On a Stability Property of the Generalized Spherical Radon Transform

Abstract

In this note, we study the operator norm of the generalized spherical Radon transform, defined by a smooth measure on the underlying incidence variety. In particular, we prove that for small perturbations of the measure, the spherical Radon transform remains an isomorphism between the corresponding Sobolev spaces.

Mathematical Subject Classifications (2010): 44A12, 53C65

1 Introduction and Background

Throughout the note, we fix a Euclidean space V =  ℝ ^d + 1, and consider the Euclidean spheres X = S ^d ⊂ V, and Y = S ^d ⊂ V ^∗. For p ∈ Y, C _p  ⊂ X will denote the copy of S ^d − 1 ⊂ X given by C _p  = { q ∈ X : ⟨q, p⟩ = 0}. Let σ _d − 1(q) denote the SO(d)-invariant probability measure on C _p . The set C _q  ⊂ Yand the measure σ _d − 1(p) on it are defined similarly. Then the spherical Radon transform is defined as follows:

$$\displaystyle\begin{array}{rcl} & & \mathcal{R} : {C}^{\infty }(X) \rightarrow {C}^{\infty }(Y ) \\ & & \mathcal{R}f(p) =\displaystyle\int _{C_{p}}f(q)d\sigma _{d-1}(q).\end{array}$$

Let σ be the unique SO(d + 1)-invariant probability measure on the incidence variety Z = { (q, p) ∈ X ×Y: ⟨q, p⟩ = 0}. Assume one is given a smooth, not necessarily positive measure dμ on Z, given by μ(q, p)σ where μ ∈ C ^∞(Z), and which satisfies μ( ± q, ± p) = μ(q, p) (call such μ symmetric). Introduce $\mathcal{R}_{\mu } : {C}^{\infty }(X) \rightarrow {C}^{\infty }(Y )$ by

$$(\mathcal{R}_{\mu }f)(p) =\displaystyle\int _{C_{p}}f(q)\mu (q,p)d\sigma _{d-1}(q).$$

Introduce also the dual Radon transform $\mathcal{R}_{\mu }^{T} : {C}^{\infty }(Y ) \rightarrow {C}^{\infty }(X)$ which is formally adjoint to $\mathcal{R}_{\mu }$ and given by

$$(\mathcal{R}_{\mu }^{T}g)(q) =\displaystyle\int _{ C_{q}}g(p)\mu (q,p)d\sigma _{d-1}(p).$$

Let $L_{s}^{2}(\mathbb{P}X)$ and $L_{s}^{2}(\mathbb{P}Y )$ denote the Sobolev space of even functions on X and Y, respectively. It is well known (see [2]) that the spherical Radon transform extends to an isomorphism of Sobolev spaces:

$$\mathcal{R} : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{ s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )$$

for every $s \in \mathbb{R}$. For general μ as above, $\mathcal{R}_{\mu }$ is a Fourier integral operator of order $\frac{d-1} {2}$ (see [3, 5]), and so extends to a bounded map $\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )$. We look for conditions on μ so that this is again an isomorphism.

It follows from Guillemin’s theorem on general Radon transforms associated to double fibrations [1], that $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu } : {C}^{\infty }(\mathbb{P}X) \rightarrow {C}^{\infty }(\mathbb{P}X)$ is an elliptic pseudo-differential operator of order d − 1 for all smooth, positive, symmetric measures μ on Z (for completeness, this is verified in the Appendix). The dependence of the principal symbol of $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu }$ on μ was investigated in [5]. In this note, we analyze the dependence on μ of the operator norm of $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+(d-1)}^{2}(\mathbb{P}X)$. We then give a sufficient condition on a perturbation μ of μ₀ so that $\mathcal{R}_{\mu } : {C}^{\infty }(\mathbb{P}X) \rightarrow {C}^{\infty }(\mathbb{P}Y )$ remains an isomorphism. Namely, we prove the following

Theorem. The set of C ^∞ measures μ on Z for which $\mathcal{R}_{\mu } : {C}^{\infty }(\mathbb{P}X) \rightarrow {C}^{\infty }(\mathbb{P}Y )$ is an isomorphism, is open in the C ^2d+1 (Z) topology.

2 Bounding the Norm of $\mathcal{R}_{\nu }^{T}\mathcal{R}_{\mu }$

We start by recalling an equivalent description of the Radon transform. Consider the double fibration

Let σ _X  = π_∗σ and σ _Y  = ρ_∗σ be the rotation-invariant probability measures on X and Y, respectively. Then for f ∈ C ^∞(X), $(\mathcal{R}f)\sigma _{Y } = \rho _{{\ast}}(\sigma ({\pi }^{{\ast}}f))$. For smooth symmetric measures dμ, dν on Z, given by μ(q, p)σ and ν(q, p)σ we can define $\mathcal{R}_{\mu } : {C}^{\infty }(X) \rightarrow {C}^{\infty }(Y )$, $\mathcal{R}_{\nu }^{T} : {C}^{\infty }(Y ) \rightarrow {C}^{\infty }(X)$ respectively by

$$(\mathcal{R}_{\mu }f)\sigma _{Y } = \rho _{{\ast}}(\mu \sigma ({\pi }^{{\ast}}f))$$

and

$$(\mathcal{R}_{\nu }^{T}g)\sigma _{ X} = \pi _{{\ast}}(\nu \sigma ({\rho }^{{\ast}}g)).$$

It follows from [1] that both $\mathcal{R}_{\mu }$ and $\mathcal{R}_{\nu }^{T}$ are Fourier integral operators of order $\frac{d-1} {2}$. Thus we restrict to even functions and consider $\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )$ and $\mathcal{R}_{\nu }^{T} : L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y ) \rightarrow L_{s+(d-1)}^{2}(\mathbb{P}X)$.

As before, q will denote a point in X and p a point in Y. We will often write q ∈ p instead of ⟨q, p⟩ = 0 ⇔ q ∈ C _p  ⇔ p ∈ C _q . In the following, the functions f, g are even. We also assume d ≥ 2.

Proposition 1.

The Schwartz kernel of $\mathcal{R}_{\nu }^{T}\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+(d-1)}^{2}(\mathbb{P}X)$ is

$$K(q^{\prime},q) = \frac{c_{d}} {\sin \text{dist}(q^{\prime},q)}\alpha (q,q^{\prime})$$

that is,

$$\mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }f(q^{\prime}) =\displaystyle\int _{X}f(q)K(q^{\prime},q)d\sigma _{X}(q).$$

Here c _d is a constant, and α(q,q′) is the average over all p ∈ Y s.t. q,q′∈ p of μ(q,p)ν(q′,p). More precisely,

$$\alpha (q,q^{\prime}) =\displaystyle\int _{SO(d-1)}\mu (q,Mp_{0})\nu (q^{\prime},Mp_{0})dM$$

where SO(d − 1) ={ g ∈ SO(d + 1) : gq = q,gq′ = q′}, $C_{p_{0}}$ is any fixed copy of S ^d through q,q′, and dM is the Haar probability measure on SO(d − 1).

Proof.

