Proof of Lemma 2.1 and Proposition 2.2

Let us first recall the following well-known Hardy-type inequalities (see, for instance, [6, Sec. 1.2, Examples 1 and 2]).

Lemma A.1

First, we assume \(d\ge 3\). Then, for any \(\varphi \in C_c^\infty (}^d)\),

$$\begin \int |\nabla \varphi |^2\ge \int h_d(x)|\varphi |^2, \end$$

(37)

where

$$\begin h_d(x):=\left( \frac\right) ^2\frac. \end$$

Then, we assume \(d=2\). Then, for any \(R>0\) and \(\varphi \in C_c^\infty (D(0,R))\),

$$\begin \int |\nabla \varphi |^2\ge \int h_(x)|\varphi |^2, \end$$

(38)

where \(D(0,R)\subset }^2\) is the open disk of radius R centered at 0 and

$$\begin h_(x):=\frac\right) }. \end$$

We can now proceed with the proof of Lemma 2.1. In case A, \(V\ge 0\) and the result is obvious (with \(C=0\)). Let us consider case B. Recall that we denote by \(X=\\) the set of singular points of V. Let us treat separately the cases \(d\ge 3\) and \(d=2\).

In case \(d\ge 3\), we have, by translating the inequality (37),

$$\begin\int |\nabla u|^2 \ge \int h_d(x-x_i)|u|^2\end$$

for all \(1\le i \le N\), hence

$$\begin \int |\nabla u|^2 \ge \int h(x)|u|^2\,, \end$$

(39)

with

$$\begin h(x):=\frac \sum _^N h_d(x-x_i)\,. \end$$

(40)

From the behavior of V at each \(x_i\), given in Definition 1.2, \(\lim _h(x)+V(x)=+\infty \). It follows that \(h+V\) is bounded from below. We can then take \(C=-\inf _}^d}(h+V)\).

In case \(d=2\), we choose for each \(1\le i\le N\) a radius \(r_i>0\), such that the open disks \(D(x_i,2r_i)\) are pairwise disjoint. In addition, we choose smooth functions \(\chi _i\), compactly supported in \(D(x_i,2r_i)\), such that \(\chi _i\le 1\) in \(}^2\) and \(\chi _i=1\) in \( D(x_i,r_i)\). We have pointwise

$$\begin|\nabla (\chi _i u)|^2\le 2\chi _i^2|\nabla u|^2+2|\nabla \chi _i|^2|u|^2\le 2|\nabla u|^2+2|\nabla \chi _i|^2|u|^2,\end$$

and therefore

$$\begin\int |\nabla u|^2\ge \frac\int |\nabla (\chi _i u)|^2-\int |\nabla \chi _i|^2|u|^2.\end$$

Using inequality (38),

$$\begin\int |\nabla u|^2\ge \int \left( \frac\chi _i^2h_(x-x_i)-|\nabla \chi _i|^2\right) |u|^2.\end$$

Summing over all i’s, we find

$$\begin \int |\nabla u|^2\ge \int h(x)|u|^2\,, \end$$

(41)

where

$$\begin h(x):=\frac\sum _^N\left( \frac\chi _i^2h_(x-x_i)-|\nabla \chi _i|^2\right) . \end$$

(42)

As in the case \(d\ge 3\), \(\lim _h(x)+V(x)=+\infty \) for all \(1\le i\le N\), and therefore, \(h+V\) is bounded from below and we can take \(C=-\inf _}^2}(h+V)\).

The analysis leading to Proposition 2.2 rather classical (see, for instance, [1]). For the sake of completeness and the convenience of the reader, let us recall the main steps of the proof. In cases A and B, the potential V belongs to \(L^p_(}^d)\) for some \(p>d/2\). Indeed, it suffices to choose p such that \(p>d/2\) and \(\,p<d\) for all \(1\le i \le N\). According to [1, Th. 3.2, pp. 44-45, Eq. (3.16)], we can then apply Persson’s formula to compute the bottom of the essential spectrum of \(H_V\):

$$\begin \inf \sigma _}()= \sup _\inf _ \varphi \in C_c^\infty (}^d\setminus K)\\ \varphi \ne 0 \end} \frac\varphi \rangle }, \end$$

(43)

where K ranges over all compact subsets of \(}^d\). It follows immediately that

$$\begin \inf \sigma _}()\ge \sup _ \inf _}^d\setminus K}V. \end$$

In case A, \(V(x)\rightarrow +\infty \) as \(|x|\rightarrow +\infty \), and thus

$$\begin \inf \sigma _}()=+\infty , \end$$

that is to say \(\sigma _}()=\emptyset \). Since \(\) is lower semi-bounded, \(\sigma ()\) is a sequence of eigenvalues with finite multiplicities tending to \(+\infty \).

