Fredholm boundary problems with first-order Calderón defects

A boundary parametrix can be Fredholm even when its Calderón matrix is only a projection to leading order. The point is to keep the first lower-order defect in every formula. That defect gains one derivative, compact Sobolev inclusion handles it at the base level, and a separate cancellation makes the boundary right error smoothing.

This lesson builds the full argument. It starts with the original two measurement scales, proves extension regularity, derives both ordered Green errors, constructs the projected boundary inverse, checks all five anisotropic error maps, and proves regularity for solutions and dual obstructions. The final result is a Fredholm realization whose index is independent of Sobolev level and lower-order terms.

Named prerequisites are Inverting mixed symbols without changing their scales, Composing symbols with two independent orders, Sobolev mapping with normal and tangential weights, Finite inverse expansions for normal polynomials, Cauchy data from jumps and residues, Solving an elliptic system from compatible boundary measurements, and Finite defects under perturbation. We use each result only with its stated hypotheses.

1. The operator, the two weights, and the boundary spaces

Let $X$ be a compact smooth manifold with boundary $Y$, and fix an inward collar coordinate $t\geq0$. Let $E\to X$ and $F\to X$ have the same finite rank. Near $Y$, the generalized interior operator has the ordered normal form

$$P^b=\sum_{a=0}^{m}P_a(y,t,D_y)D_t^a, \qquad \operatorname{ord}_{D_y}P_a\leq m-a, \qquad P_m(y,t):E_{(y,t)}\longrightarrow F_{(y,t)} \text{ invertible}. \tag{GF1}$$

The tangential entries are classical pseudodifferential operators. The interior term $P^i$ is supported away from the collar, and $P=P^b+P^i$ has an elliptic extension to a boundaryless neighborhood $\widehat X$. In a collar chart its complete left symbol $p(x,\eta,\zeta)$ and inverse symbol $\tau(x,\eta,\zeta)$ use the numerical symbol weights

$$R=1+|(\eta,\zeta)|, \qquad T=1+|\eta|, \qquad \|\partial_x^\beta\partial_\eta^\alpha\partial_\zeta^r a\| \leq C_{\beta\alpha r}R^{u-r}T^{v-|\alpha|} \tag{GF2}$$

for $a\in S^{u,v}$. The Sobolev norms retain the distinct quadratic weights

$$Q=(1+|\eta|^2+\zeta^2)^{1/2}, \qquad h=(1+|\eta|^2)^{1/2}, \qquad \|u\|_{(s,t)}^2=(2\pi)^{-n}\int Q^{2s}h^{2t}|\widehat u|^2. \tag{GF3}$$

The complete constructions in Sections 14–18 and the full-extension correction in Section 23 give proper global operators $\Tau=\operatorname{Op}(\tau)$, $R_E$, and $R_F$, with matrix multiplication in the displayed order, such that

$$\Tau P=I+R_E, \qquad P\Tau=I+R_F, \qquad r_E,r_F\in S^{-1,0}. \tag{GF4}$$

Section 20 proves the stronger finite decompositions MH1–MH19 and their one-sided estimates, including boundary-supported inputs. For the first inverse, take N=1 in the actual finite construction:

$$\begin{aligned} r^+\Tau e^+&:\bar H_{(s,t)}\longrightarrow\bar H_{(s+m,t)},\\ r^+R_Ee^+,\ r^+R_Fe^+&:\bar H_{(s,t)}\longrightarrow\bar H_{(s+1,t)} \qquad(s\geq0), \end{aligned} \tag{GF5}$$

and, whenever $w\in H_{(-m,r)}(\widehat X)$ is supported on $Y$,

$$\begin{aligned} r^+\Tau w&\in\bar H_{(\nu,r-\nu)},\\ r^+R_Ew,\ r^+R_Fw&\in\bar H_{(\nu+1-m,r-\nu)} \end{aligned} \qquad(\nu\in\mathbb R). \tag{GF6}$$

All constants in (GF5)–(GF6) are finite combinations of the source $R,T$ seminorms; the spaces themselves are the $Q,h$ spaces (GF3).

Write $\gamma_k u=(D_t^ku)|_Y$, $0\leq k<m$, and $\gamma u=(\gamma_0u,\ldots,\gamma_{m-1}u)$. For real $r$, set

$$\mathcal C^r=\bigoplus_{k=0}^{m-1}H^{r-k-1/2}(Y,E), \qquad \mathcal D^r=\bigoplus_{j=1}^{J}H^{r-m_j-1/2}(Y,G_j). \tag{GF7}$$

Let $B_j=\sum_{k<m}B_{jk}\gamma_k$, where $B_{jk}\in\Psi^{m_j-k}(Y;E,G_j)$, and write $\mathcal B=(B_{jk})$. The complementing condition says that the principal boundary matrix $b$ is a fiberwise bijection from the stable Cauchy bundle onto $\bigoplus_jG_j$.

