In the present paper, we study spectral estimates for the logarithmic Laplacian \(L_{{\tiny \Delta \,}}\!= \log (-\Delta )\), which is a (weakly) singular integral operator with Fourier symbol \(2\log |\eta |\) and arises as formal derivative \(\partial _s \Big |_{s=0} (-\Delta )^s\) of fractional Laplacians at \(s= 0\). The study of \(L_{{\tiny \Delta \,}}\!\) has been initiated recently in [5], where its relevance for the study of asymptotic spectral properties of the family of fractional Laplacians in the limit \(s \rightarrow 0^+\) has been discussed. In particular, it has been proved in [5, Theorem 1.5] that the principal Dirichlet eigenvalue of \(L_{{\tiny \Delta \,}}\!\) in a bounded Lipschitz domain is given as a right derivative of principal Dirichlet eigenvalues of fractional Laplacians \((-\Delta )^s\) at \(s=0\), whereas the corresponding principal eigenfunction arises as \(L^2\)-limit of corresponding Dirichlet eigenfunctions of \((-\Delta )^s\). Extensions of these results to higher eigenvalues and eigenfunctions were obtained afterwards in [8] together with uniform convergence and continuity results for these eigenfunctions. Further motivation for the study of \(L_{{\tiny \Delta \,}}\!\) is given in [11], where it has been shown that this operator allows to characterize the s-dependence of solution to fractional Poisson problems for the full range of exponents \(s \in (0,1)\). The logarithmic Laplacian also arises in the geometric context of the 0-fractional perimeter, which has been studied recently in [6].

For matters of convenience, we state our results for the operator \(\mathcal H= \frac{1}{2}L_{{\tiny \Delta \,}}\!\) which corresponds to the quadratic form

$$\begin{aligned} \varphi \mapsto (\varphi ,\varphi )_{log} := \frac{1}{(2\pi )^{d}} \, \int _{\mathbb R^d} \log (|\xi |)\, |\widehat{\varphi }(\xi )|^2\, d\xi . \end{aligned}$$

Here and in the following, we let \(\widehat{\varphi }\) denote the Fourier transform

$$\begin{aligned} \xi \mapsto \widehat{\varphi }(\xi )= \int _{{\mathbb R}^d} e^{-ix \xi } \varphi (x)\,dx \end{aligned}$$

of a function \(\varphi \in L^2({\mathbb R}^d)\). Let \(\Omega \subset \mathbb R^d\) be an open set of finite measure, and let \({\mathbb H}(\Omega )\) denote the closure of \(C^\infty _c(\Omega )\) with respect to the norm

$$\begin{aligned} \varphi \mapsto \Vert \varphi \Vert _{*}^2:= \int _{\mathbb R^d} \log (e + |\xi |)\, |\widehat{\varphi }(\xi )|^2\, d\xi . \end{aligned}$$

Then \((\cdot ,\cdot )_{log}\) defines a closed, symmetric and semibounded quadratic form with domain \({\mathbb H}(\Omega ) \subset L^2(\Omega )\), see Sect. 2 below. Here and in the following, we identify \(L^2(\Omega )\) with the space of functions \(u \in L^2({\mathbb R}^d)\) with \(u \equiv 0\) on \({\mathbb R}^d \setminus \Omega \). Let

$$\begin{aligned} \mathcal H : {\mathcal D}(\mathcal H) \subset L^2(\Omega ) \rightarrow L^2(\Omega ) \end{aligned}$$

be the unique self-adjoint operator associated with the quadratic form. The eigenvalue problem for \(\mathcal H\) then writes as

$$\begin{aligned} \left\{ \begin{aligned} \mathcal H \varphi&= \lambda \varphi ,&\qquad \text {in}\quad \Omega ,\\ \varphi&= 0,&\qquad \text {on}\quad {\mathbb R}^d \setminus \Omega . \end{aligned} \right. \end{aligned}$$

We understand (1.3) in weak sense, i.e.

$$\begin{aligned} \varphi \in {\mathbb H}(\Omega ) \quad \text {and}\quad (\varphi ,\psi )_{log}= \lambda \int _{\Omega }\varphi (x)\psi (x)\,dx \quad \text {for all}\quad \psi \in {\mathbb H}(\Omega ). \end{aligned}$$

As noted in [5, Theorem 1.4], there exists a sequence of eigenvalues

$$\begin{aligned} \lambda _1(\Omega )< \lambda _2(\Omega ) \le \dots , \qquad \lim _{k \rightarrow \infty } \lambda _k(\Omega ) = \infty \end{aligned}$$

and a corresponding complete orthonormal system of eigenfunctions. We note that the discreteness of the spectrum is a consequence of the fact that the embedding \({\mathbb H}(\Omega ) \hookrightarrow L^2(\Omega )\) is compact. In the case of bounded open sets, the compactness of this embedding follows easily by Pego’s criterion [18]. In the case of unbounded open sets of finite measure, the compactness can be deduced from [10, Theorem 1.2] and estimates for \(\Vert \cdot \Vert _*\), see Corollary 2.3 below.

In Sect. 2, using the results from [5] and [7], we discuss properties of functions from \({\mathbb {H}}(\Omega )\). In particular, we show that \(e^{ix\xi }\big |_{x\in \Omega } \in {\mathbb {H}}(\Omega )\), \(\xi \in \mathbb R^d\), provided \(\Omega \) is an open bounded sets with Lipschitz boundary.

In Sect. 3 we obtain a sharp upper bound for the Riesz means and for the number of eigenvalues \(N(\lambda )\) of the operator \(\mathcal H\) below \(\lambda \). Here we use technique developed in papers [3, 4, 13] and [14]. In [12] it was noticed that such technique could be applied for a class of pseudo-differential operators with Dirichlet boundary conditions in domains of finite measure without any requirements on the smoothness of the boundary.

We discuss lower bounds for \(\lambda _1(\Omega )\) in Sect. 4. In Theorem 4.1 we present an estimate that is valid for arbitrary open sets of finite measure. For sets with Lipschitz boundaries, H.Chen and T.Weth [5] have proved a Faber-Krahn inequality for the operator \(\mathcal H\) that reduces the problem to the estimate of \(\lambda _1(B)\), where B is a ball satisfying \(|B| = |\Omega |\), see Corollary 4.3. In Theorem 4.4 we find an estimate for \(\lambda _1(B_d)\), where \(B_d\) is the unit ball, that is better in lower dimensions than the one obtained in Theorem 4.1. We also compare our results with bounds resulting from previously known spectral inequalities obtained in [1] and [2].

In Sect. 5 we obtain asymptotic lower bounds using the coherent states transformation approach given in [9]. It allows us to derive, in Sect. 6, asymptotics for the Riesz means of eigenvalues in Theorem 6.1 and for \(N(\lambda )\) in Corollary 6.2. Here \(\Omega \subset {\mathbb R}^N\) is an arbitrary open set of finite measure without any additional restrictions on the boundary.

Finally in Sect. 7 we obtain uniform bounds on the Riesz means of the eigenvalues using the fact that for bounded open sets with Lipschitz boundaries we have \(e^{ix\xi }\big |_{x\in \Omega } \in \mathcal D(\mathcal H)\).