In case B, \(V(x)\rightarrow 0\) as \(|x|\rightarrow +\infty \) and we obtain

$$\begin \sigma _}()\subset [0,+\infty ). \end$$

We can easily check the reverse inclusion by constructing appropriate Weyl sequences. Indeed, let us set, for \(\xi \in }^d\),

$$\begin \varphi _(x):=C_n\chi \left( \frac\right) \exp (i \xi \cdot x), \end$$

where \(\chi \) is a smooth function in \(}^d\) with compact support, \(x_n\in }^d\), \(R_n>0\), and \(C_n>0\) is defined by \(\Vert \varphi _\Vert =1\). By a suitable choice of sequences \(|x_n|\rightarrow +\infty \) and \(R_n\rightarrow +\infty \), we can ensure that \(\langle \varphi _,\varphi _\rangle \rightarrow |\xi |^2\) and \((\varphi _)\) converges to 0 weakly in \(L^2(}^d)\), which implies \(|\xi |^2 \in \sigma _()\). We have shown that \(\sigma _}()=[0,+\infty )\).

For the characterization of the form domain, we refer the reader to [8, Th. 8.2.1] for case A and [8, Th. 8.2.3] for case B. Note that in the latter case, the result is given for \(d\ge 3\). The proof can be extended to \(d=2\) using inequality (41).

Proofs of lemmas 3.1 and 3.2 1.1 Proof of lemma 3.1

Recall that the side-length of the cubes in the partition is \(\rho (1+2\delta )\) with \(\rho = \lambda ^\). Also, we have the following conditions on V:

$$\begin V(x)&\ge C_1 |x|^a \, , \end$$

(44)

$$\begin V(x)&\le C_2 |x|^b \, , \end$$

(45)

$$\begin |\nabla V(x)|&\le C_3 |x|^c \, , \end$$

(46)

with \(c < \fraca\).

Recall also that

$$\begin M_\lambda := \sum _}^d} \lambda ^ \sup _} \left[ \lambda - V + C\lambda ^ \right] _+^} \end$$

and

$$\begin W_\lambda := \int _}^d}(\lambda -V)_+^}. \end$$

We also define

$$\begin m_\lambda := \sum _}^d} \lambda ^ \inf _} \left[ \lambda - V \right] _+^} \end$$

and

$$\beginA_\lambda := M_\lambda - W_\lambda .\end$$

Since \(A_\lambda \le M_\lambda -m_\lambda \), it suffices to show that \(A_\lambda \le C(d,V,\delta ) + \frac + \frac - 1 }}\) in order to prove lemma 3.1.

In this section, C will denote constants which may change from line to line but which never depend on \(\lambda \). However, we may write down the dependency of C on different parameters to emphasize this point. The same remark applies to the “O” and “o” occurring in the formulas.

We have the following estimates for \(A_\lambda \):

$$\begin&A_\lambda \le M_\lambda - m_\lambda \le \lambda ^ \sum _}^d} \left[ \sup _} \left[ \lambda - V +\lambda ^ \right] _+^} - \inf _} \left[ \lambda - V \right] _+^} \right] \nonumber \\&\le \lambda ^ \sum _}^d}\nonumber \\&\left[ \sup _} \left[ \lambda - V \right] _+^}- \inf _} \left[ \lambda - V \right] _+^} + C(d,\delta )\lambda ^\sup _}\left[ \lambda - V + C\lambda ^\right] _+^-1} \right] \, , \end$$