2. Elliptic regularity on the extension

Let $u\in H^{s+m-1}_{\mathrm{comp}}(\widehat X,E)$ and $Pu\in H^s_{\mathrm{comp}}(\widehat X,F)$. Choose chart cutoffs $\chi\prec\eta$, so $\eta=1$ on a neighborhood of $\operatorname{supp}\chi$. With $[P,\eta]=P\eta-\eta P$, (GF4) gives on that chart

$$\chi u =\chi\Tau(\eta Pu)+\chi\Tau([P,\eta]u) -\chi R_E(\eta u)+G_\chi u. \tag{GF8}$$

Here $G_\chi=0$: apply the exact first identity of (GF4) to $\eta u$ and then multiply by $\chi$. Section 23 proves that tangentially separated mixed kernels can remain singular in the normal coordinate; their contribution stays in the full commutator. The commutator has generalized order $m-1$: each normal commutator lowers the normal differential degree once, and each tangential commutator lowers the tangential pseudodifferential order once. U030 and (GF5) therefore give

$$\|\chi u\|_{H^{s+m}} \leq C_{\chi,\eta,K} \bigl(\|\eta Pu\|_{H^s}+\|u\|_{H^{s+m-1}}\bigr). \tag{GF9}$$

The lower norm in (GF9) retains the entire input, including distant tangential components. The exact counterexample ML1–ML5 in Section 23 disproves the stronger estimate with only $\eta u$ in that norm. In the overlap, invert the complete symbol of $P^b+P^i$; ellipticity of $P$ alone does not prove ellipticity of $P^i$ there. Use ordinary interior charts only where the collar summand vanishes. ML6–ML8 gives both ordered full-extension errors. A finite partition of unity over a fixed compact $K\supset\operatorname{supp}u$ proves

$$u\in H^{s+m}_{\mathrm{comp}}(\widehat X), \qquad \|u\|_{H^{s+m}} \leq C_K\bigl(\|Pu\|_{H^s}+\|u\|_{H^{s+m-1}}\bigr). \tag{GF10}$$

This proves extension regularity in the generalized class. Repeating (GF10) raises any distributional solution with smooth right side through all real Sobolev orders, one full derivative at a time.

3. The boundary source and two exact Green identities

For a Cauchy vector $U=(U_0,\ldots,U_{m-1})$, define the exact boundary distribution

$$\mathcal JU=\frac1i\sum_{a=1}^{m}P_a(y,t,D_y) \sum_{r=0}^{a-1}U_{a-1-r}\otimes D_t^r\delta_0. \tag{GF11}$$

The coefficient $P_a(y,t,D_y)$ acts after the delta derivative. Expanding that multiplication gives every normal derivative of $P_a$ as in U025 (BF20); none is removed. For smooth $u$, distributional differentiation of its zero extension gives

$$P(e^+u)=e^+(Pu)+\mathcal J\gamma u. \tag{GF12}$$

Define

$$\begin{gathered} V=r^+\Tau e^+, \qquad K=r^+\Tau\mathcal J, \qquad Q=\gamma K,\\ E=r^+R_Ee^+, \qquad F=r^+R_Fe^+, \qquad H=r^+R_F\mathcal J. \end{gathered} \tag{GF13}$$

Apply $r^+\Tau$ to (GF12), then use the first identity in (GF4). Apply $P$ to the two terms in (GF13), then use the second identity in (GF4) and the fact that $r^+\mathcal JU=0$. This proves the exact identities

$$u+Eu=V(Pu)+K\gamma u, \qquad PV=I+F, \qquad PK=H. \tag{GF14}$$

The boundary source estimate is obtained without deleting any delta coefficient. If $U_k\in H^{r-k-1/2}(Y)$, the coefficient of $D_t^j\delta_0$ in (GF11) lies in $H^{r-m+j+1/2}$. The full jet representation and cross-term Gram norm MH14–MH17 in Section 20 gives

$$\|\mathcal JU\|_{(-m,r)} \leq C_r\sum_{k=0}^{m-1}\|U_k\|_{H^{r-k-1/2}}. \tag{GF15}$$

Combining (GF6) and (GF15) controls $K$, $E K$, $V H$, and all later boundary-source terms for arbitrary real output split.

4. The generalized Calderón matrix

Write the full homogeneous principal polynomial, without identifying it with its leading normal coefficient, as

$$\mathfrak p(y,\eta,\zeta) =\sum_{a=0}^{m}p_{a,m-a}(y,\eta)\zeta^a. \tag{GF16a}$$

Sections 21 and 24 prove that the entries of $Q$ satisfy $Q_{kl}\in\Psi^{k-l}(Y;E,E)$, including the full source and every output derivative. PC1–PC7 compares the actual inverse with the entire homogeneous polynomial, retaining the lower-order difference. Its weighted principal symbol is

$$q_{kl}(y,\eta)=\frac1{2\pi i}\int_{\Gamma_+} \sum_{j=0}^{m-1-l}\zeta^{k+j} \mathfrak p(y,\eta,\zeta)^{-1} p_{j+l+1,m-j-l-1}(y,\eta)\,d\zeta. \tag{GF16}$$

The inverse stands on the left of the coefficient. The jump argument CD23–CD26 applies to this full matrix polynomial and proves that $q^2=q$, with range equal to the Cauchy data of the inward-decaying frozen solutions. Consequently

$$Q^2-Q\in\Psi^{-1}_{\mathrm{wt}}(Y;E^{\oplus m},E^{\oplus m}), \tag{GF17}$$

where weighted order $-1$ means that the $(k,l)$ entry lies in $\Psi^{k-l-1}$. No smoothing conclusion follows from the leading-symbol identity.

There is also an exact operator formula for this defect. Put $u=KU$ in the first identity of (GF14). Since $PKU=HU$, take Cauchy traces and obtain

$$Q^2-Q=\gamma E K-\gamma V H. \tag{GF18}$$

Every term on the right gains one weighted derivative by (GF5), (GF6), and (GF15). This verifies the order in (GF17) from the actual ordered errors.

Put $u=Vf$ in the same identity. Since $PVf=f+Ff$, cancellation of $Vf$ gives

$$Q\gamma Vf=\gamma E Vf-\gamma V Ff. \tag{GF19}$$

Thus, for $s\geq0$,

$$Q\gamma V:\bar H_{(s,t)}(X)\longrightarrow \bigoplus_{k=0}^{m-1}H^{s+t+m-k+1/2}(Y,E). \tag{GF20}$$

The extra derivative in (GF20) comes from both ordered errors in (GF19); it is not obtained by tracing $Vf$ alone.