We close this introduction with some remarks comparing the logarithmic Laplacian \(\mathcal H = \frac{1}{2}L_{{\tiny \Delta \,}}\!\) with the spectral-theoretic logarithm \(\log (-\Delta _D)\) of the Dirichlet Laplacian \(-\Delta _D\) on a bounded Lipschitz domain \(\Omega \) with form domain \(H^1_0(\Omega )\). Clearly, the eigenvalues of \(\log (-\Delta _D)\) are merely given as \(\log \lambda _k^D(\Omega )\), where \(\lambda _k^D(\Omega )\), \(k \in {\mathbb N}\) denote the Dirichlet eigenvalues of \(-\Delta \) on \(\Omega \). Comparing these eigenvalues with the eigenvalues \(\lambda _k(\Omega )\) of \(\mathcal H\), we note that

$$\begin{aligned} \lambda _k(\Omega ) \le \frac{1}{2}\log \lambda _k^D(\Omega ) \qquad \text {for}\quad k \in {\mathbb N}. \end{aligned}$$

Indeed, this follows by combining [17, Theorem 5] with [8, Theorem 1.1(i)]. On the other hand, as we shall see in Corollary 6.2 below, the Weyl asymptotics of \(\lambda _k(\Omega )\) as \(k \rightarrow \infty \) are, up to first order, the same as those for \(\frac{1}{2}\log \lambda _k^D(\Omega )\), the latter being a consequence of Weyl’s classical result for the Dirichlet Laplacian. We also stress the obvious fact that the eigenfunctions of \(\log (-\Delta _D)\) are the same as those of the Dirichlet Laplacian, while those of \(\mathcal H = \frac{1}{2}L_{{\tiny \Delta \,}}\!\) differ significantly due to a much weaker boundary regularity.

Preliminaries and basic properties of eigenvalues

As before, let \((\cdot ,\cdot )_{log}\) denote the quadratic form defined in (1.1), and let, for an open set \(\Omega \subset {\mathbb R}^d\), \({\mathbb H}(\Omega )\) denote the closure of \(C^\infty _c(\Omega )\) with respect to the norm \(\Vert \cdot \Vert _*\) defined in (1.2).

Lemma 2.1

Let \(\Omega \subset {\mathbb R}^d\) be an open set of finite measure. Then \((\cdot ,\cdot )_{log}\) defines a closed, symmetric and semibounded quadratic form with domain \({\mathbb H}(\Omega ) \subset L^2(\Omega )\).


Obviously, the form \((\cdot ,\cdot )_{log}\) is symmetric. For functions \(\varphi \in C^\infty _c(\Omega )\), we have

$$\begin{aligned} (2\pi )^{d} \Vert \varphi \Vert _2^2 =\Vert \widehat{\varphi }\Vert _2^2 \le \Vert \varphi \Vert _*^2. \end{aligned}$$

Moreover, with \(c_1:= \log (e+2)+ \sup \limits _{t \ge 2}\frac{\log (e+t)}{\log t}\) we have

$$\begin{aligned} \frac{\Vert \varphi \Vert _{*}^2}{c_1}&\le \Vert \widehat{\varphi }\Vert _{2}^2+ \int _{|\xi | \ge 2}\ln |\xi | |\widehat{\varphi }(\xi )|^2 \, d\xi \nonumber \\&\le (2\pi )^d \bigl ( \Vert \varphi \Vert _{2}^2 + (\varphi ,\varphi )_{log}\bigr ) - \int _{|\xi | \le 2}\ln |\xi | |\widehat{\varphi }(\xi )|^2 \, d\xi \nonumber \\&\le (2\pi )^d \bigl ( \Vert \varphi \Vert _{2}^2 + (\varphi ,\varphi )_{log}\bigr ) + \bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))} \Vert \widehat{\varphi }\Vert _\infty ^2 \end{aligned}$$


$$\begin{aligned} \Vert \widehat{\varphi }\Vert _\infty ^2 \le \Vert \varphi \Vert _1^2 \le |\Omega |\, \Vert \varphi \Vert _2^2. \end{aligned}$$


$$\begin{aligned} (\varphi ,\varphi )_{log}&\ge \frac{\Vert \varphi \Vert _*^2}{(2\pi )^d c_1}- \left( 1+ \frac{|\Omega |\, \bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))}}{(2\pi )^{d}}\right) \Vert \varphi \Vert _2^2 \\&\ge \left( \frac{1}{c_1}\,-\,1\,-\,\frac{|\Omega |\, \bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))}}{(2\pi )^{d}}\right) \Vert \varphi \Vert _2^2. \nonumber \end{aligned}$$

In particular, \((\varphi ,\varphi )_{log}\) is semibounded. Moreover, it follows from (2.4) and the completeness of \(({\mathbb H}(\Omega ),\Vert \cdot \Vert _*)\) that the form \((\varphi ,\varphi )_{log}\) is closed on \({\mathbb H}(\Omega )\). \(\square \)

Lemma 2.2

Let \(\Omega \subset {\mathbb R}^d\) be an open set of finite measure. Then

$$\begin{aligned} \varphi \mapsto \Vert \varphi \Vert _{**}^2:= \int \!\!\! \int _{|x-y|\le 1} \frac{(\varphi (x)-\varphi (y))^2}{|x-y|^d}\,dxdy \end{aligned}$$

defines an equivalent norm to the norm \(\Vert \cdot \Vert _*\) defined in (1.2) on \(C^\infty _c(\Omega )\).


Let \(\varphi \in C^\infty _c(\Omega )\). By [7, Lemma 2.7], we have

$$\begin{aligned} \Vert \varphi \Vert _2 \le c_2 \Vert \varphi \Vert _{**} \qquad \text {with a constant}\quad c_2>0\quad \text {independent of }\varphi . \end{aligned}$$

In particular, \(\Vert \cdot \Vert _{**}\) defines a norm on \(C^\infty _c(\Omega )\). Next we note that, by [5, Theorem 1.1(ii) and Eq. (3.1)],

$$\begin{aligned} (\varphi ,\varphi )_{log} = \frac{1}{2}\int _{{\mathbb R}^d}[L_{{\tiny \Delta \,}}\!\varphi (x)]\varphi (x)\,dx = \kappa _d \Vert \varphi \Vert _{**}^2 - \int _{\mathbb R^d} [j * \varphi ] \varphi \,dx + \zeta _d \Vert \varphi \Vert _2^2 \end{aligned}$$


$$\begin{aligned} \kappa _d:= \frac{\pi ^{- \frac{d}{2}} \Gamma (d/2)}{4}, \qquad \zeta _d:= \log 2 + \frac{1}{2}\left( \psi \left( d/2\right) -\gamma \right) \end{aligned}$$


$$\begin{aligned} j: {\mathbb R}^d \setminus \{0\} \rightarrow {\mathbb R}, \qquad j(z)= 2 \kappa _d 1_{{\mathbb R}^d \setminus B_d}(z)|z|^{-d}. \end{aligned}$$

Here \(\psi := \frac{\Gamma '}{\Gamma }\) is the Digamma function and \(\gamma = -\Gamma '(1)\) is the Euler-Mascheroni constant. Consequently, we have

$$\begin{aligned} \Bigl |(\varphi ,\varphi )_{log}- \kappa _d \Vert \varphi \Vert _{**}^2\Bigr |&\le \Vert j\Vert _\infty \Vert \varphi \Vert _1^2 + \zeta _d \Vert \varphi \Vert _2^2 \nonumber \\&\le \Bigl (\Vert j\Vert _\infty |\Omega | +\zeta _d\Bigr )\Vert \varphi \Vert _2^2. \end{aligned}$$

As a consequence of (2.1) and (2.8), we find that

$$\begin{aligned} \Vert \varphi \Vert _{**}^2&\le \frac{1}{\kappa _d}\Bigl [ (\varphi ,\varphi )_{log} + \bigl (\Vert j\Vert _\infty |\Omega |+\zeta _d\bigr ) \Vert \varphi \Vert _2^2 \Bigr ]\\&\le \frac{1}{(2\pi )^d \kappa _d}\Bigl (1 + \Vert j\Vert _\infty |\Omega |+\zeta _d\Bigr )\Vert \varphi \Vert _*^2. \end{aligned}$$