Fix some q′ ∈ X, and p ₀ ∈ Ys.t. q′ ∈ p ₀. Let SO(d) ⊂ SO(d + 1) be the stabilizer of q′ ∈ X. For g ∈ C ^∞(Y) we may write

$$\mathcal{R}_{\nu }^{T}g(q^{\prime}) =\displaystyle\int _{ p\ni q^{\prime}}g(p)\nu (q^{\prime},p)d\sigma _{d-1}(p) =\displaystyle\int _{SO(d)}g(Mp_{0})\nu (q^{\prime},Mp_{0})dM$$

where dM is the Haar probability measure on SO(d). Then taking

$$g(p) = \mathcal{R}_{\mu }f(p) =\displaystyle\int _{\tilde{q}\in p}f(\tilde{q})\mu (\tilde{q},p)d\sigma _{d-1}(\tilde{q})$$

we get

$$\displaystyle\begin{array}{rcl} \mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }f(q^{\prime})& =& \displaystyle\int _{SO(d)}\Big{(}\displaystyle\int _{\tilde{q}\in Mp_{0}}f(\tilde{q})\mu (\tilde{q},Mp_{0})d\sigma _{d-1}(\tilde{q})\Big{)}\nu (q^{\prime},Mp_{0})dM \\ & =& \displaystyle\int _{SO(d)}\Big{(}\displaystyle\int _{\tilde{q}\in p_{0}}f(M\tilde{q})\mu (M\tilde{q},Mp_{0})d\sigma _{d-1}(\tilde{q})\Big{)}\nu (q^{\prime},Mp_{0})dM \\ & =& \displaystyle\int _{\tilde{q}\in p_{0}}d\sigma _{d-1}(\tilde{q})\displaystyle\int _{SO(d)}f(M\tilde{q})\mu (M\tilde{q},Mp_{0})\nu (q^{\prime},Mp_{0})dM.\end{array}$$

Denote $\theta = \text{dist}(\tilde{q},q^{\prime})$, and S _θ ^d − 1 = { q : dist(q′, q) = θ}. Let dσ _d − 1 ^θ(q) denote the rotationally invariant probability measure on S _θ ^d − 1. The inner integral may be written as

$$\displaystyle\int _{S_{\theta }^{d-1}}f(q)\alpha (q,q^{\prime})d\sigma _{d-1}^{\theta }(q).$$

Here α(q, q′) = ∫ _{SO(d − 1)}μ(q, Mp ₀)ν(q′, Mp ₀)dM with SO(d − 1) = Stab(q) ∩ Stab(q ₀) is just the average of μ(q, p)ν(q′, p) over all (d − 1)-dimensional spheres C _p containing both q and q′. Then

$$\mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }f(q^{\prime}) =\displaystyle\int _{\tilde{q}\in p_{0}}d\sigma _{d-1}(\tilde{q})\displaystyle\int _{S_{\theta }^{d-1}}f(q)\alpha (q,q^{\prime})d\sigma _{d-1}^{\theta }(q)$$

and since the inner integral only depends on $\theta = \text{dist}(\tilde{q},q^{\prime})$, this may be rewritten as

$$c_{d}\displaystyle\int _{0}^{\pi /2}d{\theta \sin }^{d-2}\theta \displaystyle\int _{ S_{\theta }^{d-1}}f(q)\alpha (q,q^{\prime})d\sigma _{d-1}^{\theta }(q).$$

Finally, dσ _d  = c _d sin^d − 1θdθdσ _d − 1 ^θ, and so

$$\mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }f(q^{\prime}) = c_{d}\displaystyle\int _{X}\frac{1} {\sin \theta }f(q)\alpha (q,q^{\prime})d\sigma _{d}(q).$$

We conclude that the Schwartz kernel is

$$K(q^{\prime},q) = \frac{c_{d}} {\sin \text{dist}(q^{\prime},q)}\alpha (q,q^{\prime}).$$

We proceed to estimate the norm of $\mathcal{R}_{\nu }^{T}\mathcal{R}_{\mu }$. Our main tool will be the following proposition proved in Sect. C

Proposition.

Consider a pseudodifferential operator P of order m

$$P : L_{s+m}^{2}({\mathbb{R}}^{n}) \rightarrow L_{ s}^{2}({\mathbb{R}}^{n})$$

between Sobolev spaces with x−compactly supported symbol p(x,ξ) in $K \subset {\mathbb{R}}^{n}$ s.t.

$$\vert D_{x}^{\alpha }p(x,\xi )\vert \leq C_{ \alpha 0}{(1 + \vert \xi \vert )}^{\alpha }.$$

There exists a constant C(n,s) such that

$$\|P\|_{L_{s+m}^{2}({\mathbb{R}}^{n})\rightarrow L_{s}^{2}({\mathbb{R}}^{n})} \leq C(n,s)\sup\limits_{\vert \alpha \vert \leq n+\lfloor \vert s\vert \rfloor +1}C_{\alpha 0}\vert K\vert.$$

Proposition 2.

The norm of $\mathcal{R}_{\nu }^{T}\mathcal{R}_{\mu } : L_{-(d-1)}^{2}(\mathbb{P}X) \rightarrow L_{0}^{2}(\mathbb{P}X)$ is bounded from above by

$$\|\mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }\| \leq C\displaystyle\sum _{j+k=0}^{2d+1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }$$

for some constant C dependent on the double fibration.

Proof.

First introduce coordinate charts. Choose a partition of unity χ _i (q′) corresponding to a covering of X by charts U _i , and a function $\rho : [0,\infty ) \rightarrow \mathbb{R}_{+}$ with support in [0, 1] s.t. ρ(r) = 1 for $r \leq \frac{1} {2}$. Write

$$\displaystyle\begin{array}{rcl} K(q^{\prime},q)& =& \displaystyle\sum _{i}K_{i}(q^{\prime},q) + L_{i}(q^{\prime},q) \\ K_{i}(q^{\prime},q)& =& \chi _{i}(q^{\prime})\rho (\sin \text{dist}(q^{\prime},q))K(q^{\prime},q) \\ \end{array}$$

and

$$L_{i}(q^{\prime},q) = \chi _{i}(q^{\prime})(1 - \rho (\sin \text{dist}(q^{\prime},q)))K(q^{\prime},q).$$

Let $\mathcal{R}_{\nu }^{T}\mathcal{R}_{\mu } =\sum _{i}T_{K_{i}} + T_{L_{i}}$ be the corresponding decomposition for the operators.