5. Boundary inversion while the defect remains visible

Conjugate the Cauchy and target components by the invertible scalar order reductions of U025 (BF6)–(BF7). After this conjugation, $Q$ and $\mathcal B$ have order zero and the complementing condition says

$$b:q(E^{\oplus m})\longrightarrow\bigoplus_jG_j \quad\text{is an isomorphism for every }(y,\eta)\ne0. \tag{GF21}$$

Because $BQ^2$ has principal symbol $bq$, the standard symbol recursion produces $T$ such that

$$\mathcal BQ^2T-I\in\Psi^{-\infty}. \tag{GF22}$$

Set $S=QT$. Then

$$\mathcal BQS-I\in\Psi^{-\infty}, \qquad QS-S=(Q^2-Q)T\in\Psi^{-1}_{\mathrm{wt}}. \tag{GF23}$$

The column $C=(\mathcal B,I-Q)^t$ has injective principal symbol: if $(I-q)v=0$ and $bv=0$, then $v\in\operatorname{ran}q$ and (GF21) forces $v=0$. A full left symbol recursion gives a row $(T',T'')$ satisfying

$$T'\mathcal B+T''(I-Q)-I\in\Psi^{-\infty}. \tag{GF24}$$

Define $S''=T''(I-Q)$. The principal symbols of $T'$ and $S$ agree: for $g=bqv$, (GF24) gives $\sigma(T')g=qv$, while (GF22) gives $\sigma(S)g=qv$. Hence $T'-S$ is one weighted order lower. It follows that

$$\begin{aligned} R_0&=I-S\mathcal B-S''\in\Psi^{-1}_{\mathrm{wt}},\\ S''Q&=T''(I-Q)Q\in\Psi^{-1}_{\mathrm{wt}},\\ R_1&=R_0+S''Q\in\Psi^{-1}_{\mathrm{wt}}. \end{aligned} \tag{GF25}$$

Undoing the order reductions preserves every entry degree:

$$S_{kj}\in\Psi^{k-m_j}, \qquad S''_{kl}\in\Psi^{k-l}, \qquad (R_i)_{kl}\in\Psi^{k-l-1}. \tag{GF26}$$

For $U=\gamma u$ and $g=\mathcal BU$, (GF25) is the exact decomposition

$$U=Sg+S''(I-Q)U+R_1U. \tag{GF27}$$

Taking the trace of the first identity in (GF14) gives $(I-Q)U=\gamma V(Pu)-\gamma Eu$. The order-minus-one terms in (GF25)–(GF27) have been retained; none has been renamed as smoothing.

6. The generalized local parametrix and all five errors

Define

$$\mathscr L(f,g)=(I+KS''\gamma)Vf+KSg. \tag{GF28}$$

Substitute (GF27) and $(I-Q)U=\gamma Vf-\gamma Eu$ into (GF14). The exact left identity is

$$u=\mathscr L(Pu,Bu)+\mathscr Ku, \qquad \mathscr K=-E-KS''\gamma E+KR_1\gamma. \tag{GF29}$$

Using $PV=I+F$ and $PK=H$ gives the interior row of the right identity:

$$P\mathscr L(f,g)=f+\mathscr K_1f+\mathscr K_2g, \qquad \mathscr K_1=F+HS''\gamma V, \qquad \mathscr K_2=HS. \tag{GF30}$$

The boundary row is

$$B\mathscr L(f,g)=g+\mathscr K_3f+\mathscr K_4g, \tag{GF31}$$

where

$$\mathscr K_3=\mathcal B(I+QS'')\gamma V, \qquad \mathscr K_4=\mathcal BQS-I\in\Psi^{-\infty}. \tag{GF32}$$

The one-derivative gain in $\mathscr K_3$ is an exact cancellation. Since $S''=I-S\mathcal B-R_0$,

$$\begin{aligned} \mathcal B(I+QS'') &=\mathcal B+\mathcal BQ-\mathcal BQS\mathcal B-\mathcal BQR_0\\ &=\mathcal BQ+D_{-1}, \qquad D_{-1}=-\mathscr K_4\mathcal B-\mathcal BQR_0. \end{aligned} \tag{GF33}$$

The $(j,k)$ entry of $D_{-1}$ has order at most $m_j-k-1$. Therefore (GF19) controls the $\mathcal BQ\gamma V$ term and the order drop controls $D_{-1}\gamma V$.

We now verify every required parametrix-error mapping. The trace of $Vf\in\bar H_{(s+m,t)}$ has $k$-th component in $H^{s+t+m-k-1/2}$. Equations (GF20) and (GF33) therefore give

$$\mathscr K_3:\bar H_{(s,t)}\longrightarrow \bigoplus_jH^{s+t+m-m_j+1/2}(Y,G_j) \qquad(s\geq0). \tag{GF34}$$

For $g_j\in H^{t+m-m_j-1/2}$, (GF26) gives $(Sg)_k\in H^{t+m-k-1/2}$. Hence (GF15) puts $\mathcal JSg$ in $H_{(-m,t+m)}$. Use (GF6) for $R_F$ with $\nu=s+m$ to obtain

$$\mathscr K_2: \bigoplus_jH^{t+m-m_j-1/2}(Y,G_j) \longrightarrow\bar H_{(s+1,t-s)}(X,F) \qquad(s,t\in\mathbb R). \tag{GF35}$$

For $u\in\bar H_{(s,t)}$, $s\geq m$, the first term of (GF29) belongs to $\bar H_{(s+1,t)}$ by (GF5). Both $S''\gamma Eu$ and $R_1\gamma u$ have $k$-th component in $H^{s+t-k+1/2}$; (GF15) and (GF6), with $\nu=s+1$, put their images under $K$ in the same interior space. Thus

$$\mathscr K:\bar H_{(s,t)}\longrightarrow\bar H_{(s+1,t)} \qquad(s\geq m). \tag{GF36}$$

The identical calculation with $F$ and $HS''\gamma V$ gives

$$\mathscr K_1:\bar H_{(s,t)}\longrightarrow\bar H_{(s+1,t)} \qquad(s\geq0). \tag{GF37}$$

Finally, (GF23) gives

$$\mathscr K_4:\mathcal D'(Y,\mathcal G)\longrightarrow C^\infty(Y,\mathcal G). \tag{GF38}$$

Equations (GF34)–(GF38) prove all five requested estimates, including the absence of $s$ from the input exponent in (GF35).