(47)

where we have used the inequality \((a+b)^k \le a^k + k b (a+b)^\) for any \(a,b,k \in }^+\), with \(k\ge 1\). We now bound the oscillation of \([\lambda - V]^}\) over a cube using farthest distance between two points and the bounds on the gradient, recalling that the diameter of the cubes is less than or equal to \(C(d) \lambda ^\):

$$\begin&\lambda ^ \sum _}^d} \left[ \sup _} \left[ \lambda - V \right] _+^}- \inf _} \left[ \lambda - V \right] _+^} \right. \nonumber \\&\quad \left. + C(d,\delta )\lambda ^\sup _}\left[ \lambda - V + C\lambda ^\right] _+^-1} \right] \nonumber \\&\le \lambda ^ \sum _}^d} \left[ C(d) \lambda ^ \sup _}|\nabla V| \sup _}\left[ \lambda - V \right] _+^-1} \right. \nonumber \\&\quad \left. + C(d,\delta )\lambda ^\sup _}\left[ \lambda - V + C\lambda ^\right] _+^-1} \right] \, . \end$$

(48)

To estimate the last term, we use the inequality \((a+b)^k \le 2^k a^k + 2^k b^k\):

$$\begin&\lambda ^ \sum _}^d} \left[ C(d) \lambda ^ \sup _}|\nabla V| \sup _}\left[ \lambda - V \right] _+^-1} \right. \nonumber \\&\quad \left. + C(d,\delta )\lambda ^\sup _}\left[ \lambda - V + C\lambda ^\right] _+^-1} \right] \nonumber \\&\le \lambda ^\sum _}^d} \left[ C(d) \lambda ^ \sup _}|\nabla V| \sup _}\left[ \lambda - V \right] _+^-1} \right. \nonumber \\&\quad \left. + C(d,\delta ) \lambda ^ \sup _} \left[ \lambda - V \right] _+^-1} \right] \nonumber \\&\quad + \sum _ z \in }^d\\ \sup \limits _} V \le \lambda + C\lambda ^ \end} C(d,\delta )\,. \end$$

(49)

We now assume that \(m< \frac\) so that \(\lambda ^ = o(\lambda )\). As we will see later, we can find such an m for suitable a, b and c.

Since \(V(x) \ge C_1 |x|^a\), the second sum in eq. (49) can be estimated by the number of cubes of side-length \(\lambda ^\) in a ball of radius \(C(d,V) \lambda ^}\). We can estimate it directly:

$$\begin \sum _ z \in }^d\\ \sup \limits _} V \le \lambda + C\lambda ^ \end} C(d,\delta ) \le C(d,V,\delta ) \lambda ^+ dm}\,. \end$$

(50)

Now, by the definition of \(J_\), we get that for all exponents \(M >0\), there exists a constant C(d, M) such that

$$\begin |J_| \sup \limits _}} |x|^M \le C(d,M) \int _}} |x|^M \,. \end$$

(51)

We can now bound the first sum in inequality (49) using the bounds on V and \(\nabla V\):

$$\begin&\lambda ^\sum _}^d} \left[ C(d) \lambda ^ \sup _}|\nabla V| \sup _}\left[ \lambda - V \right] _+^-1} + C(d,\delta ) \lambda ^ \sup _} \left[ \lambda - V \right] _+^-1} \right] \nonumber \\&\le C(d,V,\delta ) \int \limits _)} \left[ \lambda ^ ^c + \lambda ^ \right] \left[ \lambda - C_1 ^a \right] ^-1} \,dx \end$$

(52)

$$\begin&\le C(d,V,\delta ) \left[ \lambda ^ + \frac + \frac - 1} + \lambda ^+ \frac-1} \right] \,. \end$$

(53)

In order to balance the two terms on the right-hand side of inequality (53), we need that

$$\begin -m + \frac = 2m. \end$$

Therefore, putting \(m = \frac\) and combining eqs. (50) and (53) gives us the following estimate for \(A_\lambda \):

$$\begin A_\lambda \le C(d,V,\delta ) \left[ \lambda ^ + \frac + \frac - 1 } + \lambda ^ + \frac} \right] \,. \end$$

(54)

Now, in the case \(d=2\), the exponents in the two terms are equal. In the case \(d\ge 3\), since we assumed that \(m < \frac\), the term on the left dominates and we obtain that for any dimension,

$$\begin A_\lambda \le C(d,V,\delta ) \lambda ^ + \frac + \frac - 1 } \, . \end$$

(55)

This completes the proof of lemma 3.1.

