \rightarrow H^{\tau}$ , which coincides with formula (3.3) and (3.4) on page 10. Taking $\nu=\nu_k$ in the first formula on page 12, we have
$k = N(\nu_k) \sim cv_k^{\frac{d}{2(r-\tau)}}(1+O(\nu_k^{-\frac{1}{2(r-\tau)}})).$
Question: The operator $B_U^{-1}$ is a self-adjoint elliptic operator with smooth coefficients (defined on closed manifold), the eigenvalues are irrelevant to the definition function space of the operator $B_U^{-1}$ ?

Proof of Lemma 3, Page 15

$\ldots$ and we can write
$B_0 = B_0 ((A^* A)B_1 - K_2) = B_1 - B_0 K_2.$

Notes:
Here, we assume that $t, t_0 > 0$ . By my understanding, the equality means that for $f\in H^{r+2t_0}\subset H^{r}$ we have
$B_0(f) = B_0((A^* A)B_1)(f) - B_0 K_2(f) = B_0 A^* A B_1(f) - B_0 K_2 (f) .$
Since $B_1(f) \in H^{r}$ , we obtain $B_0 A^* A B_1(f) = B_1 (f)$ and
$B_0(f) = B_1(f) + B_0K_2(f)$
holds for every $f\in H^{r+2t_0}$ . Because for any $r\in \mathbb{R}$ , we can deduce that the corresponding $B_0 : C^{\infty} \rightarrow C^{\infty}$ is continuous. Hence, we find that
$B = B_1 \, \text{mod} \, \Psi^{-\infty}$
holds on $H^{r}(N)$ for any $r\in \mathbb{R}$ . That is to say,
$B = B_1 \, \text{mod} \, \Psi^{-\infty}$
holds on space $\mathcal{D}'$ , which implies $B\in H\Psi^{2t_0, 2t}$ .