First we will bound the norm of the diagonal terms, i.e., the operators defined by K _i . Fix i, and choose some point q′ ∈ U _i . Introduce polar coordinates (r, ψ) around q′ so that ψ ∈ S ₁ ^d − 1(q′) and r = sinθ for $r \leq \frac{1} {2}$, θ = dist(q, q′). Note that α(q′, (r, ψ)) = α(q′, (r, − ψ)). By Proposition 5, the corresponding symbol is

$$p_{1}(q^{\prime},\xi ) = \chi _{i}(q^{\prime})\displaystyle\int _{0}^{1}\displaystyle\int _{{ S}^{d-1}} \frac{\alpha (q^{\prime},r\psi )} {r} {e}^{-i\langle \xi,\psi \rangle }\rho (r){r}^{d-1}drd\psi.$$

For a given q′, introduce spherical coordinates ψ = (ϕ, ϕ₁, …, ϕ _d − 2) 0 ≤ ϕ ≤ π, on S ₁ ^d − 1(q′) in such a way that cosϕ = ψ₁; Take ξ₀ = (1, 0, …, 0). Then

$$p_{1}(q^{\prime},T\xi _{0}) = C_{d}\displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{\pi }m(q^{\prime},r,\phi ){e{}^{-iTr\cos \phi }\cos }^{d-2}\phi d\phi $$

where m(q′, r, ϕ) = χ _i (q′)∫ ₀ ^π …∫ ₀ ^π ∫ ₀ ^2π dϕ₁ …dϕ _d − 2α(q′, r, ϕ, ϕ₁, …, ϕ _d − 2). We then have m(q′, r, π ∕ 2 + ϕ) = m(q′, r, π ∕ 2 − ϕ). Take t = cosϕ and M(q′, r, t) = m(q′, r, arccost). Then M(q′, r, t) = M(q′, r, − t) and

$$p_{1}(q^{\prime},T\xi _{0}) = C_{d}\displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ -1}^{1}M(q^{\prime},r,t){e}^{-iTrt}{(1 - {t}^{2})}^{\frac{d-3} {2} }dt.$$

Since M is even, we may write

$$p_{1}(q^{\prime},T\xi _{0}) = 2\displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}M(q^{\prime},r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt$$

and so for all multi-indices α

$$D_{q^{\prime}}^{\alpha }p_{ 1}(q^{\prime},T\xi _{0}) = 2\displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}D_{ q^{\prime}}^{\alpha }M(q^{\prime},r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt.$$

Define as in Appendix A

$$I(d,\rho,m) =\displaystyle\int _{ 0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}M(q^{\prime},r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt$$

then we can write

$$D_{q^{\prime}}^{\alpha }p_{ 1}(q^{\prime},T\xi _{0}) = 2I(d,\rho,D_{q^{\prime}}^{\alpha }m(q^{\prime},r,\phi ))$$

and conclude by Proposition 3 that

$$\displaystyle\begin{array}{rcl} \vert p_{1}(q^{\prime},T\xi _{0})\vert & \leq & \frac{C} {{T}^{d-1}}\displaystyle\sum _{\begin{array}{c} a + b \leq d \\ a \leq d/2,b \leq d - 1\end{array} }\sup\frac{{\partial }^{a+b}} {\partial {r}^{a}\partial {\phi }^{b}}m \\ &\leq & \frac{C} {{T}^{d-1}}\displaystyle\sum _{j+k\leq d}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty } \\ \end{array}$$

and similarly for all multi-indices α

$$\vert D_{q^{\prime}}^{\alpha }p_{ 1}(q^{\prime},T\xi _{0})\vert \leq \frac{C} {{T}^{d-1}}\displaystyle\sum _{j+k\leq \vert \alpha \vert +d}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

It is also immediate that

$$\vert p_{1}(q^{\prime},T\xi _{0})\vert \leq C_{d}\displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{\pi }\vert m(q^{\prime},r,\phi )\vert d\phi \leq C\|\mu \|_{ \infty }\|\nu \|_{\infty }$$

and similarly

$$\vert D_{q^{\prime}}^{\alpha }p_{ 1}(q^{\prime},T\xi _{0})\vert \leq C\displaystyle\sum _{j+k\leq \vert \alpha \vert }\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

So we can write

$$\vert D_{q^{\prime}}^{\alpha }p_{ 1}(q^{\prime},\xi )\vert \leq \frac{C} {{(1 + \vert \xi \vert )}^{d-1}}\displaystyle\sum _{j+k\leq \vert \alpha \vert +d}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }$$

for some universal constant C = C(d). Then choosing s = 0 and | α |  = d + 1 in Proposition 4 we get that

$$\|T_{K_{i}}\|_{L_{-(d-1)}^{2}(\mathbb{P}X)\rightarrow L_{0}^{2}(\mathbb{P}X)} \leq C\displaystyle\sum _{j+k=0}^{2d+1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

Now we bound the norm of the off-diagonal term, namely the sum of operators corresponding to L _i . They constitute a smoothing operator T _L (μ, ν); its Schwartz kernel k(q′, q) = (1 − ρ(dist(q′, q)))K(q′, q) is a smooth function in both arguments. Denoting by ∇ ^j : C ^∞(X) → (T ^∗ X)^⊗ j the j-th derivative obtained from the Levi-Civita connection,

$$\displaystyle\begin{array}{rcl} & & \left \|\displaystyle\int _{X}d\sigma _{X}(q)f(q)k(q^{\prime},q)\right \|_{L_{d-1}^{2}(\mathbb{P}X)}^{2} \\ & & \quad =\displaystyle\int _{X}{\vert }\nabla _{q^{\prime}}^{d-1}\displaystyle\int _{ X}d\sigma _{X}(q)f(q)k(q^{\prime},q){{\vert }}^{2}d\sigma _{ X}(q^{\prime}) \\ & & \quad =\displaystyle\int _{X}{\vert }\displaystyle\int _{X}d\sigma _{X}(q)f(q)\nabla _{q^{\prime}}^{d-1}k(q^{\prime},q){{\vert }}^{2}d\sigma _{ X}(q^{\prime}) \\ & & \quad \leq \displaystyle\int _{X}d\sigma _{X}(q^{\prime})\left (\displaystyle\int _{X}\vert f{(q)}^{2}\vert d\sigma _{ X}(q)\displaystyle\int _{X}\vert \nabla _{q^{\prime}}^{d-1}k(q^{\prime},q){\vert }^{2}d\sigma _{ X}(q)\right ) \\ & & \quad =\| f\|_{{L}^{2}(X)}^{2}\displaystyle\int _{ X}\vert \nabla _{q^{\prime}}^{d-1}k(q^{\prime},q){\vert }^{2}d\sigma _{ X}(q) \\ \end{array}$$

and

$$\sup\limits_{q^{\prime}}\sqrt{\displaystyle\int _{X } \vert \nabla _{q^{\prime} }^{d-1 }k(q^{\prime}, q){\vert }^{2 } d\sigma _{X } (q)} \leq C\displaystyle\sum _{j+k\leq d-1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

So

$$\|T_{L}(\mu,\nu )\|_{L_{0}^{2}(\mathbb{P}X)\rightarrow L_{d-1}^{2}(\mathbb{P}X)} \leq C\displaystyle\sum _{j+k\leq d-1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

It is easy to see that the adjoint operator $T_{L}{(\mu,\nu )}^{{\ast}} : L_{(d-1)}^{2}{(\mathbb{P}X)}^{{\ast}}\rightarrow L_{0}^{2}{(\mathbb{P}X)}^{{\ast}}$ equals T _L (ν, μ) after the isomorphic identification $L_{s}^{2}{(\mathbb{P}X)}^{{\ast}}\simeq L_{-s}^{2}(\mathbb{P}X)$ for s = 0, d − 1. Since the bound above is symmetric in μ, ν we conclude

$$\|T_{L}(\mu,\nu )\|_{L_{-(d-1)}^{2}(\mathbb{P}X)\rightarrow L_{0}^{2}(\mathbb{P}X)} \leq C\displaystyle\sum _{j+k\leq d-1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

Finally

$$\|\mathcal{R}_{\nu }^{T}\mathcal{R}_{ \mu }\| \leq \|\displaystyle\sum _{i}T_{K_{i}}\| +\| T_{L}\| \leq C\displaystyle\sum _{j+k=0}^{2d+1}\|{D}^{j}\mu \|_{ \infty }\|{D}^{k}\nu \|_{ \infty }.$$

Theorem 1.

