We mainly consider the case of an unbounded operator

$$\displaystyle \begin{aligned}P=P_0+V, \end{aligned}$$

where

*P*_{0} is an elliptic differential operator on

*X* and

*V* ∈

*L*^{∞}(

*X*). The underlying Hilbert space is

\(\mathcal {H}=L^2(X)=H^0\) and we will view

*P* as an operator from

*H*^{m} to

*H*^{0}. The dual of

*H*^{m} is

*H*^{−m} and we shall keep in mind the variational point of view with the triple

$$\displaystyle \begin{aligned}H^m\subset H^0\subset H^{-m}.\end{aligned}$$

Consider

Open image in new window . For

*u* ∈

*H*^{m}, we have

Notice that when *P* is injective, then by ellipticity and compactness, we have Open image in new window for some *C* > 0 and we get the conclusion of the proposition also when *w* = 0.

The operator

Open image in new window induces a compact self-adjoint operator

*H*^{0} →

*H*^{0}. The range consists of all

*u* ∈

*H*^{m} such that

*Pu* ∈

*H*^{m}. The spectral theorem for compact self-adjoint operators tells us that there is an orthonormal basis of eigenfunctions

*e*_{1},

*e*_{2}, … in

*H*^{0} such that

where

Open image in new window when

*j* → +

*∞*. Clearly

*e*_{j} ∈

*H*^{m},

*Pe*_{j} ∈

*H*^{m}, so we can apply

Open image in new window to (

17.A.2) and get

which we write as

Here

Open image in new window , so we have found an orthonormal basis

*e*_{1},

*e*_{2}, … ∈

*H*^{0} with

*e*_{j},

*Pe*_{j} ∈

*H*^{m} such that

It is easy to check that

\(t_j^2\) are independent of

*w*, that

*e*_{j} can be chosen independent of

*w*, and we have

$$\displaystyle \begin{aligned}\mu _j(w)^2=\frac{1}{t_j^2+w}. \end{aligned}$$

From Proposition 17.A.1 and its proof we know that Open image in new window is self-adjoint as a bounded operator: *H*^{m} → *H*^{−m}. Consider Open image in new window as a closed unbounded operator Open image in new window with domain Open image in new window . Then Open image in new window , which is dense in Open image in new window and hence dense in *L*^{2}. Open image in new window (or equivalently Open image in new window ) is closed: If Open image in new window , *u*_{j} → *u*, *v*_{j} → *v* in *L*^{2}, then *v*_{j} → *v* in *H*^{−m}, *u*_{j} → *u* in *H*^{m}, hence Open image in new window and since *v* ∈ *L*^{2}, we get \(u\in \mathcal {D}_{\mathrm {sa}}\). Similar arguments show that Open image in new window is self-adjoint. We also know that Open image in new window has a purely discrete spectrum and that \(\{ e_j\}_{j=1}^\infty \) is an orthonormal basis of eigenfunctions.

We have the max-min principle

where

*L* varies in the set of closed subspaces of

*H*^{0} that are also contained in

*H*^{m}. Similarly, from (

17.A.3), we have the mini-max principle

where

*L* varies in the set of closed subspaces of

*H*^{0}. When 0∉

*σ*(

*P*), so that

*P* :

*H*^{m} →

*H*^{0} is bijective, we can extend (

17.A.7) to the case

*w* = 0 and then, as we have seen,

\(\mu _j(0)^2=t_j^{-2}\).