1.2 Proof of lemma 3.2

Let \(R_\Omega \) be the Rayleigh quotient for the Dirichlet Laplacian on \(\Omega \) and \(Q_\Omega (f)\) the modified Rayleigh quotient for \(H = -\Delta +V\):

$$\begin Q_\Omega (f) = \frac+<V f,f>_}}. \end$$

Recall that \(J_\) denotes the hypercubes \(\rho (z_1-\frac, z_1+\frac) \times \rho (z_2 - \frac, z_2+\frac) \ldots \rho (z_d - \frac, z_d + \frac )\). Since they do not overlap, we can use them to do a Dirichlet bracketing for the eigenvalue count. More specifically, we use the following facts. First,

$$\beginN(\lambda )=\max \}^d) \text \text Q_}^d}(f)<\lambda \text \text f\in V\setminus \\}\,, \end$$

with V a vector space. Second,

$$\begin N(\lambda )\ge \sum _}^d}}_z(\lambda )\,, \end$$

where

$$\begin }_z(\lambda )=\max \left\) \text \text Q_}(f)<\lambda \text \text f\in V\setminus \\right\} . \end$$

Finally, we obviously have, for all \(z\in }^d\),

$$\begin }_z(\lambda )\ge N_z(\lambda ), \end$$

where \( N_z(\lambda )=\max \left\) \text \text R_}(f)<\lambda -\sup _} V \text \text f\in V\ \right\} \).

We recognize \(N_z(\lambda )\) as the eigenvalue count below \(\lambda -\sup _}V\) for the Dirichlet Laplacian on the hypercube \(J_\). We can use the explicit formula found in [9, lemma 2.4] to get:

$$\begin N_z(\lambda )= |J_| (2\pi )^w_d \inf \limits _}(\lambda -V)_+^} + O\left( |\partial J_| \inf \limits _}(\lambda -V)_+^ - \frac}\right) \, , \end$$

(56)

where \(w_d\) is the volume of the unit ball in \(}^d\).

Recall that we want to show that \(N(\lambda ) \ge (2\pi )^w_d W_\lambda + O( \lambda ^ + \frac + \frac - \frac })\). By summing eq. (56) over all cubes, we obtain the following estimate:

$$\begin N(\lambda )\ge & (2\pi )^w_d\sum _}^d} |J_| \inf \limits _}(\lambda -V)_+^} + O\left( \sum _}^d}|\partial J_| \inf \limits _}(\lambda -V)_+^ - \frac}\right) \nonumber \\ \end$$

(57)

$$\begin\ge & (2 \pi )^w_d m_\lambda + O\left( \sum _}^d}|\partial J_| \inf \limits _}(\lambda -V)_+^ - \frac}\right) \, , \end$$

(58)

where \(m_\lambda \) is the same as in sect. B.1. Since \(m_\lambda = W(\lambda ) + O(A_\lambda )\), by choosing \(m=c/3a\) the first term in (58) is equal to \((2\pi )^w_d W_\lambda + O(\lambda ^ + \frac + \frac - 1 })\) by inequality (55).

We now bound the second term:

$$\begin&\sum _}^d}|\partial J_| \inf \limits _}(\lambda -V)_+^ - \frac}\le }\sum _}^d}\lambda ^ \inf \limits _}(\lambda -V)_+^ -\frac} \nonumber \\&\le \lambda ^m \int \limits _)} (\lambda -C_1 |x|^a)_+^ -\frac}\,dx \end$$

(59)

$$\begin&\le C(d,V) \lambda ^ + \frac + \frac - \frac }\,. \end$$

(60)

Now, since \(2c <3a\), then \(\frac - \frac > \frac -1\), and we have the final estimate for \(N(\lambda )\):

$$\begin N(\lambda ) \ge (2\pi )^w_d W_\lambda + O \left( \lambda ^ + \frac + \frac - \frac }\right) \,. \end$$

(61)

This completes the proof of lemma 3.2.

Proofs of lemmas 4.3 and 4.4 1.1 Proof of lemma 4.3

Recall that the diameter of the hypercubes \(B_\) in the partition is less than \(C(d) \sup \limits _}|x|^q\) and their volume is less than \(C(d) \sup \limits _}|x|^\).