Assume d ≥ 2, and let μ ₀ ∈ C ^∞ (Z) be such that $\mathcal{R}_{\mu _{0}} : {C}^{\infty }(\mathbb{P}X) \rightarrow {C}^{\infty }(\mathbb{P}Y )$ is an isomorphism. Then there exists ε ₀ > 0 (depending on the double fibration), such that if $\|\mu - \mu _{0}\|_{{C}^{2d+1}(Z)} < \epsilon _{0}$ then $\mathcal{R}_{\mu } : {C}^{\infty }(\mathbb{P}X) \rightarrow {C}^{\infty }(\mathbb{P}Y )$ is an isomorphism (for all s).

Proof.

Since $\mathcal{R}_{\mu _{0}}^{T}\mathcal{R}_{\mu _{0}} : L_{-(d-1)}^{2}(\mathbb{P}X) \rightarrow L_{0}^{2}(\mathbb{P}X)$ (and likewise for Y) is elliptic, it is an isomorphism. Let us verify that both of the maps $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu } : L_{-(d-1)}^{2}(\mathbb{P}X) \rightarrow L_{0}^{2}(\mathbb{P}X)$ and $\mathcal{R}_{\mu }\mathcal{R}_{\mu }^{T} : L_{-(d-1)}^{2}(\mathbb{P}Y ) \rightarrow L_{0}^{2}(\mathbb{P}Y )$ remain an isomorphism for small perturbations μ of μ₀ in the C ^2d + 1(Z) norm:

$$\displaystyle\begin{array}{rcl} \|\mathcal{R}_{\mu }^{T}\mathcal{R}_{ \mu } -\mathcal{R}_{\mu _{0}}^{T}\mathcal{R}_{ \mu _{0}}\|& =& \|(\mathcal{R}_{\mu _{0}}^{T} + \mathcal{R}_{ \mu -\mu _{0}}^{T})(\mathcal{R}_{ \mu _{0}} + \mathcal{R}_{\mu -\mu _{0}}) -\mathcal{R}_{\mu _{0}}^{T}\mathcal{R}_{ \mu _{0}}\| \\ & \leq & \|\mathcal{R}_{\mu _{0}}^{T}\mathcal{R}_{ \mu -\mu _{0}}\| +\| \mathcal{R}_{\mu -\mu _{0}}^{T}\mathcal{R}_{ \mu _{0}}\| +\| \mathcal{R}_{\mu -\mu _{0}}^{T}\mathcal{R}_{ \mu -\mu _{0}}\| \\ \end{array}$$

so by Corollary 2, there is an ε₀ > 0 s.t. all norms are indeed small when $\|\mu - \mu _{0}\|_{{C}^{2d+1}(Z)} < \epsilon _{0}$. The operator $\mathcal{R}_{\mu }\mathcal{R}_{\mu }^{T}$ is treated identically, and we take the minimal of the two ε₀.

Since both $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu }$ and $\mathcal{R}_{\mu }\mathcal{R}_{\mu }^{T}$ are elliptic operators, the dimension of the kernel and cokernel are independent of s. It follows that $\mathcal{R}_{\mu }^{T}\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+d-1}^{2}(\mathbb{P}X)$ and $\mathcal{R}_{\mu }\mathcal{R}_{\mu }^{T} : L_{s}^{2}(\mathbb{P}Y ) \rightarrow L_{s+d-1}^{2}(\mathbb{P}Y )$ are isomorphisms for all s. In particular, $\text{Ker}(\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )) = 0$ and $\text{Coker}(\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )) = 0$, implying by the open mapping theorem that $\mathcal{R}_{\mu } : L_{s}^{2}(\mathbb{P}X) \rightarrow L_{s+\frac{d-1} {2} }^{2}(\mathbb{P}Y )$ is an isomorphism for all s. The result follows.

Remark 1.

It is unlikely that the result is sharp. For instance, in the case of d = 1 one only needs $\|\mu - 1\|_{{C}^{0}}$ to be small to conclude that $\mathcal{R}_{\mu }$ is an isomorphism, while the statement (although non-applicable for d = 1) would suggest bounding the C ³-norm.

3 Appendix

3.1 A Some Integral Estimates

Fix some real T > 0. For an integer d ≥ 2, a smooth function $\rho : [0,\infty ) \rightarrow \mathbb{R}$ compactly supported in [0, 1), and smooth functions $m(r,\phi ),n(r,\phi ) : [0,\infty ) \times [0,\pi ]\mathbb{R}$ we define the integrals

$$I(d,\rho,m) =\displaystyle\int _{ 0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}M(r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt$$

and

$$J(d,\rho,m) =\displaystyle\int _{ 0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}N(r,t)\sin (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt$$

where M(r, t) = m(r, ϕ) and N(r, t) = n(r, ϕ) for t = cosϕ. We assume m is even w.r.t. $\frac{\pi } {2}$, namely $m(r, \frac{\pi } {2} + \phi ) = m(r, \frac{\pi } {2} - \phi )\;\Longleftrightarrow\;M(r,t) = M(r,-t)$; while n is odd, i.e. $n(r, \frac{\pi } {2} + \phi ) = -n(r, \frac{\pi } {2} - \phi )\;\Longleftrightarrow\;N(r,t) = -N(r,-t).$

Proposition 3.

There exists a constant C = C(d,ρ) such that for d ≥ 2 and all even functions m

$$\vert I(d,\rho,m)\vert \leq \frac{C} {{T}^{d-1}}\displaystyle\sum _{\begin{array}{c} a + b \leq d \\ a \leq d/2,b \leq d - 1 \end{array} }\sup {\vert } \frac{{\partial }^{a+b}} {\partial {r}^{a}\partial {\phi }^{b}}m{\vert }$$

and for odd functions n

$$\vert J(d,\rho,n)\vert \leq \frac{C} {{T}^{d-1}}\displaystyle\sum _{\begin{array}{c} a + b \leq d \\ a \leq d/2,b \leq d - 1 \end{array} }\sup {\vert } \frac{{\partial }^{a+b}} {\partial {r}^{a}\partial {\phi }^{b}}n{\vert }.$$

Proof.