Moreover, by (2.2), (2.3), (2.6) and (2.8) we have

$$\begin{aligned}&\frac{\Vert \varphi \Vert _{*}^2}{c_1} \le (2\pi )^d \bigl (\Vert \varphi \Vert _{2}^2 + (\varphi ,\varphi )_{log}\bigr ) + \bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))} |\Omega | \Vert \varphi \Vert _2^2\\&\quad \le (2\pi )^d \Bigl ( \kappa _d \Vert \varphi \Vert _{**}^2 + \bigl (1+ \Vert j\Vert _\infty |\Omega | +\zeta _d\bigr )\Vert \varphi \Vert _2^2 \Bigr ) + \bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))} |\Omega | \Vert \varphi \Vert _2^2\\&\quad \le c_3 \Vert \varphi \Vert _{**}^2 \end{aligned}$$

with \(c_3 = (2\pi )^d \kappa _d + c_2\bigl [(2\pi )^d \bigl (1+ \Vert j\Vert _\infty |\Omega |+\zeta _d\bigr ) +\bigl \Vert \ln |\cdot | \bigr \Vert _{L^1(B_2(0))}|\Omega | \bigr ]\). Hence the norms \(\Vert \cdot \Vert _{*}\) and \(\Vert \cdot \Vert _{**}\) are equivalent on \(C^\infty _c(\Omega )\). \(\square \)

Corollary 2.3

Let \(\Omega \subset {\mathbb R}^d\) be an open set of finite measure. Then the embedding \({\mathbb H}(\Omega ) \hookrightarrow L^2(\Omega )\) is compact.


Let \(\tilde{\mathbb H}(\Omega )\) be defined as the space of functions \(\varphi \in L^2({\mathbb R}^d)\) with \(\varphi \equiv 0\) on \({\mathbb R}^d \setminus \Omega \) and

$$\begin{aligned} \int \!\!\! \int _{|x-y|\le 1} \frac{(\varphi (x)-\varphi (y))^2}{|x-y|^d}\,dxdy <\infty . \end{aligned}$$

By [10, Theorem 1.2], the Hilbert space \((\tilde{\mathbb H}(\Omega ),\Vert \cdot \Vert _{**})\) is compactly embedded in \(L^2(\Omega )\). Since, by Lemma 2.2, the norms \(\Vert \cdot \Vert _*\) and \(\Vert \cdot \Vert _{**}\) are equivalent on \(C^\infty _c(\Omega )\), the space \({\mathbb H}(\Omega )\) is embedded in \(\tilde{\mathbb H}(\Omega )\). Hence the claim follows. \(\square \)

Corollary 2.4

Let \(\Omega \subset {\mathbb R}^d\) be a bounded open set with Lipschitz boundary.

  1. (i)

    The space \({\mathbb H}(\Omega )\) is equivalently given as the set of functions \(\varphi \in L^2({\mathbb R}^d)\) with \(\varphi \equiv 0\) on \({\mathbb R}^d \setminus \Omega \) and

    $$\begin{aligned} \int \!\!\! \int _{|x-y|\le 1} \frac{(\varphi (x)-\varphi (y))^2}{|x-y|^d}\,dxdy <\infty . \end{aligned}$$
  2. (ii)

    \({\mathbb H}(\Omega )\) contains the characteristic function \(1_\Omega \) of \(\Omega \) and also the restrictions of exponentials \(x \mapsto 1_\Omega (x) \, e^{ix \xi }\), \(\xi \in {\mathbb R}^d\).


(i) Let, as in the proof of Corollary 2.3, \(\tilde{\mathbb H}(\Omega )\) be the space of functions \(\varphi \in L^2({\mathbb R}^d)\) with \(\varphi \equiv 0\) on \({\mathbb R}^d \setminus \Omega \) and with (2.9), endowed with the norm \(\Vert \cdot \Vert _{**}\). Since \(\Omega \subset {\mathbb R}^d\) be a bounded open set with Lipschitz boundary, it follows from [5, Theorem 3.1] that \(C_0^\infty (\Omega ) \subset \tilde{\mathbb H}(\Omega )\) is dense. Hence the claim follows from Lemma 2.2.

(ii) follows from (i) and a straightforward computation. \(\square \)

Next we note an observation regarding the scaling properties of the eigenvalues \(\lambda _k(\Omega )\).

Lemma 2.5

Let \(\Omega \subset {\mathbb R}^d\) be a bounded open set with Lipschitz boundary, and let

$$\begin{aligned} R\Omega := \{R x\,:\, x \in \Omega \}. \end{aligned}$$

Then we have

$$\begin{aligned} \lambda _k(R \Omega ) = \lambda _k(\Omega ) - \log R \qquad \text {for all}\quad k \in {\mathbb N}. \end{aligned}$$


Since \(C_0^\infty (\Omega ) \subset {\mathbb H}(\Omega )\) is dense, it suffices to note that

$$\begin{aligned} (\varphi _R,\varphi _R)_{log} = (\varphi ,\varphi )_{log} - \log R \Vert \varphi \Vert _{L^2({\mathbb R}^d)}^2 \qquad \text {for}\quad \varphi \in C^\infty _c({\mathbb R}^d) \end{aligned}$$

with \(\varphi _R \in C^\infty _c({\mathbb R}^d)\) defined by \(\varphi _R(x)= R^{-\frac{d}{2}}\varphi (\frac{x}{R})\), whereas \(\Vert \varphi _R\Vert _{L^2({\mathbb R}^d)}= \Vert \varphi \Vert _{L^2({\mathbb R}^d)}\). Since

$$\begin{aligned} \widehat{\varphi _R}= R^{\frac{d}{2}} \widehat{\varphi }(R \,\cdot \,) \end{aligned}$$

we have

$$\begin{aligned}&(\varphi _R,\varphi _R)_{log}\\&\quad = \frac{1}{(2\pi )^{d}} \, \int _{\mathbb R^d} \log (|\xi |)\, |\widehat{\varphi _R}(\xi )|^2\, d\xi = \frac{R^{d}}{(2\pi )^{d}} \, \int _{\mathbb R^d} \log (|\xi |)\, |\widehat{\varphi }(R \xi )|^2\, d\xi \\&\quad = \frac{1}{(2\pi )^{d}} \, \int _{\mathbb R^d} \bigl (\log (|\xi |)-\log R\bigr )\, |\widehat{\varphi }(\xi )|^2\, d\xi =(\varphi ,\varphi )_{log} - \log R \Vert \varphi \Vert _{L^2({\mathbb R}^d)}^2, \end{aligned}$$

as stated in (2.10). \(\square \)

An upper trace bound

Throughout this section, we let \(\Omega \subset {\mathbb R}^d\) denote an open set of finite measure. Let \(\{\varphi _k\}\) and \(\{\lambda _k\}\) be the orthonormal in \(L^2(\Omega )\) system of eigenfunctions and the eigenvalues of the operator \(\mathcal H\) respectively. In what follows we denote

$$\begin{aligned} (\lambda - t)_+ = {\left\{ \begin{array}{ll} \lambda - t, &{} \mathrm{if} \quad t <\lambda , \\ 0, \quad &{} \mathrm{if} \quad t \ge \lambda . \end{array}\right. } \end{aligned}$$

Then we have

Theorem 3.1

For the eigenvalues of the problem (1.3) and any \(\lambda \in \mathbb R\) we have

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \le \frac{1}{(2\pi )^{d}}\, |\Omega |\, e^{d\lambda } \, |B_d|\, d^{-1}, \end{aligned}$$

where \(|B_d|\) is the measure of the unit ball in \(\mathbb R^d\).