Now assume

$$\displaystyle \begin{aligned} P:H^m\to H^0\text{ is a Fredholm operator of index }0. \end{aligned} $$

(17.A.8)

The discussion above applies also to

*PP*^{∗} when

*P* is viewed as an operator

*H*^{0} →

*H*^{−m} so that

*P*^{∗} :

*H*^{m} →

*H*^{0}. Put

Then as in (

17.A.5) we have an orthonormal basis

*f*_{1},

*f*_{2}, … in

*H*^{0} with

*f*_{j},

*P*^{∗}*f*_{j} ∈

*H*^{m} such that

Write

*t*_{j}(

*P*) =

*t*_{j}, so that

*t*_{j}(

*P*^{∗}) =

*t*_{j} by the proposition. When

*P* has a bounded inverse let

*s*_{1}(

*P*^{−1}) ≥

*s*_{2}(

*P*^{−1}) ≥⋯ be the singular values of the inverse (as a compact operator in

*L*^{2}). We have

$$\displaystyle \begin{aligned} s_j(P^{-1})=\frac{1}{t_j(P)}. \end{aligned} $$

(17.A.11)

Let 1 ≤

*N* <

*∞* and let

*R*_{+} :

*H*^{m} →

**C**^{N},

*R*_{−} :

**C**^{N} →

*H*^{0} be bounded operators. Assume that

$$\displaystyle \begin{aligned} \mathcal{P}=\left(\begin{array}{ccc}P &R_-\\ R_+ &0 \end{array}\right): H^m\times {\mathbf{C}}^N\to H^0\times {\mathbf{C}}^N \end{aligned} $$

(17.A.12)

is bijective with a bounded inverse

$$\displaystyle \begin{aligned} \mathcal{E}=\left(\begin{array}{ccc}E &E_+\\E_- &E_{-+} \end{array}\right). \end{aligned} $$

(17.A.13)

Recall that

*P* has a bounded inverse precisely when

*E*_{−+} does, and when this happens we have the relations,

$$\displaystyle \begin{aligned} P^{-1}=E-E_+E_{-+}^{-1}E_-,\quad E_{-+}^{-1}=-R_+P^{-1}R_-. \end{aligned} $$

(17.A.14)

Cf. Sects.

3.2,

5.3, Chap.

6, Sect.

8.1 and Chap.

13. Recall ([

48] and Proposition

8.2.2) that if

\(A,B:\mathcal {H}_1\to \mathcal {H}_2\) and

\(C:\mathcal {H}_2\to \mathcal {H}_3\) are bounded operators, where

\(\mathcal {H}_j\) are complex Hilbert spaces then we have the general estimates

$$\displaystyle \begin{aligned} s_{n+k-1}(A+B)\le s_n(A)+s_k(B), \end{aligned} $$

(17.A.15)

$$\displaystyle \begin{aligned} s_{n+k-1}(CA)\le s_n(C)s_k(A), \end{aligned} $$

(17.A.16)

In particular, for

*k* = 1, we get

$$\displaystyle \begin{aligned}s_n(CA)\le \Vert C\Vert s_n(A),\ s_n(CA)\le s_n(C)\Vert A\Vert,\ s_n(A+B)\le s_n(A)+\Vert B\Vert . \end{aligned}$$

Applying this to the second part of (

17.A.14), we get

$$\displaystyle \begin{aligned}s_k(E_{-+}^{-1})\le \Vert R_-\Vert \Vert R_+\Vert s_k(P^{-1}),\quad 1\le k\le N \end{aligned}$$

whence

$$\displaystyle \begin{aligned} t_k(P)\le \Vert R_-\Vert \Vert R_+\Vert t_k(E_{-+}),\quad 1\le k\le N. \end{aligned} $$

(17.A.17)

By a perturbation argument, we see that this holds also in the case when

*P*,

*E*_{−+} are non-invertible.

Similarly, from the first part of (

17.A.14) we get

$$\displaystyle \begin{aligned}s_k(P^{-1})\le \Vert E\Vert +\Vert E_+\Vert \Vert E_-\Vert s_k(E_{-+}^{-1}), \end{aligned}$$

leading to

$$\displaystyle \begin{aligned} t_k(P)\ge \frac{t_k(E_{-+})}{\| E\| t_k(E_{-+})+\Vert E_+\Vert\Vert E_-\Vert}. \end{aligned} $$

(17.A.18)

Again this can be extended to the non-necessarily invertible case by means of small perturbations.