Induction on d. Start by verifying the bounds for d = 2. To bound

$$I(2,\rho,m) =\displaystyle\int _{ 0}^{1}{(1 - {t}^{2})}^{-\frac{1} {2} }\displaystyle\int _{0}^{1}\rho (r)M(r,t)\cos (Trt)dr$$

we first integrate the inner integral by parts:

$$\displaystyle\int _{0}^{1}M(r,t)\rho (r)\cos (Trt)dr = - \frac{1} {Tt}\displaystyle\int _{0}^{1}\sin (Trt)(M(r,t)\rho ^{\prime}(r) + \frac{\partial M} {\partial r} (r,t)\rho (r))dr$$

Let us bound separately

$$\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}\sin (Trt)M(r,t)\rho ^{\prime}(r)dr$$

and

$$\displaystyle\begin{array}{rcl} \displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}\sin (Trt)\frac{\partial M} {\partial r} (r,t)\rho (r)dr.& & \\ \end{array}$$

Now

$$\displaystyle\begin{array}{rcl} {\vert }\displaystyle\int _{\frac{1} {2} }^{1}\frac{\sin (Trt)M(r,t)dt} {t\sqrt{1 - {t}^{2}}} {\vert }\leq C\sup \vert m\vert & & \\ \end{array}$$

and since $\frac{\partial M} {\partial t} = -\frac{\partial m} {\partial \phi } \frac{1} {\sqrt{1-{t}^{2}}}$,

$$\displaystyle\begin{array}{rcl} & & \displaystyle\int _{0}^{\frac{1} {2} } \frac{\sin (Trt)M(r,t)dt} {t\sqrt{1 - {t}^{2}}} \\ & & = Si(Trt) \frac{M(r,t)} {\sqrt{1 - {t}^{2}}}{\vert }_{0}^{\frac{1} {2} } \\ & & \qquad +\displaystyle\int _{ 0}^{\frac{1} {2} }Si(Trt)\Bigg{(}\frac{\partial m} {\partial \phi } (r,t) \frac{1} {{\sqrt{1 - {t}^{2}}}^{2}} + M(r,t) \frac{t} {{(1 - {t}^{2})}^{3/2}}\Bigg{)}dt\end{array}$$

Now since Si is bounded, it follows that

$${\vert }\displaystyle\int _{0}^{\frac{1} {2} } \frac{\sin (Trt)M(r,t)dt} {t\sqrt{1 - {t}^{2}}} {\vert }\leq C\left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial \phi } \right \vert \right )$$

Thus

$$\displaystyle\begin{array}{rcl} & & {\vert }\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}\sin (Trt)M(r,t)\rho ^{\prime}(r)dr{\vert } \\ & &\quad \leq C\left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial \phi } \right \vert \right )\displaystyle\int _{0}^{1}dr\vert \rho ^{\prime}(r)\vert \leq C\left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial \phi } \right \vert \right )\end{array}$$

Similarly,

$${\vert }\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}\sin (Trt)\frac{\partial M} {\partial r} (r,t)\rho (r)dr{\vert }\leq C\left (\sup \left \vert \frac{\partial m} {\partial r} \right \vert +\sup \left \vert \frac{{\partial }^{2}m} {\partial r\partial \phi }\right \vert \right ).$$

Tracing back,

$$\vert I(2,\rho,m)\vert \leq \frac{C} {T}\left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial r} \right \vert +\sup \left \vert \frac{\partial m} {\partial \phi } \right \vert +\sup \left \vert \frac{{\partial }^{2}m} {\partial r\partial \phi }\right \vert \right ).$$

Next we bound | J(2, ρ, n) | 

$$J(2,\rho,n) =\displaystyle\int _{ 0}^{1}{(1 - {t}^{2})}^{-1/2}dt\displaystyle\int _{ 0}^{1}N(r,t)\rho (r)\sin (Trt)dr.$$

Integrate the inner integral by parts:

$$\displaystyle\begin{array}{rcl} \displaystyle\int _{0}^{1}N(r,t)\rho (r)\sin (Trt)dr& =& - \frac{1} {Tt}\displaystyle\int _{0}^{1}(1 -\cos (Trt))(N(r,t)\rho ^{\prime}(r) \\ & & +\frac{\partial N} {\partial r} (r,t)\rho (r))dr.\end{array}$$

Let us bound separately

$$\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}(1 -\cos (Trt))N(r,t)\rho ^{\prime}(r)dr$$

and

$$\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}(1 -\cos (Trt))\frac{\partial N} {\partial r} (r,t)\rho (r)dr.$$

Now

$${\vert }\displaystyle\int _{\frac{1} {2} }^{1}\frac{(1 -\cos (Trt))N(r,t)dt} {t\sqrt{1 - {t}^{2}}} {\vert }\leq C\sup \vert n\vert $$

and since $\frac{\partial N} {\partial t} = -\frac{\partial n} {\partial \phi } \frac{1} {\sqrt{1-{t}^{2}}}$ and N(r, 0) = 0 we get that $\vert \frac{\partial } {\partial t}N(r,t)\vert \leq C\sup \vert \frac{\partial n} {\partial \phi }\vert$ for $0 \leq t \leq \frac{1} {2}$ so $\vert N(r,t)\vert \leq C\sup \vert \frac{\partial n} {\partial \phi }\vert t$ and

$${\vert }\displaystyle\int _{0}^{\frac{1} {2} } \frac{(1 -\cos (Trt))N(r,t)dt} {t\sqrt{1 - {t}^{2}}} {\vert }\leq C\sup \left \vert \frac{\partial n} {\partial \phi }\right \vert \displaystyle\int _{0}^{1} \frac{dt} {\sqrt{1 - {t}^{2}}} = C\sup \left \vert \frac{\partial n} {\partial \phi }\right \vert.$$

Thus

$$\displaystyle\begin{array}{rcl} & & {\vert }\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}(1 -\cos (Trt))N(r,t)\rho ^{\prime}(r)dr{\vert } \\ & &\quad \leq C\left (\sup \vert n\vert +\sup \left \vert \frac{\partial n} {\partial \phi }\right \vert \right )\displaystyle\int _{0}^{1}dr\vert \rho ^{\prime}(r)\vert \leq C\left (\sup \vert n\vert +\sup \left \vert \frac{\partial n} {\partial \phi }\right \vert \right )\end{array}$$

Similarly,

$${\vert }\displaystyle\int _{0}^{1} \frac{dt} {t\sqrt{1 - {t}^{2}}}\displaystyle\int _{0}^{1}(1 -\cos (Trt))\frac{\partial N} {\partial r} (r,t)\rho (r)dr{\vert }\leq C\left (\sup \left \vert \frac{\partial n} {\partial r}\right \vert +\sup \left \vert \frac{{\partial }^{2}n} {\partial r\partial \phi }\right \vert \right )$$

and putting all together,

$$\vert J(2,\rho,n)\vert \leq \frac{C} {T}\left (\sup \vert n\vert +\sup \left \vert \frac{\partial n} {\partial r}\right \vert +\sup \left \vert \frac{\partial n} {\partial \phi }\right \vert +\sup \left \vert \frac{{\partial }^{2}n} {\partial r\partial \phi }\right \vert \right )$$

as required.