Again, let

$$\begin}_\lambda := \sum _ |B_i| \sup _} \left[ \lambda - V + C(d,\delta ) |x|^ \right] _+^}\end$$

and

$$\begin}_\lambda := \int _}^d \backslash B_}(\lambda -V)_+^}.\end$$

We also define

$$\begin}_\lambda := \sum _ |B_i| \inf _} \left[ \lambda - V \right] _+^}\end$$

and

$$\begin}_\lambda := }_\lambda - }_\lambda .\end$$

Finally, recall that

$$\begin V(x)&\ge -C_1 |x|^ \, , \end$$

(62)

$$\begin V(x)&\le -C_2 |x|^ \, , \end$$

(63)

$$\begin |\nabla V(x)|&\le C_3 |x|^ \, , \end$$

(64)

with \(c > \frac a\).

Since \(}_\lambda \le }_\lambda -}_\lambda \), it suffices to show that \(}_\lambda \le C(d,V,\delta ) |\lambda |^- \frac + \frac -1 }\) in order to prove lemma 4.3.

We now fix \(r >0\) such that all the singularities of V are contained in the ball of radius r centered at the origin (if there are no singularities, fix \(r=1\)).

In this section, C will denote constants which may change from line to line but which never depend on \(\lambda \). However, we may write down the dependency of C on different parameters to emphasize this point. The same remark applies to the “O” and “o” occurring in the formulas.

We will follow a strategy similar to the one we used for positive potentials in order to bound \(}_\lambda \). Again, recalling that the diameter of \(B_\) is less than \(C(d,\delta )\sup \limits _}|x|^\), the volume of \(B_\) is less than \(C(d,\delta ) \sup \limits _}|x|^ \) and using the inequalities \((a+b)^k \le a^k + kb(a+b)^\) and \((a+b)^k \le 2^ka^k+2^kb^k\), we obtain the following:

$$\begin }_\lambda&\le \sum _ |B_i| \left[ \sup _} \left[ \lambda - V + C(d,\delta ) |x|^ \right] _+^} - \inf _} \left[ \lambda - V \right] _+^} \right] \\&\le \sum _ |B_i| \left[ \sup _} \left[ \lambda - V \right] _+^}- \inf _} \left[ \lambda - V \right] _+^} + C(d,\delta ) \sup \limits _}|x|^\right. \nonumber \\&\quad \left. \sup _}\left[ \lambda - V + C(d,\delta ) |x|^\right] _+^-1} \right] \end$$

(65)

$$\begin&\le \sum _ |B_i| \left[ C(d,\delta ) \sup _}|x|^\sup _}|\nabla V| \sup _}\left[ \lambda - V \right] _+^-1} \right. \nonumber \\&\quad \left. + C(d,\delta ) \sup \limits _}|x|^\sup _}\left[ \lambda - V \right] _+^-1} \right] \nonumber \end$$

(66)

$$\begin&\quad + \sum _ \\ } V(x) -C(d,\delta )|x|^ \le \lambda } \end} C(d,\delta ) \end$$

(67)

Now, the last term in (67) is bounded (up to a constant factor) by the number of hypercubes \(B_i\) which intersect the region \( \left\ \le \lambda \right\} \). In order to estimate this, let Q(x) be the inverse of the volume of the hypercube that contains x. From the construction of our partition, \(Q(x) \le C |x|^\). We assume that \(2q>a\), which will be justified later by a suitable choice of q. Hence, we can bound the last term by the following (recalling that \(x_i\) is the center of \(B_i\)):

$$\begin&\sum _ \\ } V(x) -C(d,\delta )|x|^ \le \lambda } \end} C(d,\delta ) \le C(d,\delta )\sum _ \\ } -|x|^ \le C(d,\delta ,V) \lambda } \end } Q(x_i)|B_i| \end$$

(68)

$$\begin&\quad \le C(d,\delta , V)\int \limits _\right) \backslash B_0} |x|^ \end$$

(69)

$$\begin&\quad \le C(d,\delta , V) |\lambda |^ - \frac}\,. \end$$

(70)