7. One derivative at a time for solutions

Let $u\in\bar H^m$ have bounded collar support, let $Pu\in\bar H^{s-m}$, and let $B_ju\in H^{s-m_j-1/2}$, where $s\geq m$. Suppose first that $u\in\bar H^r$ for some $m\leq r<s$, and put

$$r_+=\min(s,r+1). \tag{GF39}$$

Choose $\chi\prec\eta$ in the collar and apply (GF29) to $\eta u$. The interior commutator $[P,\eta]u$ has order $m-1$, so it lies in $H^{r-m+1}\subset H^{r_+-m}$. For the boundary rows, write $\eta_a=(D_t^a\eta)|_Y$, so $\eta_0=\eta|_Y$, and let $M_{\eta_a}$ denote multiplication by $\eta_a$. The full normal Leibniz formula is

$$\begin{aligned} \gamma_k(\eta u) &=\sum_{a=0}^k\binom ka M_{\eta_a}\gamma_{k-a}u,\\ B_j(\eta u)-M_{\eta_0}B_ju &=\sum_{k=0}^{m-1}[B_{jk},M_{\eta_0}]\gamma_ku\\ &\quad+\sum_{k=0}^{m-1}\sum_{a=1}^k \binom ka B_{jk}M_{\eta_a}\gamma_{k-a}u. \end{aligned} \tag{GF40}$$

The first line follows by induction from the product rule for $D_t$; applying each $B_{jk}$ and subtracting $M_{\eta_0}\sum_kB_{jk}\gamma_ku$ gives the remaining lines in the displayed order. The first sum has coefficient order at most $m_j-k-1$. In the second sum put $l=k-a$. Its coefficient has order at most $m_j-k=m_j-l-a\leq m_j-l-1$, with every binomial factor and cutoff derivative retained. Thus each coefficient acting on $\gamma_lu\in H^{r-l-1/2}$ has output in $H^{r-m_j+1/2}\subset H^{r_+-m_j-1/2}$. The normal-cutoff terms need not vanish; they satisfy precisely the same weighted order bound. Tangentially separated terms retain their normal-diagonal kernels and the actual lower mixed orders of Sections 14, 18 and 20. The input at this step is known globally in the lower Sobolev space, so the full commutator estimates apply. The data term $\mathscr L(P(\eta u),B(\eta u))$ therefore lies in $\bar H^{r_+}$, while (GF36) puts $\mathscr K(\eta u)$ in $\bar H^{r+1}\subset\bar H^{r_+}$. Hence $\chi u\in\bar H^{r_+}$.

Starting with $r_0=m$, define $r_{a+1}=\min(s,r_a+1)$. After $\lceil s-m\rceil$ steps, $r_a=s$. If $s-m$ is not an integer, the final step has size strictly below one and all inclusions in (GF40) remain valid. A finite collar partition and ordinary interior regularity prove

$$u\in\bar H^s(X,E). \tag{GF41}$$

This proves the solution bootstrap with the fractional final step and all cutoff commutators included.

8. The dual one-step bootstrap

Assume the following relation on the whole compact manifold and its boundary:

$$\langle Pu,v\rangle_X+\sum_j\langle B_ju,h_j\rangle_Y=0 \quad\text{for every compactly supported smooth }u, \tag{GF42}$$

with

$$v\in\bar H_{(0,t)}(X,F^*\otimes\Omega_X), \qquad h_j\in H^{t+m_j+1/2-m}(Y,G_j^*\otimes\Omega_Y). \tag{GF43}$$

Insert $u=\mathscr L(0,g)$ in (GF42) and use (GF30)–(GF32). For smooth global boundary data $g$,

$$h=-\mathscr K_2'v-\mathscr K_4'h. \tag{GF44}$$

In (GF35), choose its two free parameters to be $s=-1$ and $t=-t_{\!0}-1$, where $t_{\!0}$ denotes the exponent in (GF43). Then

$$\mathscr K_2: \bigoplus_jH^{-t_{\!0}+m-m_j-3/2} \longrightarrow\bar H_{(0,-t_{\!0})}. \tag{GF45}$$

Transposition of (GF45), together with the smoothing term in (GF44), proves

$$h_j\in H^{t_{\!0}+1+m_j+1/2-m}. \tag{GF46}$$

Next insert $u=\mathscr L(f,0)$. Then

$$v=-\mathscr K_1'v-\mathscr K_3'h. \tag{GF47}$$

The pointwise inequality $h/Q\leq1$ gives the continuous embedding $H_{(0,t_{\!0})}\hookrightarrow H_{(-1,t_{\!0}+1)}$. With $s=0$, $t=-t_{\!0}-1$, (GF37) transposes to

$$\mathscr K_1':\dot H_{(-1,t_{\!0}+1)}(\overline X)\longrightarrow \dot H_{(0,t_{\!0}+1)}(\overline X). \tag{GF48}$$