Extending the eigenfunction \(\varphi _k\) by zero outside \(\Omega \) and using the Fourier transform we find

$$\begin{aligned}&\sum _{k}(\lambda - \lambda _k)_+ = \sum _{k}\left( \lambda (\varphi _k, \varphi _k) - (\mathcal H\varphi _k, \varphi _k) \right) _+ \\&\quad = \frac{1}{(2\pi )^{d}}\, \left( \sum _k \int _{\mathbb R^d} \left( \lambda - \log (|\xi |) \right) \, |\widehat{\varphi _k}(\xi )|^2 \, d\xi \right) _+\\&\quad \le \frac{1}{(2\pi )^{d}}\, \int _{\mathbb R^d} \left( \lambda - \log (|\xi |) \right) _+ \, \sum _k |\widehat{\varphi _k}(\xi )|^2 \, d\xi . \end{aligned}$$

Using that \(\{\varphi _k\}\) is an orthonormal basis in \(L^2(\Omega )\) and denoting \(e_\xi = e^{-i (\cdot ,\xi )}\)we have

$$\begin{aligned} \sum _k |\widehat{\varphi _k}(\xi )|^2 = \sum _k |(e_\xi , \varphi _k)|^2 = \Vert e_\xi \Vert ^2_{L^2(\Omega )} = |\Omega |, \end{aligned}$$

and finally obtain

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+&\le \frac{1}{(2\pi )^{d}}\, |\Omega |\, \int _{\mathbb R^d} \left( \lambda - \log (|\xi |) \right) _+ \\&= \frac{1}{(2\pi )^{d}}\, |\Omega |\, e^{d\lambda } \, \int _{|\xi |\le 1} \log (|\xi |^{-1}) \, d\xi . \end{aligned}$$

We complete the proof by computing the last integral. \(\square \)

Let \(\eta >\lambda \) and let us consider the function

$$\begin{aligned} \psi _\lambda (t) = \frac{1}{\eta - \lambda } (\eta - t)_+. \end{aligned}$$

Denote by \(\chi \) the step function

$$\begin{aligned} \chi _\lambda (t) = {\left\{ \begin{array}{ll} 1, \quad &{} \mathrm{if} \quad t<\lambda ,\\ 0,\quad &{} \mathrm{if} \quad t \ge \lambda , \end{array}\right. } \end{aligned}$$

and let

$$\begin{aligned} N(\lambda ) = \# \{k:\, \lambda _k<\lambda \}, \end{aligned}$$

be the number of the eigenvalues below \(\lambda \) of the operator \(\mathcal H\).

Then by using the previous statement we have

$$\begin{aligned} N(\lambda ) \le \frac{1}{\eta - \lambda } \, \sum _k (\eta - \lambda _k)_+ \le \frac{1}{\eta - \lambda } \, \frac{1}{(2\pi )^{d}}\, |\Omega |\, e^{d\eta } \, |B_d|\, d^{-1}. \end{aligned}$$

Minimising the right hand side w.r.t. \(\eta \) we find \(\eta = \lambda + \frac{1}{d} \) and thus obtain the following

Corollary 3.2

For the number \(N(\lambda )\) of the eigenvalues of the operator \(\mathcal H\) below \(\lambda \) we have

$$\begin{aligned} N(\lambda ) \le e^{\lambda d +1} \frac{1}{(2\pi )^{d}}\, |\Omega | \, |B_d|. \end{aligned}$$

A lower bound for \(\lambda _1(\Omega )\)

In this section, we focus on lower bounds for the first eigenvalue \(\lambda _1= \lambda _1(\Omega )\). From Corollary 3.2, we readily deduce the following bound.

Theorem 4.1

Let \(\Omega \subset {\mathbb R}^d\) be an open set of finite measure. Then we have

$$\begin{aligned} \lambda _1(\Omega ) \ge \frac{1}{d} \log \frac{(2\pi )^{d}}{e |\Omega | \, |B_d|}. \end{aligned}$$

In particular, if \(|\Omega | \le \frac{(2\pi )^{d}}{e\, |B_d|}\), then the operator \(\mathcal H\) does not have negative eigenvalues.


If \(\lambda < \frac{1}{d} \log \frac{(2\pi )^{d}}{e |\Omega | \, |B_d|}\), then \(N(\lambda )<1\) by (3.2), and therefore \(N(\lambda )=0\). Consequently, \(\mathcal H\) does not have eigenvalues below \(\frac{1}{d} \log \frac{(2\pi )^{d}}{e |\Omega | \, |B_d|}\). \(\square \)

Remark 1

Note that the inequalities (3.1), (3.2) and (4.1) hold for any open set \(\Omega \) of finite measure without any additional conditions on its boundary.

In the following, we wish to improve the bound given in Theorem 4.1 in low dimensions d for open boundary sets with Lipschitz boundary. We shall use the following Faber-Krahn type inequality.

Theorem 4.2

([5, Corollary 1.6]) Let \(\rho >0\). Among all bounded open sets \(\Omega \) with Lipschitz boundary and \(|\Omega | = \rho \), the ball \(B=B_r(0)\) with \(|B|=\rho \) minimizes \(\lambda _1(\Omega )\).

Corollary 4.3

For every open bounded sets \(\Omega \) with Lipschitz boundary we have

$$\begin{aligned} \lambda _1(\Omega ) \ge \lambda _1(B_d) + \frac{1}{d}\log \frac{|B_d|}{|\Omega |}, \end{aligned}$$

and equality holds if \(\Omega \) is a ball.


The result follows by combining Theorem 4.2 with the identity

$$\begin{aligned} \lambda _1(B_r(0)) = \lambda _1(B_d) + \log \frac{1}{r}\qquad \text {for}\quad r>0, \end{aligned}$$

which follows from the scaling property of \(\lambda _1\) noted in Lemma 2.5. \(\square \)

Corollary 4.3 gives a sharp lower bound, but it contains the unknown quantity \(\lambda _1(B_d)\). By Theorem 4.1, we have

$$\begin{aligned} \lambda _1(B_d)&\ge \frac{1}{d} \log \frac{(2\pi )^{d}}{e |B_d|^2} = \log (2\pi ) -\frac{1}{d}\bigl (1+ 2 \log |B_d|\bigr ) \nonumber \\&= \frac{2}{d} \log \Gamma \left( d/2\right) + \log 2 + \frac{2}{d} \log \frac{d}{2} -\frac{1}{d}. \end{aligned}$$

The following theorem improves this lower bound in low dimensions \(d \ge 2\).