Next consider the case d = 3. We bound I, J simultaneously. Apply integration by parts to the inner integrals:

$$\displaystyle\begin{array}{rcl} I(3,\rho,m)& =& \displaystyle\int _{0}^{1}\rho (r)rdr\displaystyle\int _{ 0}^{1}M(r,t)\cos (Trt)dt \\ & =& \frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)dr\displaystyle\int _{ 0}^{1}\frac{\partial m} {\partial \phi } (r,t) \frac{1} {\sqrt{1 - {t}^{2}}} \frac{\sin (Trt)} {r} dt \\ & & + \frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)M(r,1)\frac{\sin (Tr)} {r} dr \\ \end{array}$$

and similarly

$$\displaystyle\begin{array}{rcl} J(3,\rho,n)& =& -\frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)dr\displaystyle\int _{ 0}^{1}\frac{\partial n} {\partial \phi }(r,t) \frac{1} {\sqrt{1 - {t}^{2}}} \frac{\cos (Trt)} {r} dt \\ & & -\frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)\left (N(r,1)\frac{\cos (Tr)} {r} -\frac{N(r,0)} {r} \right )dr\end{array}$$

These first summands are

$$\frac{1} {T}\displaystyle\int _{0}^{1}\rho (r)dr\displaystyle\int _{ 0}^{1}\frac{\partial m} {\partial \phi } (r,t) \frac{1} {\sqrt{1 - {t}^{2}}}\sin (Trt)dt = \frac{1} {T}J\left (2,\rho, \frac{\partial m} {\partial \phi } \right )$$

and

$$-\frac{1} {T}\displaystyle\int _{0}^{1}\rho (r)dr\displaystyle\int _{ 0}^{1}\frac{\partial n} {\partial \phi }(r,t) \frac{1} {\sqrt{1 - {t}^{2}}}\cos (Trt)dt = -\frac{1} {T}I\left (2,\rho, \frac{\partial n} {\partial \phi }\right )$$

the second summand for I

$$\displaystyle\begin{array}{rcl} & & \frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)M(r,1)\frac{\sin (Tr)} {r} dr \\ & & \quad = \frac{1} {T}\displaystyle\int _{0}^{1}\rho (r)M(r,1)\sin (Tr)dr \\ & & \quad = -\frac{1} {T}\rho (r)M(r,1)\frac{\cos (Tr)} {T} {\vert }_{0}^{1} + \frac{1} {{T}^{2}}\displaystyle\int _{0}^{1}(\rho ^{\prime}(r)M(r,1) \\ & & \qquad + \rho (r) \frac{\partial } {\partial r}M(r,1))\cos (Tr)dr \\ \end{array}$$

is bounded by

$$\frac{C} {{T}^{2}}\left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial r} \right \vert \right ).$$

Similarly since N(r, 0) = 0, also the second summand for J

$${\vert }\frac{1} {T}\displaystyle\int _{0}^{1}r\rho (r)N(r,1)\frac{\cos (Tr)} {r} dr{\vert }\leq \frac{C} {{T}^{2}}\left (\sup \vert n\vert +\sup \left \vert \frac{\partial n} {\partial r}\right \vert \right )$$

thus we showed that

$$\vert I(3,\rho,m)\vert \leq \frac{C(3,\rho )} {T} J(2,\rho, \frac{\partial m} {\partial \phi } ) + \frac{C(3,\rho )} {{T}^{2}} \left (\sup \vert m\vert +\sup \left \vert \frac{\partial m} {\partial r} \right \vert \right )$$

and

$$\vert J(3,\rho,n)\vert \leq \frac{C(3,\rho )} {T} I(2,\rho, \frac{\partial n} {\partial \phi }) + \frac{C(3,\rho )} {{T}^{2}} \left (\sup \vert n\vert +\sup \left \vert \frac{\partial n} {\partial r}\right \vert \right )$$

and plugging the already proved estimates for d = 2 concludes the case d = 3.

Finally, for d > 3 we will apply induction. Again consider both integrals simultaneously. Start by integrating by parts the inner integral: the boundary term is zero (for J since n is odd), so

$$\displaystyle\begin{array}{rcl} & & \displaystyle\int _{0}^{1}M(r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt \\ & & \quad = \frac{1} {Tr}\displaystyle\int _{0}^{1}\sin (Trt)\left (\frac{\partial m} {\partial \phi } {(1 - {t}^{2})}^{\frac{d-4} {2} } + (d - 3)M(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\right )dt \\ \end{array}$$

$$\displaystyle\begin{array}{rcl} & & \displaystyle\int _{0}^{1}N(r,t)\sin (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt \\ & & \quad = - \frac{1} {Tr}\displaystyle\int _{0}^{1}\cos (Trt)\left (\frac{\partial n} {\partial \phi }{(1 - {t}^{2})}^{\frac{d-4} {2} } + (d - 3)N(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\right )dt.\end{array}$$

Thus

$$\displaystyle\begin{array}{rcl} I(d,\rho,m)& =& \displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}M(r,t)\cos (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt \\ & =& \frac{1} {T}\displaystyle\int _{0}^{1}\rho (r){r}^{d-3}dr\displaystyle\int _{ 0}^{1}\sin (Trt)\frac{\partial m} {\partial \phi } {(1 - {t}^{2})}^{\frac{d-4} {2} }dt \\ & & +\frac{C_{d}} {T} \displaystyle\int _{0}^{1}\rho (r){r}^{d-3}\displaystyle\int _{ 0}^{1}M(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\sin (Trt)dt. \end{array}$$

and

$$\displaystyle\begin{array}{rcl} J(d,\rho,n)& =& \displaystyle\int _{0}^{1}\rho (r){r}^{d-2}dr\displaystyle\int _{ 0}^{1}N(r,t)\sin (Trt){(1 - {t}^{2})}^{\frac{d-3} {2} }dt \\ & =& -\frac{1} {T}\displaystyle\int _{0}^{1}\rho (r){r}^{d-3}dr\displaystyle\int _{ 0}^{1}\cos (Trt)\frac{\partial n} {\partial \phi }{(1 - {t}^{2})}^{\frac{d-4} {2} }dt \\ & & -\frac{C_{d}} {T} \displaystyle\int _{0}^{1}\rho (r){r}^{d-3}\displaystyle\int _{ 0}^{1}N(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\cos (Trt)dt.\end{array}$$

The first terms are $\frac{1} {T}J(d - 1,\rho, \frac{\partial m} {\partial \phi } )$ and $-\frac{1} {T}I(d - 1,\rho, \frac{\partial n} {\partial \phi })$, respectively.