For the same parameters, (GF34) has boundary output exponent $-t_{\!0}+m-m_j-1/2$; its dual is exactly the exponent of $h_j$ in (GF43). Therefore

$$\mathscr K_3': \bigoplus_jH^{t_{\!0}+m_j+1/2-m} \longrightarrow \dot H_{(0,t_{\!0}+1)}(\overline X). \tag{GF49}$$

Equations (GF47)–(GF49) prove

$$v\in\bar H_{(0,t+1)}, \qquad h_j\in H^{t+1+m_j+1/2-m}. \tag{GF50}$$

The dual of a restriction space is the supported Sobolev space with negated exponents. Since the normal order of $v$ is zero, its zero extension is bounded into $H_{(0,t_{\!0})}$ and has support in $\overline X$. PC8 proves its embedding into $\dot H_{(-1,t_{\!0}+1)}(\overline X)$. Thus (GF48)–(GF49) act on that supported zero extension and their output is restricted back to $X$. Section 24 proves every pairing, density and extension-independence step in PC8–PC9. The global relation authorizes both parametrix columns; no arbitrary isolated-patch smoothing claim is used.

Iterating (GF50) makes every $h_j$ smooth and gives unlimited tangential regularity to $v$. After finitely many steps the second mixed exponent is nonnegative, so $v\in L^2$. Tests away from $Y$ show $P^\dagger v=0$ in the interior. The boundary relation can be written

$$P^\dagger e^+v =-\sum_{k=0}^{m-1}(-1)^k \left(\sum_jB_{jk}^\dagger h_j\right)\otimes D_t^k\delta_0. \tag{GF51}$$

Move the normal derivatives of $P^\dagger$ to the right. Its leading normal coefficient is the invertible map $(-1)^mP_m^t$. Exactly as in BF40, comparison from $D_t^{m-1}\delta_0$ downward uniquely determines smooth exterior jets $\gamma_0w,\ldots,\gamma_{m-1}w$ that cancel (GF51). Recursively impose $\gamma_r(P^\dagger w)=0$ for every $r\geq0$; at each stage the new jet has coefficient $(-1)^mP_m^t$, while the tangential pseudodifferential coefficients act on already known smooth fields and therefore preserve smoothness. The convergent Borel construction BF41 realizes these jets on the exterior collar. Then

$$W=e^+v+e^-w, \qquad P^\dagger W\in C^\infty \tag{GF52}$$

near $Y$. The full transpose is elliptic and has the same mixed kernel estimates. Applying the corrected full-operator (GF10) repeatedly to $P^\dagger$ proves $W$, and hence $v$, smooth up to the boundary.

9. Closed range and the finite defect spaces

At the base level put

$$\mathcal Y^m=\bar H^0(X,F)\oplus \bigoplus_jH^{m-m_j-1/2}(Y,G_j), \qquad A_m u=(Pu,Bu). \tag{GF53}$$

Equations (GF28)–(GF29) give bounded $\mathscr L:\mathcal Y^m\to\bar H^m$ and

$$I=\mathscr L A_m+\mathscr K, \qquad \mathscr K:\bar H^m\longrightarrow\bar H^{m+1}. \tag{GF54}$$

The inclusion $\bar H^{m+1}\hookrightarrow\bar H^m$ is compact. If the unit ball of $\ker A_m$ were not compact, choose a sequence in it with no convergent subsequence. Formula (GF54) makes that sequence equal to its $\mathscr K$-image, which does have a convergent subsequence. Hence $\ker A_m$ is finite-dimensional.

Let $u_n\perp\ker A_m$ and suppose $A_mu_n$ is Cauchy. The sequence $u_n$ is bounded. Otherwise, after passing to a subsequence with $\|u_n\|\to\infty$, set $w_n=u_n/\|u_n\|$. Then $A_mw_n\to0$, while compactness gives a convergent subsequence of $\mathscr Kw_n$. Equation (GF54) makes $w_n$ converge to a unit vector $w\in\ker A_m$, contradicting $w_n\perp\ker A_m$. From boundedness, compactness of $\mathscr K$, convergence of $\mathscr LA_mu_n$, and (GF54), a subsequence of $u_n$ converges strongly. Applying the same argument to differences shows the whole sequence is Cauchy. Thus $\operatorname{ran}A_m$ is closed.

For the cokernel use the full right identity (GF30)–(GF32) on the base data space:

$$A_m\mathscr L=I+\mathscr R,\qquad \mathscr R= \begin{pmatrix} \mathscr K_1&\mathscr K_2\\ \mathscr K_3&\mathscr K_4 \end{pmatrix}:\mathcal Y^m\longrightarrow\mathcal Y^m. \tag{GF54R}$$

Every block is compact in this specific space. With $\mathcal D^m$ as in (GF7), (GF34), (GF35), and (GF37) at $s=t=0$, and (GF38), give the factorizations

$$\begin{aligned} \mathscr K_1&:\bar H^0(X,F)\longrightarrow \bar H^1(X,F)\hookrightarrow\bar H^0(X,F),\\ \mathscr K_2&:\mathcal D^m\longrightarrow \bar H^1(X,F)\hookrightarrow\bar H^0(X,F),\\ \mathscr K_3&:\bar H^0(X,F)\longrightarrow \mathcal D^{m+1}\hookrightarrow\mathcal D^m,\\ \mathscr K_4&:\mathcal D^m\longrightarrow \mathcal D^{m+1}\hookrightarrow\mathcal D^m. \end{aligned} \tag{GF54M}$$

The first arrows are bounded. The second arrows are compact Sobolev inclusions on the compact $X$ or $Y$, with a finite sum over the boundary bundles. The inclusion $\bar H^1(X,F)\hookrightarrow\bar H^0(X,F)$ gains a full derivative, including the normal direction. Tangential gain alone would not justify this step. Write the matrix as the sum of its four blocks composed with the bounded coordinate projections and injections of $\mathcal Y^m$. Composition with bounded maps and finite addition preserve compactness, so $\mathscr R$ is compact.