Theorem 4.4

For \(d \ge 2\), we have

$$\begin{aligned} \lambda _1(B_d) \ge \log \bigl (2 \sqrt{d +2}\bigr ) - \frac{2^{d+1} |B_d|^2 (d +2)^{\frac{d}{2}} }{d (2\pi )^{2d}}. \end{aligned}$$


Let \(u \in L^2(B_d)\) be radial with \(\Vert u\Vert _{L^2}=1\). Then \(\widehat{u}\) is also radial, and

$$\begin{aligned} |\widehat{u}(\xi )|&=|\widehat{u}(s)|= s^{1-\frac{d}{2}}\left| \int _{0}^{1} u(r)J_{\frac{d}{2}-1}(rs)r^{\frac{d}{2}}dr\right| \\&\le s^{1-\frac{d}{2}} \left( \int _0^{1}r^{d-1} u^2(r)\,dr\right) ^{1/2} \left( \int _0^{1}rJ_{\frac{d}{2}-1}^2(sr) \,dr\right) ^{1/2}\\&=\frac{s^{1-\frac{d}{2}}}{\sqrt{|S^{d-1}|}} \left( s^{-2} \int _0^{s} \tau J_{\frac{d}{2}-1}^2(\tau ) \,d\tau \right) ^{1/2} \\&=\frac{s^{-\frac{d}{2}}}{\sqrt{|S^{d-1}|}} \left( \int _0^{s} \tau J_{\frac{d}{2}-1}^2(\tau ) \,d\tau \right) ^{1/2}\qquad \text {for}\quad \xi \in {\mathbb R}^d \quad \text {with}\quad s = |\xi |. \end{aligned}$$


$$\begin{aligned} |S^{d-1}| |\widehat{u}(s)|^2 \le s^{-d} \int _0^{s} \tau J_{\frac{d}{2}-1}^2(\tau ) \,d\tau . \end{aligned}$$

In the case where, in addition, u is a radial eigenfunction of (1.3) corresponding to \(\lambda _1\) in \(\Omega = B_d\), it follows that, for every \(\lambda \in {\mathbb R}\),

$$\begin{aligned}&(2\pi )^{d}[\lambda -\lambda _1] = \int _{{\mathbb R}^d} (\lambda -\ln |\xi |)|\widehat{u}(\xi )|^2\,d\xi \le \int _{{\mathbb R}^d} (\lambda -\ln |\xi |)_+|\widehat{u}(\xi )|^2\,d\xi \\&\quad =|S^{d-1}| \int _0^\infty s^{d-1} (\lambda -\ln s)_+|\widehat{u}(s)|^2\,ds \le \int _0^\infty \frac{(\lambda -\ln s)_+}{s} \int _0^{s} \!\!\!\tau J_{\frac{d}{2}-1}^2(\tau ) \,d\tau \,ds\\&\quad = \int _0^\infty \tau J_{\frac{d}{2}-1}^2(\tau ) \int _{\tau }^\infty \frac{(\lambda -\ln s)_+}{s} \,ds d\tau = \int _0^{e^\lambda } \tau J_{\frac{d}{2}-1}^2(\tau ) \int _{\tau }^{e^\lambda } \frac{\lambda -\ln s}{s} \,ds d\tau \\&\quad = \int _0^{e^\lambda } \tau J_{\frac{d}{2}-1}^2(\tau ) \int _{\ln \tau }^{\lambda }(\lambda - s) \,ds d\tau = \int _0^{e^\lambda } \tau J_{\frac{d}{2}-1}^2(\tau ) \int _{0}^{\lambda - \ln \tau } s \,ds d\tau \\&\quad = \frac{1}{2} \int _0^{e^\lambda } \tau J_{\frac{d}{2}-1}^2(\tau ) \bigl (\lambda - \ln \tau \bigr )^2 \,d\tau = \frac{e^{2\lambda }}{2} \int _0^{1} \tau J_{\frac{d}{2}-1}^2(e^\lambda \tau ) \ln ^2 \tau \,d\tau . \end{aligned}$$

We now use the following estimate for Bessel functions of the first kind:

$$\begin{aligned} J_\nu (x) \le \frac{x^\nu }{2^\nu \Gamma (\nu +1)} \quad \text {for}\quad \,\nu > \sqrt{3}-2, \,0 \le x < 2 \sqrt{2(\nu +2)}. \end{aligned}$$

A proof of this elementary estimate is given in the Appendix. We wish to apply (4.5) with \(\nu = \frac{d}{2}-1\). This gives

$$\begin{aligned} e^{2\lambda } J_{\frac{d}{2}-1}^2(r_0 e^\lambda \tau ) \le e^{d\lambda } \frac{\tau ^{d-2}}{2^{d-2}\Gamma ^2 (\frac{d}{2})}= \frac{d^2 |B_d|^2e^{d\lambda }}{(2\pi )^{d}} \tau ^{d-2} \qquad \text {for}\quad \tau \in [0,1] \end{aligned}$$

if \(d \ge 2\) and \(e^\lambda \le 2 \sqrt{d +2}\), i.e., if

$$\begin{aligned} d \ge 2 \quad \text {and}\quad \lambda \le \log \bigl (2 \sqrt{d +2}\bigr ). \end{aligned}$$

Here we used that \(|B_d|= \frac{2}{d} \frac{\pi ^{\frac{d}{2}}}{\Gamma (d/2)}\). Consequently, if (4.6) holds, we find that

$$\begin{aligned} (2\pi )^{d}[\lambda -\lambda _1] \le \frac{d^2 |B_d|^2e^{d\lambda }}{(2\pi )^{d}}\int _0^{1} \tau ^{d-1} \ln ^2 \tau \,d\tau , \end{aligned}$$


$$\begin{aligned} \int _0^{1} \tau ^{d-1} \ln ^2 \tau d\tau = - \frac{2}{d} \int _0^1 \tau ^{d-1} \ln \tau d\tau = \frac{2}{d^2} \int _{0}^1 \tau ^{d-1}d\tau = \frac{2}{d^3}. \end{aligned}$$


$$\begin{aligned} (2\pi )^{d}[\lambda -\lambda _1] \le \frac{2|B_d|^2}{d (2\pi )^{d}}e^{d\lambda }, \quad \text {i.e.,}\quad \lambda _1 \ge \lambda - \frac{2|B_d|^2 }{d (2\pi )^{2d}} e^{d\lambda }. \end{aligned}$$

Inserting the value \(\lambda = \log \bigl (2 \sqrt{d +2}\bigr )\) from (4.6), we deduce that

$$\begin{aligned} \lambda _1= \lambda _1(B_d) \ge \log \bigl (2 \sqrt{d +2}\bigr ) - \frac{2^{d+1} |B_d|^2 (d +2)^{\frac{d}{2}} }{d(2\pi )^{2d}}, \end{aligned}$$

as claimed. \(\square \)

Remark 2

It seems instructive to compare the lower bounds given in (4.3) and (4.4) with other bounds obtained from spectral estimates which are already available in the literature. We first mention Beckner’s logarithmic estimate of uncertainty [2, Theorem 1], which implies thatFootnote 1

$$\begin{aligned} (\varphi ,\varphi )_{log} \ge \int _{{\mathbb R}^d} \left[ \psi \left( d/4\right) + \log \frac{2}{|x|}\right] \varphi ^2(x) dx \ge \left[ \psi \left( d/4\right) + \log 2\right] \Vert \varphi \Vert _2^2 \end{aligned}$$

for functions \(\varphi \in C^\infty _c(B_d)\) and therefore

$$\begin{aligned} \lambda _1(B_d) \ge \psi \left( d/4\right) + \log 2 . \end{aligned}$$

Here, as before, \(\psi = \frac{\Gamma '}{\Gamma }\) denotes the Digamma function. Next we state a further lower bound for \((\varphi ,\varphi )_{log}\) which follows from [5, Proposition 3.2 and Lemma 4.11]. We have

$$\begin{aligned} (\varphi ,\varphi )_{log} \ge \zeta _d \Vert \varphi \Vert _2^2 \qquad \text {for}\quad \varphi \in C^\infty _c(B_d), \end{aligned}$$

where \(\zeta _d\) is given in (2.7), i.e.,

$$\begin{aligned} \zeta _d = \log 2 + \frac{1}{2}\left( \psi (d/2)-\gamma \right) = \left\{ \begin{aligned}&- \gamma + \sum _{k=1}^{\frac{d-1}{2}} \frac{1}{2k-1},&\qquad d \quad \text {odd,}\\&\log 2 - \gamma + \sum _{k=1}^{\frac{d-2}{2}} \frac{1}{k},&\qquad d \quad \text {even.} \end{aligned} \right. \end{aligned}$$