Generalizing Sect. 3.2 (as in [56]), we get a natural construction of a Grushin problem associated to a given operator. Let *P* = *P*^{0} : *H*^{m} → *H*^{0} be a Fredholm operator of index 0 as above. Choose *N* so that *t*_{N+1}(*P*^{0}) is strictly positive. In the following we sometimes write *t*_{j} instead of *t*_{j}(*P*^{0}) for short.

Recall that

\(t_j^2\) are the first eigenvalues both for

*P*^{0}^{∗}*P*^{0} and

*P*^{0}*P*^{0}^{∗}. Let

*e*_{1}, …,

*e*_{N} and

*f*_{1}, …,

*f*_{N} be corresponding orthonormal systems of eigenvectors of

*P*^{0}^{∗}*P*^{0} and

*P*^{0}*P*^{0}^{∗}, respectively. They can be chosen so that

$$\displaystyle \begin{aligned} P^0e_j=t _jf_j,\quad {P^0}^*f_j=t_je_j. \end{aligned} $$

(17.A.19)

Define

*R*_{+} :

*L*^{2} →

**C**^{N} and

*R*_{−} :

**C**^{N} →

*L*^{2} by

$$\displaystyle \begin{aligned}R_+u(j)=(u|e_j),\quad R_-u_-=\sum_1^Nu_-(j)f_j.\end{aligned} $$

(17.A.20)

It is easy to see that the Grushin problem

$$\displaystyle \begin{aligned} \left\{ \begin{array}{ll}P^0u+R_-u_-=v,\\ R_+u=v_+, \end{array} \right. \end{aligned} $$

(17.A.21)

has a unique solution (

*u*,

*u*_{−}) ∈

*L*^{2} ×

**C**^{N} for every (

*v*,

*v*_{+}) ∈

*L*^{2} ×

**C**^{N}, given by

$$\displaystyle \begin{aligned} \left\{\begin{array}{ll} u=E^0v+E_+^0v_+,\\ u_-=E_-^0v+E_{-+}^0v_+, \end{array}\right.\end{aligned} $$

(17.A.22)

where

$$\displaystyle \begin{aligned}\begin{cases} E^0_+v_+=\sum_1^Nv_+(j)e_j,& E^0_-v(j)=(v|f_j),\\ E^0_{-+}=-\mathrm{diag}\,(t _j),& \Vert E^0\Vert \le {1\over t_{N+1}}. \end{cases} \end{aligned} $$

(17.A.23)

*E*^{0} can be viewed as the inverse of

*P*^{0} as an operator from the orthogonal space (

*e*_{1},

*e*_{2}, …,

*e*_{N})

^{⊥} to (

*f*_{1},

*f*_{2}, …,

*f*_{N})

^{⊥}.

We notice that in this case the norms of *R*_{+} and *R*_{−} are equal to 1, so (17.A.17) tells us that \(t_k(P^0)\le t_k(E^0_{-+})\) for 1 ≤ *k* ≤ *N*, but of course the expression for \(E^0_{-+}\) in (17.A.23) implies equality.

Let

\(Q\in \mathcal {L}(H^0,H^0) \) and put

*P*^{δ} =

*P*^{0} −

*δQ* (where we sometimes put a minus sign in front of the perturbation for notational convenience). We are particularly interested in the case when

*Q* =

*Q*_{ω}*u* =

*qu* is the operator of multiplication by a function

*q*. Here

*δ* > 0 is a small parameter. Choose

*R*_{±} as in (

17.A.20). Then if

*δ* <

*t*_{N+1} and ∥

*Q*∥≤ 1, the perturbed Grushin problem

$$\displaystyle \begin{aligned} \left\{ \begin{array}{ll}P^\delta u+R_-u_-=v,\\ R_+u=v_+, \end{array} \right. \end{aligned} $$

(17.A.24)

is well posed and has the solution

$$\displaystyle \begin{aligned} \left\{\begin{array}{ll} u=E^\delta v+E_+^\delta v_+,\\ u_-=E_-^\delta +E_{-+}^\delta v_+, \end{array} \right.\end{aligned} $$