We also need compactness of its transpose on the continuous dual, with the original pairings and densities. Here is a proof. For any $\varepsilon>0$, take a finite $\varepsilon$-net $y_1,\ldots,y_N$ for the image of the unit ball under $\mathscr R$. Let $\Pi_\varepsilon$ be the orthogonal projection in the Hilbert space $\mathcal Y^m$ onto their span. The distance-minimizing property of orthogonal projection gives $\|\mathscr R-\Pi_\varepsilon\mathscr R\|\leq\varepsilon$. The transpose of $\Pi_\varepsilon\mathscr R$ has finite rank, and $\|T'\|=\|T\|$ for a bounded map $T$: the inequality in one direction follows from the dual norm, and the reverse follows by taking a unit functional attaining the norm of each nonzero vector $Tx$ in a Hilbert space. Hence $\mathscr R'$ is an operator-norm limit of finite-rank maps. Such a limit is compact, because the image of the unit ball has, for every positive radius, a finite net obtained from a sufficiently close finite-rank map; its closure is complete and totally bounded, therefore compact.

Let

$$Z=\{z\in(\mathcal Y^m)':\langle A_mu,z\rangle=0 \text{ for every }u\in\bar H^m(X,E)\}. \tag{GF54Z}$$

This is a closed linear subspace of the dual. Its elements are the pairs $(v,h)$ in (GF42) at $t=0$, with $$v\in\bar H^0(X,F^*\otimes\Omega_X),\qquad h_j\in H^{-m+m_j+1/2}(Y,G_j^*\otimes\Omega_Y).$$ Transposing (GF54R), with no reversal of the block composition omitted, gives

$$\mathscr L'A_m'=I+\mathscr R',\qquad z=-\mathscr R'z\quad(z\in Z). \tag{GF54D}$$

The closed unit ball of $Z$ is contained in the compact closure of the image of the dual unit ball under $-\mathscr R'$, and it is closed there. It is therefore compact. The fixed-point lemma proved in Exercise 5 now makes $Z$ finite-dimensional. Section 8 separately proves its elements smooth; the tangential bootstrap in that section is used for regularity, not compact inclusion. Since the range is closed, the quotient $\mathcal Y^m/\operatorname{ran}A_m$ is a Hilbert space isomorphic to its orthogonal complement. Continuous functionals vanishing on the range identify its dual with $Z$, so its dimension equals $\dim Z$. Consequently $A_m$ is Fredholm.

10. Every Sobolev level and the index

For $s\geq m$, define

$$\mathcal Y^s=\bar H^{s-m}(X,F)\oplus \bigoplus_jH^{s-m_j-1/2}(Y,G_j), \qquad A_s:\bar H^s(X,E)\longrightarrow\mathcal Y^s. \tag{GF55}$$

Let $Z$ be the finite-dimensional smooth annihilator found at level $m$. Necessity in

$$\operatorname{ran}A_s =\left\{(f,g)\in\mathcal Y^s: \langle f,v\rangle+\sum_j\langle g_j,h_j\rangle=0 \text{ for every }(v,h)\in Z\right\} \tag{GF56}$$

follows by density. Conversely, data in the right side belong to $\mathcal Y^m$, so base-level closed range gives $u\in\bar H^m$ with $A_mu=(f,g)$. The solution bootstrap (GF41) gives $u\in\bar H^s$, proving sufficiency. A nullvector has zero smooth data, so the same bootstrap makes it smooth. Hence the kernel and annihilator are the same spaces at every $s\geq m$, and

$$\operatorname{ind}A_s=\dim\ker A_m-\dim Z \qquad(s\geq m). \tag{GF57}$$

Changing $P$ by generalized order at most $m-1$, or changing the $(j,k)$ boundary entry by order at most $m_j-k-1$, maps into a Sobolev space one order above the target. Compact Sobolev inclusion makes the resulting change of $A_s$ compact, so (GF57) is unchanged. A sufficiently small change of the finite coefficient and symbol seminorms gives a small operator-norm change $\bar H^s\to\mathcal Y^s$. Fredholm operators form an open set and their index is locally constant. Uniform invertibility of the interior principal symbol and of (GF21) on the compact cosphere bundle also persists under such a change. This proves the generalized Fredholm and index-stability theorem.

11. A scalar check that detects the retained defect

Take $m=1$, one boundary component, and suppose the stable symbol is the scalar $q=1$, while the constructed Calderón operator is

$$Q=I+A_{-1}, \qquad A_{-1}\in\Psi^{-1}(Y), \qquad A_{-1}\notin\Psi^{-\infty}(Y). \tag{GF58}$$

Then

$$Q^2-Q=A_{-1}+A_{-1}^2\in\Psi^{-1}, \tag{GF59}$$

and the leading order-minus-one symbol is that of $A_{-1}$. With $B=I$, choose a two-sided parametrix $T$ of $Q^2$ and set $S=QT$. Then $BQS-I=Q^2T-I$ is smoothing, while

$$QS-S=(Q^2-Q)T \tag{GF60}$$

generally remains order $-1$. This model checks the distinction used in (GF23)–(GF25): the boundary right error can be smoothing even though the Calderón and range defects are only one order lower.

12. Exercises with complete solutions

Exercise 1. Recover every entry order after the order reduction

Assume the source Cauchy component $k$ has weight $k$, the target component $j$ has weight $m_j$, and the conjugated inverse has order zero. Prove that the original entries of $S$, $S''$, and $R_1$ have the orders asserted in (GF26).