Inequality (4.8) implies that

$$\begin{aligned} \lambda _1(B_d) \ge \zeta _d. \end{aligned}$$

The latter inequality can also be derived from a lower bound of Bañuelos and Kulczycki for the first Dirichlet eigenvalue \(\lambda _1^\alpha (B_d)\) of the fractional Laplacian \((-\Delta )^{\alpha /2}\) in \(B_d\). In [1, Corollary 2.2], it is proved that

$$\begin{aligned} \lambda _1^\alpha (B_d) \ge 2^\alpha \frac{\Gamma (1+\frac{\alpha }{2}) \Gamma (\frac{d+\alpha }{2})}{\Gamma (\frac{d}{2})}\qquad \text {for}\quad \alpha \in (0,2). \end{aligned}$$

Combining this inequality with the characterization of \(\lambda _1(B_d)\) given in [5, Theorem 1.5], we deduce that

$$\begin{aligned} \lambda _1(B_d)= \lim _{\alpha \rightarrow 0^+}\frac{\lambda _1^\alpha (B_d)-1}{\alpha }\ge \frac{d}{d\alpha }\Big |_{\alpha =0}\, 2^\alpha \frac{\Gamma (1+\frac{\alpha }{2}) \Gamma (\frac{d+\alpha }{2})}{\Gamma (\frac{d}{2})} = \zeta _d, \end{aligned}$$

as stated in (4.9).

We briefly comment on the quality of the lower bounds obtained here in low and high dimensions. In low dimensions \(d \ge 2\), (4.4) is better than the bounds (4.3), (4.7) and (4.9). In dimension \(d=1\) where the bound (4.4) is not available, the bound (4.3) yields the best value. The following table shows numerical values of the bounds \(b_1(d)\), \(b_2(d)\), \(b_3(d)\) resp. \(b_4(d)\) given by (4.3), (4.4), (4.7), (4.9), respectively.

d 1 2 3 4 5 6 7 8 9 10
\(b_1(d)\) \(-0,55\) 0, 19 0, 55 0, 79 0, 97 1, 12 1, 25 1, 36 1, 46 1, 55
\(b_2(d)\) \(\quad /\) 1, 28 1, 48 1, 59 1, 67 1, 73 1, 79 1, 84 1, 89 1, 94
\(b_3(d)\) \(-3.53\) \(-1,27\) \(-0,39\) 0, 12 0, 47 0, 73 0, 94 1, 12 1, 27 1, 40
\(b_4(d)\) \(-0,58\) 0, 12 0, 42 0, 62 0, 76 0, 87 0, 96 1, 03 1, 10 1, 16

To compare the bounds in high dimensions, we consider the asymptotics as \(d \rightarrow \infty \). Since \(\frac{\log \Gamma (t)}{t} = \log t - 1 + o(t)\) as \(t \rightarrow \infty \), the bound (4.3) yields

$$\begin{aligned} \lambda _1(B_d) \ge \log d - 1 + o(1) \qquad \text {as}\quad d \rightarrow \infty , \end{aligned}$$

whereas (4.4) obviously gives

$$\begin{aligned} \lambda _1(B_d) \ge \log \sqrt{d+2} + \log 2 + o(1) \qquad \text {as}\quad d \rightarrow \infty , \end{aligned}$$

Moreover, from (4.7) and the fact that

$$\begin{aligned} \psi (t) = \log t + o(1)\qquad \text {as}\quad t \rightarrow \infty , \end{aligned}$$

we deduce that

$$\begin{aligned} \lambda _1(B_d) \ge \log d - \log 2 + o(1) \qquad \text {as}\quad d \rightarrow \infty , \end{aligned}$$

Finally, (4.8) and (4.12) yield

$$\begin{aligned} \lambda _1(B_d) \ge \log \sqrt{d} + \log 2 -\frac{\gamma }{2} + o(1) \qquad \text {as} \quad d \rightarrow \infty . \end{aligned}$$

So (4.13) provides the best asymptotic bound as \(d \rightarrow \infty \).

Numerical computations indicate that the bound (4.4) is better than the other bounds for \(2 \le d \le 21\), and (4.7) is the best among these bounds for \(d \ge 22\).

An asymptotic lower trace bound

Throughout this section, we let \(\Omega \subset {\mathbb R}^d\) denote an open set of finite measure. In this section we prove the following asymptotic lower bound. A similar statement was obtained in [9] for the Dirichlet boundary problem for a fractional Laplacian.

Theorem 5.1

For the eigenvalues of the problem (1.3) and any \(\lambda \in \mathbb R\) we have

$$\begin{aligned} \liminf _{\lambda \rightarrow \infty } e^{-d\lambda } \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{1}{(2\pi )^{d}}\, |\Omega |\, \, |B_d|\, d^{-1}. \end{aligned}$$


Let us fix \(\delta >0\) and consider

$$\begin{aligned} \Omega _\delta = \{ x\in \Omega : \, \mathrm{dist}(x, \mathbb R^d \setminus \Omega ) >\delta \}. \end{aligned}$$

Since \(\delta \) is arbitrary it suffices to show the lower bound (5.1), where \(\Omega \) is replaced by \(\Omega _\delta \). Let \(g\in C_0^\infty (\mathbb R^d)\) be a real-valued even function, \(\Vert g\Vert _{L^2(\mathbb R^d)} = 1\) with support in \(\{x\in \mathbb R^d: \, |x| \le \delta /2\}\). For \(\xi \in \mathbb R^d\) and \(x\in \Omega _\delta \) we introduce the “coherent state”

$$\begin{aligned} e_{\xi ,y}(x) = e^{-i\xi x} g(x-y). \end{aligned}$$

Note that \(\Vert e_{\xi ,y}\Vert _{L^2(\mathbb R^d)} = 1\). Using the properties of coherent states [15, Theorem 12.8] we obtain

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{1}{(2\pi )^{d}}\, \int _{\mathbb R^d} \int _{\Omega _\delta } (e_{\xi ,y}, (\lambda - \mathcal H)_+ e_{\xi ,y})_{L^2(\Omega )} \, dy d\xi . \end{aligned}$$

Since \(t \mapsto (\lambda -t)_+\) is convex then applying Jensen’s inequality to the spectral measure of \(\mathcal H\) we obtain

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{1}{(2\pi )^{d}}\, \int _{\mathbb R^d} \int _{\Omega _\delta } \left( \lambda - (\mathcal H e_{\xi ,y}, e_{\xi ,y})_{L^2(\Omega )} \right) _+ \, dy d\xi . \end{aligned}$$

Next we consider the quadratic form

$$\begin{aligned}&\left( \mathcal H e_{\xi ,y}, e_{\xi ,y} \right) _{L^2(\Omega )} = \frac{1}{(2\pi )^d} \, \int _{\mathbb R^d} \int _{\Omega } \int _{\Omega } e^{i(x-z)(\eta -\xi )} g(x-y)g(z-y) \log (|\eta |) \, dz dx d\eta \\&\quad = \frac{1}{(2\pi )^d} \, \int _{\mathbb R^{d}} \int _{\Omega } \int _{\Omega } e^{i(x-z)\rho } g(x-y)g(z-y) \log (|\xi -\rho |) \, dz dx d\rho \\&\quad = \frac{1}{(2\pi )^d} \, \int _{\mathbb R^{d}} \int _{\Omega } \int _{\Omega } e^{i(x-z)\rho } g(x-y)g(z-y) \left( \log |\xi | + \log \left( \left| \xi -\rho \right| /|\xi | \right) \right) \, dz dx d\rho \\&\quad = \log |\xi | + R(y,\xi ). \end{aligned}$$