In the second term, first change the order of integration:

$$\displaystyle\begin{array}{rcl} & & \frac{C_{d}} {T} \displaystyle\int _{0}^{1}\rho (r){r}^{d-3}\displaystyle\int _{ 0}^{1}M(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\sin (Trt)dt \\ & & \quad = \frac{C_{d}} {T} \displaystyle\int _{0}^{1}t{(1 - {t}^{2})}^{\frac{d-5} {2} }dt\displaystyle\int _{0}^{1}M(r,t)\sin (Trt){r}^{d-3}\rho (r)dr.\end{array}$$

Now apply integration by parts to the inner integral. Since d − 3 > 0 and ρ(1) = 0, again there is no boundary term:

$$\displaystyle\begin{array}{rcl} & & \displaystyle\int _{0}^{1}M(r,t)\sin (Trt){r}^{d-3}\rho (r)dr \\ & & =\displaystyle\int _{ 0}^{1}dr\frac{\cos (Trt)} {Tt} ({r}^{d-3}\rho (r)\frac{\partial M} {\partial r} (r,t) + ({r}^{d-3}\rho ^{\prime}(r) \\ & & \quad + (d - 3){r}^{d-4}\rho (r))M(r,t)) \\ \end{array}$$

Thus

$$\displaystyle\begin{array}{rcl} & & \frac{C_{d}} {T} \displaystyle\int _{0}^{1}\rho (r){r}^{d-3}\displaystyle\int _{ 0}^{1}M(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\sin (Trt)dt \\ & & \quad = \frac{C_{d}} {{T}^{2}} \displaystyle\int _{0}^{1}{(1 - {t}^{2})}^{\frac{d-5} {2} }dt\displaystyle\int _{0}^{1}dr\cos (Trt){r}^{d-3}\rho (r)\frac{\partial M} {\partial r} (r,t) \\ & & \qquad + \frac{C_{d}} {{T}^{2}} \displaystyle\int _{0}^{1}{(1 - {t}^{2})}^{\frac{d-5} {2} }dt\displaystyle\int _{0}^{1}dr\cos (Trt)({r}^{d-3}\rho ^{\prime}(r) \\ & & \qquad + (d - 3){r}^{d-4}\rho (r))M(r,t) \\ & & \quad = \frac{C_{d}} {{T}^{2}} I\left (d - 2,r\rho (r), \frac{\partial m} {\partial r} \right ) + \frac{C_{d}} {{T}^{2}} I(d - 2,\rho (r) + r\rho ^{\prime}(r),m) \\ \end{array}$$

and the corresponding term for J:

$$\displaystyle\begin{array}{rcl} & & -\frac{C_{d}} {T} \displaystyle\int _{0}^{1}\rho (r){r}^{d-3}\displaystyle\int _{ 0}^{1}N(r,t)t{(1 - {t}^{2})}^{\frac{d-5} {2} }\cos (Trt)dt \\ & & = -\frac{C_{d}} {T} \displaystyle\int _{0}^{1}t{(1 - {t}^{2})}^{\frac{d-5} {2} }dt\displaystyle\int _{0}^{1}N(r,t)\cos (Trt){r}^{d-3}\rho (r)dr \\ & & = -\frac{C_{d}} {T} \displaystyle\int _{0}^{1}t{(1 - {t}^{2})}^{\frac{d-5} {2} }dt\displaystyle\int _{0}^{1}dr\frac{\sin (Trt)} {Tt} \\ & & \quad \times \left ({r}^{d-3}\rho (r)\frac{\partial N} {\partial r} (r,t) + ({r}^{d-3}\rho ^{\prime}(r) + (d - 3){r}^{d-4}\rho (r))N(r,t)\right ) \\ & & = -\frac{C_{d}} {{T}^{2}} J(d - 2,r\rho (r), \frac{\partial n} {\partial r} ) + \frac{C_{d}} {{T}^{2}} J(d - 2,\rho (r) + r\rho ^{\prime}(r),n) \\ \end{array}$$

and we conclude by induction.

3.2 B Guillemin’s Condition

For q ∈ X, we denote by $\bar{q} \in X$ the unique point proportional to q and distinct from it. We will consider the projective space $\mathbb{P}X = \mathbb{R}{\mathbb{P}}^{d}$, $\mathbb{P}Y = \mathbb{R}{\mathbb{P}}^{d}$ and the projectivized incidence variety $\mathbb{P}Z =\{ (q,p) \in \mathbb{P}X \times \mathbb{P}Y :\langle q,p\rangle = 0\}$.Consider the projectivized double fibration

Then any two fibers F _p (ℙX) intersect transversally (since before projectivization, the only non-transversal intersection was between fibers over antipodal points). Denote N _W  ⊂ T ^∗(ℙX ×ℙY ), N _E  ⊂ T ^∗(X ×Y) the conormal bundles of W, E respectively. Since dimE = 2d − 1 and dim(X ×Y) = 2d, the fibers of N _E , N _W are one-dimensional. Recall that T _(q, p) E = { (ξ, η) ∈ T _q X ×T _p Y: ⟨q, η⟩ + ⟨ξ, p⟩ = 0}. Therefore, N _E over (q, p) ∈ E has its fiber spanned by (p, q) ∈ T _q ^∗ X ×T _p ^∗ Y. One thus has N _E  ∖ 0 ⊂ (T ^∗ X ∖ 0) ×(T ^∗ Y ∖ 0), and ρ : N _E  ∖ 0 → T ^∗ Y ∖ 0 given by ((q, p), t(p, q))↦(p, tq) is an immersion, which is two-to-one since $\rho ((q,p),t(p,q)) = \rho ((\bar{q},p),(-t)(p,\bar{q}))$. The corresponding map ρ : N _W  ∖ 0 → T ^∗ ℙY ∖ 0 is already an injective immersion. Thus Guillemin’s condition is satisfied, and we conclude

Corollary 1.

For any smooth positive measure $\mu \in {\mathcal{M}}^{\infty }(\mathbb{P}Z)$ , ℛ _μ ^T ℛ _μ : C ^∞ (ℙX) → C ^∞ (ℙX) is an elliptic pseudodifferential operator.

3.3 C Pseudo-Differential Operators

For a survey of the subject, see for instance [4].We will study the norm of a pseudodifferential linear operator $P : {C}^{\infty }({\mathbb{R}}^{n}) \rightarrow {C}^{\infty }({\mathbb{R}}^{n})$ which is given by its symbol p(x, ξ)

$$Pf(x) =\displaystyle\int d\xi {e}^{i\langle x,\xi \rangle }p(x,\xi )\hat{f}(\xi )$$

where p ∈ Sym ^m(K), i.e.,

• 1. $p \in {C}^{\infty }({\mathbb{R}}^{n} \times {\mathbb{R}}^{n})$

• 2. p has compact x − support $K \subset {\mathbb{R}}^{n}$

• 3.  | D _ξ ^β D _x ^α p(x, ξ) | ≤ C _αβ(1 +  | ξ | )^{m − | β |}

It is well known that for all $s \in \mathbb{R}$, P extends to a bounded operator between Sobolev spaces

$$P : L_{s+m}^{2}({\mathbb{R}}^{n}) \rightarrow L_{ s}^{2}({\mathbb{R}}^{n})$$

We will trace the proof of this fact to understand the dependence on p of the operator norm $\|P\|$.

Proposition 4.

There exists a constant C(n,s) such that

$$\|P\|_{L_{s+m}^{2}\rightarrow L_{s}^{2}} \leq C(n,s)\sup\limits_{\vert \alpha \vert \leq n+\lfloor \vert s\vert \rfloor +1}C_{\alpha 0}\vert K\vert $$

Proof.