Solution. The source reduction $D_a$ has diagonal entry of order $k$, and the target reduction $D_b$ has diagonal entry of order $m_j$. Undoing $\widehat S=D_a^{-1}SD_b$ gives $S=D_a\widehat S D_b^{-1}$, so

$$\operatorname{ord}S_{kj}=k-m_j. \tag{GF61}$$

For an endomorphism of the Cauchy vector, both reductions use the source weights, giving $\operatorname{ord}S''_{kl}=k-l$. A weighted order-minus-one endomorphism loses one further order, so $\operatorname{ord}(R_1)_{kl}=k-l-1$. These are exactly the degrees in (GF26); no common scalar order replaces the component degrees.

Exercise 2. Derive the cancellation in the boundary right error

Starting only from $R_0=I-S\mathcal B-S''$ and $\mathscr K_4=\mathcal BQS-I$, derive (GF33).

Solution. The first identity gives $S''=I-S\mathcal B-R_0$. Substitute it without commuting any matrix factors:

$$\begin{aligned} \mathcal B(I+QS'') &=\mathcal B+\mathcal BQ-\mathcal BQS\mathcal B-\mathcal BQR_0\\ &=\mathcal BQ-\mathscr K_4\mathcal B-\mathcal BQR_0. \end{aligned} \tag{GF62}$$

The first correction is smoothing. The second contains $R_0$, whose $(k,l)$ entry has order $k-l-1$. Thus the entire correction to $\mathcal BQ$ has entry order at most $m_j-k-1$, which is the derivative gain used in (GF34).

Exercise 3. Check the two free exponents in $\mathscr K_2$

Let $g_j\in H^{t+m-m_j-1/2}$. Use only (GF15), (GF6), and the orders of $S$ to prove (GF35).

Solution. Formula (GF61) gives $(Sg)_k\in H^{t+m-k-1/2}$. Hence (GF15) gives $\mathcal JSg\in H_{(-m,t+m)}$. The $R_F$ line of (GF6) sends a boundary-supported input in $H_{(-m,r)}$ to $H_{(\nu+1-m,r-\nu)}$. Set

$$r=t+m, \qquad \nu=s+m. \tag{GF63}$$

The output is $H_{(s+1,t-s)}$, for arbitrary real $s,t$. The input exponent contains no $s$, exactly as (GF35) states.

Exercise 4. Verify the dual gain for the boundary obstruction

In (GF35), put $s=-1$ and replace its second parameter by $-t-1$. Compute the transposed map and show that it raises the exponent of $h_j$ by one.

Solution. With those parameters, (GF35) becomes

$$\mathscr K_2: \bigoplus_jH^{-t+m-m_j-3/2} \longrightarrow H_{(0,-t)}. \tag{GF64}$$

Sobolev duality reverses both exponents. Therefore $\mathscr K_2'$ maps $H_{(0,t)}$ into $\bigoplus_jH^{t+1+m_j+1/2-m}$. Formula (GF44), together with the smoothing transpose $\mathscr K_4'$, proves the asserted one-step gain for $h$.

Exercise 5. Prove the compact fixed-point lemma used twice above

Let $X$ be a Banach space, $C:X\to X$ compact, and $M\subset X$ a linear subspace such that $x=Cx$ for every $x\in M$. Prove that $M$ is finite-dimensional.

Solution. Keep the given subspace $M$, and introduce the entire fixed space $F=\ker(I-C)$. It is a closed linear subspace of the Banach space $X$, and $M\subset F$. If $x_n\in F$ tends to $x$ in $X$, continuity gives $x=Cx$, which proves $x\in F$; it does not by itself prove membership in the original arbitrary $M$.

Let $B_F=\{x\in F:\|x\|\leq1\}$. For every $x\in B_F$, $x=Cx$, so $B_F$ is contained in $C(B_X)$. It is closed in $X$ and thus closed in the compact closure of $C(B_X)$. Therefore $B_F$ is compact.

For completeness, an infinite-dimensional normed space has a sequence of unit vectors whose pairwise distances exceed $1/2$. To construct the next vector, let $L$ be the span of the finitely many previous ones. This finite-dimensional subspace is closed and is proper. Choose $y\notin L$, put $d=\operatorname{dist}(y,L)>0$, and choose $l\in L$ with $\|y-l\|<2d$. Then $x=(y-l)/\|y-l\|$ has norm one and distance $d/\|y-l\|>1/2$ from $L$. Induction gives the asserted sequence, which cannot have a Cauchy subsequence. Applying this to $F$ would contradict compactness of $B_F$. Hence $F$ is finite-dimensional, and its subspace $M$ satisfies

$$\dim M<\infty. \tag{GF65}$$

Apply this once to $M=\ker A_m$ through (GF54), and once to the range annihilator through the compact transpose identity (GF54D).

13. Reading notes and references

The mixed symbol and Sobolev estimates used here are proved in the four named prerequisite lessons at the start. The residue projection and its stable Cauchy range come from Cauchy data from jumps and residues, while the order reductions, boundary source, exterior-jet construction, and functional-analytic index argument extend the corresponding differential proofs in Solving an elliptic system from compatible boundary measurements.

A related primary source with author-submitted TeX is Gerd Grubb, “Fractional Laplacians on domains, a development of Hörmander’s theory of μ-transmission pseudodifferential operators,” arXiv:1310.0951v5. The present lesson gives its own proofs of (GF1)–(GF65) and does not use either reference as a replacement for a displayed argument.

The remaining sections provide the complete collar, normal expansion, half-space, source, trace and freezing proofs used above. Section 23 records the exact correction of the stronger local estimate; Section 24 supplies the full principal-symbol comparison and the globally scoped dual relation.