Since \(g\in C_0^\infty (\mathbb R^d)\) we have for any \(M>0\)

$$\begin{aligned}&R(y,\xi ) \\&\quad =\frac{1}{(2\pi )^d} \, \int _{\mathbb R^{d}} \int _{\Omega } \int _{\Omega } e^{i(x-y)\rho } g(x-y) e^{i(y-z)\rho } g(z-y) \log \left( \left| \xi -\rho \right| /|\xi | \right) \, dz dx d\rho \\&\quad = \int _{\mathbb R^{d}} |\widehat{g}|^2\, \log \left( \left| \xi -\rho \right| /|\xi | \right) \, d\rho \le C_M\, \int _{\mathbb R^{d}} (1+ |\rho |)^{-M} \log \left( \left| \xi -\rho \right| /|\xi | \right) \, d\rho \\&\quad \le C\, |\xi |^{-1}. \end{aligned}$$

Therefore from (5.2) we find

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \ge (2\pi )^{-d}\, |\Omega _\delta | \, \int _{\mathbb R^d} (\lambda - \log |\xi | - C|\xi |^{-1})_+ \, d\xi . \end{aligned}$$

Let us redefine the spectral parameter \(\lambda = \ln \mu \). Then introducing polar coordinates we find

$$\begin{aligned}&\int _{\mathbb R^d} (\lambda - \log |\xi | - C|\xi |^{-1})_+ \, d\xi = \left| \mathbb S^{d-1} \right| \, \int _0^\infty \left( \ln \frac{\mu }{r} - \frac{C}{r}\right) _+ \, r^{d-1}dr\nonumber \\&\quad = \mu ^d\, \left| \mathbb S^{d-1} \right| \, \int _0^\infty \left( \ln \frac{1}{r} - \frac{C}{\mu r}\right) _+ \, r^{d-1}dr. \end{aligned}$$

The expression in the latter integral is positive if \( - r\ln r > C\mu ^{-1}\). The function \( -r\ln r \) is concave.

figure a

Its maximum is achieved at \(r=1/e\) at the value 1/e. The equation \( - r\ln r = C\mu ^{-1}\) has two solutions \(r_1(\mu )\) and \(r_2(\mu )\) such that \(r_1(\mu ) \rightarrow 0\) and \(r_2(\mu )\rightarrow 1\) as \(\mu \rightarrow \infty \). Therefore

$$\begin{aligned}&\int _0^\infty \left( \ln \frac{1}{r} - \frac{C}{\mu r}\right) _+ \, r^{d-1}dr \ge \int _{r_1(\mu )}^{r_2(\mu )} \left( \ln \frac{1}{r} - \frac{C}{\mu r}\right) \, r^{d-1}dr \nonumber \\&\quad =-\frac{1}{d}\, r^d \ln r \Big |_{r_1(\mu )}^{r_2(\mu )}+ \frac{C}{\mu (d+1)} r^{d+1} \Big |_{r_1(\mu )}^{r_2(\mu )} + \frac{1}{d^2}r^d \Big |_{r_1(\mu )}^{r_2(\mu )} \rightarrow \frac{1}{d^2} \quad \mathrm{as} \quad \mu \rightarrow \infty .\nonumber \\ \end{aligned}$$

Putting together (5.3), (5.4) and (5.5) and using \(\mu = e^\lambda \) we obtain

$$\begin{aligned} \liminf _{\lambda \rightarrow \infty } e^{-d\lambda } \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{1}{(2\pi )^{d}}\, |\Omega _\delta |\, \, |B_d|\, d^{-1}. \end{aligned}$$

Since \(\delta >0\) is arbitrary we complete the proof of Theorem 5.1. \(\square \)

Weyl asymptotics

Throughout this section, we let \(\Omega \subset {\mathbb R}^d\) denote an open set of finite measure. Combining Theorems 3.1 and 5.1 we have

Theorem 6.1

The Riesz means of the eigenvalues of the Dirichlet boundary value problem (1.3) satisfy the following asymptotic formula

$$\begin{aligned} \lim _{\lambda \rightarrow \infty } e^{-d\lambda }\, \sum _{k}(\lambda - \lambda _k)_+ = \frac{1}{(2\pi )^{d}}\, |\Omega |\, |B_d|\, d^{-1}. \end{aligned}$$

As a corollary we can obtain asymptotics of the number of the eigenvalues of the operator \(\mathcal H\).

Corollary 6.2

The number of the eigenvalues \(N(\lambda )\) of the Dirichlet boundary value problem (1.3) below \(\lambda \) satisfies the following asymptotic formula

$$\begin{aligned} \lim _{\lambda \rightarrow \infty } e^{-d\lambda } \, N(\lambda ) = \frac{1}{(2\pi )^{d}}\, |\Omega |\, |B_d|. \end{aligned}$$


In order to prove (6.2) we use two simple inequalities. If \(h>0\), then

$$\begin{aligned} \frac{(\lambda + h - \lambda _k)_+ - (\lambda - \lambda _k)_+}{h} \ge 1_{{\tiny (-\infty ,\lambda )}}(\lambda _k) \end{aligned}$$


$$\begin{aligned} \frac{(\lambda - \lambda _k)_+ - (\lambda - h- \lambda _k)_+}{h} \le 1_{{\tiny (-\infty ,\lambda )}}(\lambda _k) \end{aligned}$$

The inequality (6.3) implies, together with Theorems 3.1 and 5.1, that

$$\begin{aligned}&\limsup _{\lambda \rightarrow \infty }e^{-d\lambda }N(\lambda ) \le \limsup _{\lambda \rightarrow \infty }e^{-d\lambda } \sum _{k}\frac{(\lambda + h - \lambda _k)_+ - (\lambda - \lambda _k)_+}{h}\\&\quad \le \frac{1}{h}\Bigl [e^{dh} \limsup _{\lambda \rightarrow \infty }e^{-d(\lambda +h)} \sum _{k}(\lambda + h - \lambda _k)_+ -\liminf _{\lambda \rightarrow \infty }e^{-d\lambda } \sum _{k}(\lambda - \lambda _k)_+\Bigr ]\\&\quad \le \frac{|\Omega | |B_d|}{d(2\pi )^d}\,\frac{e^{dh}-1}{h} \qquad \text {for every}\quad h>0 \end{aligned}$$

and thus

$$\begin{aligned} \limsup _{\lambda \rightarrow \infty }e^{-d\lambda }N(\lambda ) \le \frac{|\Omega | |B_d|}{d(2\pi )^d}\lim _{h \rightarrow 0^+}\frac{e^{dh}-1}{h}= \frac{|\Omega |\, |B_d|}{(2\pi )^{d}}. \end{aligned}$$

Moreover, (6.3) implies, together with Theorems 3.1 and 5.1, that

$$\begin{aligned}&\liminf _{\lambda \rightarrow \infty }e^{-d\lambda }N(\lambda ) \ge \liminf _{\lambda \rightarrow \infty }e^{-d\lambda } \sum _{k}\frac{(\lambda - \lambda _k)_+ - (\lambda -h - \lambda _k)_+}{h}\\&\quad \ge \frac{1}{h}\Bigl [e^{dh} \liminf _{\lambda \rightarrow \infty }e^{-d \lambda } \sum _{k}(\lambda - \lambda _k)_+ -e^{-dh}\limsup _{\lambda \rightarrow \infty } e^{-d(\lambda -h)} \sum _{k}(\lambda -h - \lambda _k)_+\Bigr ]\\&\quad \ge \frac{|\Omega | |B_d|}{d(2\pi )^d}\,\frac{1-e^{-dh}}{h} \qquad \text {for every}\quad h>0 \end{aligned}$$

and therefore

$$\begin{aligned} \liminf _{\lambda \rightarrow \infty }e^{-d\lambda }N(\lambda ) \ge \frac{|\Omega | |B_d|}{d(2\pi )^d}\lim _{h \rightarrow 0^+}\frac{1-e^{-dh}}{h}= \frac{|\Omega |\, |B_d|}{(2\pi )^{d}}. \end{aligned}$$

The claim follows by combining (6.5) and (6.6). \(\square \)

Remark 3

The proof of Corollary 6.2 is a version of a Tauberian type arguments that is particularly simple due to properties of exponential functions.