(17.A.25)

where

$$\displaystyle \begin{aligned} \mathcal{E}^\delta =\left(\begin{array}{ccc}E^\delta &E_+^\delta \\ E_-^\delta &E_{-+}^\delta \end{array}\right) \end{aligned} $$

(17.A.26)

is obtained from

\(\mathcal {E}^0\) by the formula

$$\displaystyle \begin{aligned} \mathcal{E}^\delta =\mathcal{E}^0\left( 1-\delta \left(\begin{array}{ccc}Q E^0 & Q E_+^0 \\ 0&0 \end{array}\right)\right) ^{-1}. \end{aligned} $$

(17.A.27)

Using the Neumann series, we get

$$\displaystyle \begin{aligned} \begin{aligned} &E_{-+}^\delta =E_{-+}^0+E_-^0\delta Q(1-E^0\delta Q)^{-1}E_+^0 \\&=E_{-+}^0+\delta E_-^0 Q E_+^0+ \delta^2 E_-^0 Q E^0 Q E_+^0+ \delta^3 E_-^0 Q (E^0 Q )^2 E_+^0+\cdots \end{aligned} \end{aligned} $$

(17.A.28)

We also get

$$\displaystyle \begin{aligned} E^\delta =E^0(1-\delta QE^0)^{-1} =E^0+\sum_{k=1}^{\infty }\delta ^kE^0(QE^0)^k, \end{aligned} $$

(17.A.29)

$$\displaystyle \begin{aligned} E_+^\delta =(1-E^0\delta Q)^{-1}E_+^0 =E_+^0+\sum_{k=1}^{\infty }\delta ^k(E^0Q)^kE_+^0, \end{aligned} $$

(17.A.30)

$$\displaystyle \begin{aligned} E_-^\delta =E_-^0(1-\delta QE^0)^{-1} =E_-^0+\sum_{k=1}^{\infty }\delta ^kE_-^0(QE^0)^k. \end{aligned} $$

(17.A.31)

The leading perturbation in

\(E_{-+}^\delta \) is

*δM*, where

\(M =E_-^0 Q E_+^0: {\mathbf {C}}^N\to {\mathbf {C}}^N\) has the matrix

$$\displaystyle \begin{aligned} M(\omega )_{j,k}=(Q e_k|f_j), \end{aligned} $$

(17.A.32)

which in the multiplicative case reduces to

$$\displaystyle \begin{aligned} M(\omega )_{j,k}=\int q (x)e_k(x)\overline{f_j(x)}dx. \end{aligned} $$

(17.A.33)

Put

*τ*_{0} =

*t*_{N+1}(

*P*^{0}) and recall the assumption

$$\displaystyle \begin{aligned} \Vert Q\Vert \le 1. \end{aligned} $$

(17.A.34)

Then, if

*δ* ≤

*τ*_{0}∕2, the new Grushin problem is well posed with an inverse

\(\mathcal {E}^{\delta }\) given in (

17.A.26)–(

17.A.31). We get

$$\displaystyle \begin{aligned} \Vert E^\delta \Vert \le \frac{1}{1-\frac{\delta }{\tau _0}} \Vert E^0\Vert \le \frac{2}{\tau _0},\quad \| E_\pm ^\delta \| \le \frac{1}{1-\frac{\delta }{\tau _0}}\le 2, \end{aligned} $$

(17.A.35)

$$\displaystyle \begin{aligned} \Vert E_{-+}^\delta -(E_{-+}^0+\delta E_-^0QE_+^0)\Vert \le \frac{\delta ^2}{\tau _0}\frac{1}{1-\frac{\delta }{\tau_0}}\le 2\frac{\delta ^2}{\tau _0}. \end{aligned} $$

(17.A.36)

Using this in (

17.A.17), (

17.A.18) together with the fact that

\(t_k(E_{-+}^\delta ) \le 2\tau _0\),

^{2} we get

$$\displaystyle \begin{aligned} \frac{t_k(E^\delta _{-+})}{8}\le t_k(P^\delta )\le t_k(E^\delta _{-+}). \end{aligned} $$

(17.A.37)