We can combine estimates (67) and (70) as well as bounds on |V| and \(|\nabla V|\):

$$\begin }_\lambda&\le C(d,\delta , V) |\lambda |^ - \frac} + \sum _ |B_i| C(d,\delta ) \sup _}\left[ \lambda - V\right] _+^-1} \nonumber \\&\quad \left[ \sup _} |x|^ + \sup _} |x|^ \sup _}|\nabla V| \right] \end$$

(71)

$$\begin&\le C(d,\delta , V) |\lambda |^ - \frac} + \sum _ |B_i| C(d,\delta , V) \sup _} |x|^+a} \nonumber \\&\quad \left[ \sup _} |x|^+ \sup _} |x|^ \sup _} |x|^ \right] \,. \end$$

(72)

Now, by the construction of the cubes \(B_i\), we have that for every \(i\ge 1,\delta \) and exponent \(P<0\),

$$\begin\sup _} |x|^P \le C(P) \inf _} |x|^P.\end$$

Hence, we can estimate the sum on the right-hand side of 72 by an integral:

$$\begin }_\lambda&\le C(d,\delta , V) |\lambda |^ - \frac} + \int \limits _) \backslash B_0(r)} C(d,\delta , V) |x|^ +a} \left[ |x|^ + |x|^q |x|^ \right] \end$$

(73)

$$\begin&\le C(d,\delta , V) |\lambda |^ - \frac} + C(d,\delta , V) |\lambda |^-1 + \frac - \frac} + C(d,\delta , V) |\lambda |^-1 + \frac-\frac- \frac}\,. \end$$

(74)

Now, taking \(q=\frac\) balances the last two terms and we end up with the following:

$$\begin }_\lambda \le C |\lambda |^ - \frac} + C |\lambda |^- \frac + \frac -1 }\,. \end$$

(75)

Let us recall that \(c<3\) (see remark 1.3) and therefore \(q<1\). In turn, this makes sure that the conditions of lemma 4.1 are fulfilled. Now, if \(d=2\), both terms are equal. If \(d \ge 3\), then the right term dominates and in both cases

$$\begin }_\lambda \le C |\lambda |^- \frac + \frac -1 }\, . \end$$

(76)

Finally, we note that by imposing \(c > \frac a\), we fulfill the condition \(2q>a\). This completes the proof of lemma 4.3.

1.2 Proof of lemma 4.4

We want to show that \( }(\lambda ) \ge (2\pi )^w_d }_\lambda + O( \lambda ^ - \frac + \frac- \frac })\).

As in the case with positive potentials, we will use a Dirichlet bracketing argument for the number of eigenvalues. Let \(R_\Omega \) be the Rayleigh quotient for the Dirichlet Laplacian on \(\Omega \). Since the hypercubes \(B_i\) do not overlap, we have the following lower bound:

$$\begin }(\lambda ) \ge \sum _ }_(\lambda )\,, \end$$

(77)

where, similarly to sect. 3.3,

$$\begin }_i(\lambda ):=\sup \left\ \text R_(f) \le \inf \limits _(\lambda - V) \text \text f \in V, \, f \ne 0\right\} \,, \end$$

(78)

V being a vector space.

Using inequality (77) and the formula for the Dirichlet spectrum of a hypercube with remainder, we obtain the following:

$$\begin }(\lambda ) \ge \sum \limits _ |B_i|(2\pi )^w_d \inf _}(\lambda -V)_+^} + O\left( \sum \limits _ |\partial B_i| \inf \limits _ (\lambda -V)_+^-\frac} \right) \, . \end$$

(79)

We now compare the first term with \(}_\lambda \):

$$\begin \sum \limits _ |B_i| \inf _}(\lambda -V)_+^}= & \tilde + O \left( \sum _ |B_i|\sup _}(\lambda -V)_+^} - \sum _ |B_i|\inf _}(\lambda -V)_+^} \right) \nonumber \\ \end$$

(80)

$$\begin= & \tilde + O\left( \sum _ |B_i| \sup _}|\nabla V| \sup _} (\lambda - V)_+^-1} \right) \,. \end$$

(81)