All the integrals in the following are over ${\mathbb{R}}^{n}$. Start by integrating by parts:

$${\vert }\displaystyle\int dxD_{x}^{\alpha }p(x,\xi ){e}^{i\langle x,\zeta \rangle }{\vert } = {\vert }{\zeta }^{\alpha }{\vert }{\vert }\displaystyle\int dxp(x,\xi ){e}^{i\langle x,\zeta \rangle }{\vert }.$$

So

$$\displaystyle\begin{array}{rcl} \left \vert \displaystyle\int dxp(x,\xi ){e}^{i\langle x,\zeta \rangle }\right \vert & \leq & \min (\vert \zeta {\vert }^{-\vert \alpha \vert }C_{ \alpha 0}{(1 + \vert \xi \vert )}^{m}\vert K\vert,C_{ 00}{(1 + \vert \xi \vert )}^{m}\vert K\vert ) \\ & \leq & {2}^{\vert \alpha \vert }(C_{ 00} + C_{\alpha 0})\vert K\vert {(1 + \vert \xi \vert )}^{m}{(1 + \vert \zeta \vert )}^{-\vert \alpha \vert } \\ & =& C_{\alpha }\vert K\vert {(1 + \vert \xi \vert )}^{m}{(1 + \vert \zeta \vert )}^{-\vert \alpha \vert } \\ \end{array}$$

where C _α = 2^| α |(C ₀₀ + C _α0). We want to bound

$$Pu(x) =\displaystyle\int d\xi {e}^{i\langle x,\xi \rangle }p(x,\xi )\hat{u}(\xi ).$$

Take $v \in L_{-s}^{2}({\mathbb{R}}^{n})$, then

$$\displaystyle\begin{array}{rcl} (Pu,v)& =& \displaystyle\int d\zeta \hat{v}(\zeta )\hat{Pu}(\zeta ) =\displaystyle\int d\zeta \hat{v}(\zeta )\displaystyle\int dxPu(x){e}^{-i\langle x,\zeta \rangle } \\ & =& \displaystyle\int \displaystyle\int d\zeta dx\hat{v}(\zeta ){e}^{-i\langle x,\zeta \rangle }\displaystyle\int d\xi \hat{u}(\xi )p(x,\xi ){e}^{i\langle x,\xi \rangle } \\ & =& \displaystyle\int \displaystyle\int d\zeta d\xi \hat{u}(\xi )\hat{v}(\zeta )\displaystyle\int dxp(x,\xi ){e}^{i\langle x,\xi -\zeta \rangle } \\ \end{array}$$

so denoting

$$\Phi (\xi,\zeta ) = {(1 + \vert \xi \vert )}^{-m-s}{(1 + \vert \zeta \vert )}^{s}\left \vert \displaystyle\int dxp(x,\xi ){e}^{i\langle x,\xi -\zeta \rangle }\right \vert $$

we have

$$\displaystyle\begin{array}{rcl} \vert (Pu,v)\vert & \leq & \displaystyle\int \displaystyle\int d\xi d\zeta \hat{u}(\xi )\hat{v}(\zeta )\Phi (\xi,\zeta ){(1 + \vert \xi \vert )}^{m+s}{(1 + \vert \zeta \vert )}^{-s} \\ & \leq & \Big{(}\displaystyle\int d\xi \vert \hat{u}(\xi ){\vert }^{2}(1 + \vert \xi )\vert {)}^{2(m+s)}\displaystyle\int d\zeta \Phi (\xi,\zeta ){\Big{)}}^{1/2} \\ & & \times \Big{(}\displaystyle\int \displaystyle\int d\zeta \vert \hat{v}(\zeta ){\vert }^{2}{(1 + \vert \zeta \vert )}^{-2s}\displaystyle\int d\xi \Phi (\xi,\zeta ){\Big{)}}^{1/2}.\end{array}$$

Now

$$\displaystyle\begin{array}{rcl} \Phi (\xi,\zeta )& \leq & C_{\alpha }\vert K\vert {(1 + \vert \xi \vert )}^{-m-s}{(1 + \vert \zeta \vert )}^{s}{(1 + \vert \xi \vert )}^{m}{(1 + \vert \xi - \zeta \vert )}^{-\vert \alpha \vert } \\ &\leq & C_{\alpha }\vert K\vert {(1 + \vert \xi - \zeta \vert )}^{\vert s\vert -\vert \alpha \vert }.\end{array}$$

Therefore,

$$\displaystyle\int d\xi \Phi (\xi,\zeta ) \leq A(n,\vert s\vert -\vert \alpha \vert )C_{\alpha }\vert K\vert $$

where

$$A(n,l) =\displaystyle\int d\xi {(1 + \vert \xi \vert )}^{l}$$

similarly

$$\displaystyle\int d\zeta \Phi (\xi,\zeta ) \leq C(n,\vert s\vert -\vert \alpha \vert )C_{\alpha }\vert K\vert $$

implying

$$\vert (Pu,v)\vert \leq A(n,\vert s\vert -\vert \alpha \vert )C_{\alpha }\vert K\vert \|u\|_{m+s}\|v\|_{-s}$$

$$\Rightarrow \| P\|_{L_{s+m}^{2}\rightarrow L_{s}^{2}} \leq A(n,\vert s\vert -\vert \alpha \vert )C_{\alpha }\vert K\vert $$

and this holds for all α s.t. A(n, | s | − | α | ) = ∫dξ(1 +  | ξ | )^{| s | − | α |} < ∞, i.e. | s | − | α |  <  − n ⇔  | α |  > n +  | s | . We thus choose α s.t. | α |  = ⌊ | s | ⌋ + n + 1, and recall that C _α = 2^| α |(C ₀₀ + C _α0) to obtain the stated estimate.

We will also need the relation between the Schwartz kernel and the symbol.

Proposition 5.

Suppose the Schwartz kernel of P is given by K(x,y), namely

$$\langle Pf(x),g(x)\rangle =\displaystyle\int dxdyK(x,y)f(y)g(x).$$

Then the symbol p(x,ξ) of P is given by

$$p(x,\xi ) =\displaystyle\int {e}^{-i\langle y,\xi \rangle }K(x,x - y)dy.$$

Proof.

Write for smooth compactly supported f, g

$$\displaystyle\begin{array}{rcl} \langle Pf(x),g(x)\rangle & =& \displaystyle\int dxd\xi {e}^{i\langle x,\xi \rangle }p(x,\xi )\hat{f}(\xi )g(x) \\ & =& \displaystyle\int dxdyd\xi {e}^{i\langle x-y,\xi \rangle }p(x,\xi )f(y)g(x)\end{array}$$

That is, K(x, y) = ∫dξe ^{i⟨x − y, ξ⟩} p(x, ξ), and ⟨Pf, g⟩ = ∫f(y)g(x)K(x, y)dydx. Denoting by $\check{h}(x) =\int d\xi h(\xi ){e}^{i\langle x,\xi \rangle }$ the inverse Fourier transform, we can also write $K(x,y) =\check{ p}(x,\bullet )(x - y)\;\Longleftrightarrow\;K(x,x - y) =\check{ p}(x,\bullet )(y)$, so

$$p(x,\xi ) =\displaystyle\int {e}^{-i\langle y,\xi \rangle }K(x,x - y)dy$$

as claimed.