14. Patching the original collar operator with both error bundles

The unchanged collar object and its two weights

Let $X$ be compact with smooth boundary $Y$, and fix one collar $c:Y\times[0,\epsilon)\to X$ with inward coordinate $r\geq0$. Fix the source and target bundles $E,F$, their collar identifications with the pullbacks of $E_Y,F_Y$, and Hermitian norms. Write $D_r=-i\partial_r$. The operator in the collar is $$P^b=\sum_{j=0}^{m}A_j(r)D_r^j:E_Y\longrightarrow F_Y, \qquad A_j(r)\in\Psi_{\mathrm{phg}}^{m-j}(Y;E_Y,F_Y), \qquad A_m(r)=M(r), \tag{GI1}$$ where $m\geq1$, $M(r)$ is a bundle map, the families are smooth in $r\in(-1,1)$, and they vanish for $r>2/3$ in this original collar presentation. The interior term $P^i\in\Psi_{\mathrm{phg}}^m(X^\circ;E,F)$ has kernel supported in $(X\setminus X_{1/2})^2$. Thus $P=P^b+P^i$ equals $P^b$ when either operator variable is in a sufficiently narrow collar. The interior term and the extension of (GI1) away from that narrow collar are retained as parts of the original object; the construction below only uses the region where (GI1) actually represents $P$.

The original boundary operators are also retained: $$B_j u=\sum_{k=0}^{m-1}B_{jk}\gamma_k u, \quad \gamma_k u=(D_r^ku)|_{r=0}, \quad B_{jk}\in\Psi_{\mathrm{phg}}^{m_j-k}(Y;E_Y,G_j). \tag{GI1a}$$ Their transverse degree is below $m$, while the total order $m_j$ is unrestricted. They play no role in inverting $P_c$ on the full cylinder, but their exact transformation and trace targets are recorded in GI-007; a boundary condition is never inferred from the interior inverse alone.

In every chart of $Y$, put $x=(y,r)$, $\xi=(\eta,\kappa)$, and $$R(\eta,\kappa)=1+|(\eta,\kappa)|,\qquad T(\eta)=1+|\eta|. \tag{GI2}$$ The local mixed class $S^{a,b}$ has all seminorms $$\sup_{x,\eta,\kappa} \frac{\|\partial_x^\beta\partial_\eta^\gamma\partial_\kappa^\ell a(x,\eta,\kappa)\|} {R^{a-\ell}T^{b-|\gamma|}}<\infty. \tag{GI3}$$ The total-frequency order and tangential order are different. The Sobolev Hilbert norm instead has the smooth quadratic multipliers $$\|u\|_{H_{(s,t)}}^2=(2\pi)^{-\dim X}\int (1+|\xi|^2)^s(1+|\eta|^2)^t\|\widehat u(\xi)\|^2\,d\xi. \tag{GI4}$$ The mixed inverse and mixed Sobolev prerequisite lessons prove the exact identity of the vectors and equivalent seminorms/norms obtained from (GI2) and the corresponding square-root quadratic weights. The factors in (GI2) and (GI4) are never identified numerically.

For the gluing argument let $\mathscr M^a$ denote the proper fixed-collar operators defined exactly in GI-003a below: they have local symbols $S^{a,0}$ in common tangential charts and retain tangentially separated normal kernels of order $a$. Let $\mathscr E^a\subset\mathscr M^a$ denote the enhanced operators whose local symbols satisfy $$\partial_{\xi_k}a\in S^{a-1,0}\quad(1\leq k\leq\dim X). \tag{GI5}$$ The enhanced class also has tangentially separated normal kernels one total order lower, as GI-003a proves. This is an added verified property of the actual polynomial and its inverse, not an inference from $S^{a,0}$ membership. We use $\mathscr M^{a,b}$ when the second order must be stated. Local operators are compared only after compact cutoff localization. A smooth kernel of compact support is in every $\mathscr M^{a,b}$, but a tangentially smoothing operator followed by $D_r^j$ need not have a smooth full kernel.

Shrink to a closed interval $I=[-\delta,\delta]\subset(-1,1)$ so that the extension of (GI1) remains elliptic and $M(y,r):E_y\to F_y$ is invertible on $Y\times I$. This is possible: the real normal covector $(\eta,\kappa)=(0,1)$ makes the homogeneous polynomial equal to $M(y,0)$, so ellipticity gives its inverse at $r=0$; continuity and compactness give a uniform interval. Let $h:\mathbb R\to I^\circ$ be smooth, equal to $r$ for $|r|\leq\delta_0<\delta$, constant for $|r|\geq2\delta_0$, with bounded derivatives and image in a smaller closed interval inside $I$. Such a map is obtained by integrating a smooth compactly supported nonnegative derivative of total mass less than $2\delta$, then choosing its central portion to have derivative one. Set $$P_c=\sum_{j=0}^{m}A_j(h(r))D_r^j :C_c^\infty(Y\times\mathbb R,E_Y)\longrightarrow C_c^\infty(Y\times\mathbb R,F_Y). \tag{GI6}$$ Each coefficient is the original family evaluated at the displayed normal value; no matrix is commuted through a tangential operator. Equation (GI6) agrees exactly with $P$ on the smaller collar where $h(r)=r\geq0$, since $P^i$ has no kernel meeting that collar. Its $r$-derivatives are uniformly bounded because $h$ has bounded derivatives and its image is compact in $I$. Its principal symbol is the original homogeneous polynomial at $(y,h(r),\eta,\kappa)$; compactness of $Y\times h(\mathbb R)\times\{|\xi|=1\}$ gives a uniform smallest singular value $c_*>0$. This is an elliptic extension on the product cylinder, not a claim that $P$ extends elliptically across an entire topological double.