Under the assumption \(c>\fraca\), choosing \(q=\frac\) in the partition enables us to use the exact same steps as in sect. C.1 (the details are left to the reader):

$$\begin \sum \limits _ |B_i| \inf _}(\lambda -V)_+^} = }_\lambda + O(|\lambda |^- \frac + \frac -1 })\,. \end$$

(82)

We now bound the second term in (79) by noting that due to the construction of the partition, for any exponents \(q>0\) (for the size of the cubes) and \(P<0\), \(\sup _} |x|^P \le C(q,P) \inf _} |x|^P\):

$$\begin \sum \limits _ |\partial B_i| \inf \limits _ (\lambda -V)_+^-\frac}&\le C(d,V)\sum \limits _ |B_i| \sup \limits _ |x|^} \inf \limits _ (\lambda -V)_+^-\frac} \end$$

(83)

$$\begin&\le C(d,V)\sum \limits _|B_i| \inf \limits _ |x|^} \inf \limits _ (C|x|^)_+^-\frac} \end$$

(84)

$$\begin&\le C(d,V)\int \limits _) \backslash B_0} |x|^}(C|x|^)^-\frac} \end$$

(85)

$$\begin&\le C(d,V) |\lambda |^- \frac + \frac -\frac }\,. \end$$

(86)

Since \(\frac > \frac\), then \(- \frac + \frac -\frac }< - \frac + \frac -1 }\) and

$$\begin }(\lambda ) \ge (2\pi )^w_d \tilde + O(^- \frac + \frac -\frac }). \end$$

This completes the proof of lemma 4.4.

Proof of lemma 4.1

We recall lemma 4.1:

For any \(q<1\) and \(r>0\), there exist a sequence of pairs of positive numbers \((r_i,d_i)\), \(i\ge 1\) and positive constants \(Q,Q'\) such that

(1)

\(r_1=r>0\),

(2)

\(r_=r_i+d_i\),

(3)

\(\frac\in }\),

(4)

\(Qr_i^q\le d_i\le Q'r_i^q\),

(5)

\(r_i\rightarrow \infty \).

Proof

Let us assume that we are given a sequence of positive integers \((n_i)\), \(i\ge 1\) and that we define \(r_i\) and \(d_i\) recursively by

$$\begin r_1&=r\,,\\ d_i&=\frac\,,\\ r_&=r_i+d_i\,. \end$$

Then, properties (1)–(3) are automatically satisfied. It remains to choose \((n_i)\) in such a way that (4)–(5) hold. Dividing by \(r_i^q\), (4) can be rewritten

$$\begin Q\le \frac^}\le Q'. \end$$

(87)

We can define \(n_i\) by

where \(\lceil x \rceil \) is the smallest integer at least as large as x. Note that by combining this definition with the previous one, we construct the sequences \((r_i,d_i)\) and \((n_i)\) recursively, starting from \(r_1=r\). The sequence \((r_i)\) is increasing, so it either is bounded or goes to \(\infty \). If it was bounded, then we would have \(n_i\le N\) for some integer N, and therefore

$$\begin r_\ge \left( 1+\frac\right) r_i. \end$$

This implies \(r_i\rightarrow \infty \), in contradiction with the assumption. Therefore \(r_i\rightarrow \infty \), so (5) holds. Finally, we have

$$\begin \ r_i^\le n_i<r_i^+1. \end$$

We deduce

$$\begin \frac}\le \frac}\le \frac^}\le 1. \end$$

Since \(d_i=r_i/n_i\), this shows that (4) holds with \(Q=1/(1+r^)\) and \(Q'=1\) . \(\square \)

Remark D.1

For any sequence \((r_i,d_i)\) satisfying the properties of Lemma 4.1, there exist positive constants \(Q'',Q'''\) such that

$$\begin Q''\le \frac}\le Q'''. \end$$

(88)

Indeed, dividing property (4) by \(r_i \), we find

$$\begin Qr_i^\le \frac}\le Q'r_i^, \end$$

and property (5) then implies \(d_i/r_i\rightarrow 0\). We then get, from property (2),

$$\begin \frac}=1+\frac\rightarrow 1. \end$$

Since we have

$$\begin \frac\left( \frac}\right) ^q\le \frac}\le \frac\left( \frac}\right) ^q \end$$

from property (4), the desired result follows.