An exact lower trace bound

In this section we prove the following exact lower bound in the case of bounded open sets with Lipschitz boundary.

Theorem 7.1

Let \(\Omega \subset {\mathbb R}^d\), \(N \ge 2\) be an open bounded set with Lipschitz boundary, let \(\tau \in (0,1)\), and let

$$\begin{aligned} C_{\Omega ,\tau } := \frac{1}{|\Omega |(2\pi )^d} \int _{\mathbb R^d}(1+|\rho |)^\tau \log (1+|\rho |) |\widehat{1_\Omega }(\rho )|^2\,d\rho , \end{aligned}$$

where \(1_\Omega \) denotes the indicator function of \(\Omega \).

For any \(\lambda \ge 2 C_{\Omega ,\tau }\), we have

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{|\Omega |\, |B_d|}{(2\pi )^{d}\,d} \Bigl [e^{d\lambda } \,- \,a_\tau \, C_{\Omega ,\tau }\,e^{(d-\tau )\lambda } \,- \,b_\tau \, C_{\Omega ,\tau }^2\, e^{(d-2\tau )\lambda } \,-\, (d \lambda + 1) \Bigr ] \end{aligned}$$

with \(a_\tau := \frac{d(d-\tau )-1}{d-\tau }\) and \(b_\tau := 4d \tau \).

Remark 4

In the definition of \(C_{\Omega ,\tau }\), we need \(\tau <1\), otherwise the integral might not converge. In particular, if \(\Omega =B_d\) is the unit ball in \({\mathbb R}^d\), we have

$$\begin{aligned} \widehat{1_\Omega }(\rho )= (2\pi )^{\frac{d}{2}} |\rho |^{-\frac{d}{2}}J_{\frac{d}{2}}(|\rho |) \end{aligned}$$

where \(J_{\frac{d}{2}}(r)= O(\frac{1}{\sqrt{r}})\) as \(r \rightarrow \infty \). Hence the integral defining \(C_{\Omega ,\tau }\) converges if \(\tau <1\). A similar conclusion arises for cubes or rectangles, where

$$\begin{aligned} \widehat{1_\Omega }(\rho ) = f_1(\rho _1) \cdot \dots \cdot f_d(\rho _d) \end{aligned}$$

and \(f_j(s) = O(\frac{1}{s})\) as \(|s| \rightarrow \infty \), \(j=1,\dots ,d\).

On the other hand, if \(\Omega \subset {\mathbb R}^d\) is an open bounded set with Lipschitz boundary, we have

$$\begin{aligned} C_{\Omega ,\tau }<\infty \qquad \text {for}\quad \tau \in (0,1). \end{aligned}$$

Indeed, in this case, \(\Omega \) has finite perimeter, i.e., \(1_\Omega \in BV({\mathbb R}^d)\). Therefore, as noted e.g. in [16, Theorem 2.14], \(\Omega \) also has finite fractional perimeter

$$\begin{aligned} P_\tau (\Omega )= \int _{\Omega }\int _{{\mathbb R}^d \setminus \Omega } |x-y|^{-d-\tau }\,dxdy = \frac{1}{2} \int \!\! \int _{{\mathbb R}^{2d}}\frac{(1_\Omega (x)-1_\Omega (y))^2}{|x-y|^{d+\tau }}\,dxdy \end{aligned}$$

for every \(\tau \in (0,1)\). Moreover, \(P_\tau (\Omega )\) coincides, up to a constant, with the integral

$$\begin{aligned} \int _{\mathbb R^d}|\rho |^\tau |\widehat{1_\Omega }(\rho )|^2\,d\rho \end{aligned}$$

which therefore is also finite for every \(\tau \in (0,1)\). Since moreover \(1_\Omega \) and therefore also \(\widehat{1_\Omega }\) are functions in \(L^2(\mathbb R^d)\) and for every \(\varepsilon >0\) there exists \(C_\varepsilon >0\) with

$$\begin{aligned} (1+|\rho |)^\tau \log (1+|\rho |) \le C_\varepsilon \bigl (1 + |\rho |^{\tau +\varepsilon }\bigr ) \qquad \text {for}\quad \rho \in {\mathbb R}^d, \end{aligned}$$

it follows that (7.2) holds.

In the proof of Theorem 7.1, we will use the following elementary estimate.

Lemma 7.2

For \(r \ge 0\), \(s>0\) and \(\tau \in (0,1)\), we have

$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \le \frac{1}{s} \log (1+r) \qquad \text {if}\quad s \in (0,1) \end{aligned}$$


$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \le \frac{(1+r)^\tau }{s^\tau } \log (1+r) \qquad \text {if}\quad s \ge 1. \end{aligned}$$

In particular,

$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \le \max \left\{ \frac{1}{s}, \frac{1}{s^\tau } \right\} (1+r)^\tau \log (1+r) \qquad \text {for}\quad r,s>0. \end{aligned}$$

Remark 5

The obvious bound \(\log (1 + \frac{r}{s}) \le \frac{r}{s}\) will not be enough for our purposes. We need an upper bound of the form g(s)h(r) where h grows less than linearly in r.

Proof of Lemma 7.2

Let first \(s \in (0,1)\). Since

$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \Big |_{r=0} = 0 = \frac{1}{s} \log (1+r)\Big |_{r=0} \end{aligned}$$

and, for every \(r>0\),

$$\begin{aligned} \frac{d}{dr} \log \left( 1 + \frac{r}{s}\right) = \frac{1}{s+r} \le \frac{1}{s+ sr} = \frac{d}{dr} \frac{1}{s} \log (1+r), \end{aligned}$$

inequality (7.3) follows. To see (7.4), we fix \(s>1\), and we note that

$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \Big |_{r=0} = 0 = \frac{(1+r)^\tau }{s^\tau } \log (1+r)\Big |_{r=0}. \end{aligned}$$

Moreover, for \(0 < r \le s-1\), we have

$$\begin{aligned}&\frac{d}{dr} \frac{(1+r)^\tau }{s^\tau } \log (1+r)= \frac{(1+r)^{\tau -1}}{s^\tau }(1+ \tau \log (1+r))\\&\quad \ge \frac{(1+r)^{\tau -1}}{s^\tau } \ge \frac{1}{s}\ge \frac{1}{s+r}= \frac{d}{dr} \log \left( 1 + \frac{r}{s}\right) , \end{aligned}$$

so the inequality holds for \(r \le s-1\). If, on the other hand, \(r \ge s-1\), we have obviously

$$\begin{aligned} \log \left( 1 + \frac{r}{s}\right) \le \log (1 + r) \le \frac{(1+r)^\tau }{s^\tau } \log (1+r). \end{aligned}$$

We may now complete the

Proof of Theorem 7.1

For \(\xi \in {\mathbb R}^d\), we define \(f_\xi \in L^2(\mathbb R^d)\) by \(f_\xi (x)= \frac{1}{\sqrt{|\Omega |}}1_{\Omega } e^{-i x \xi }\). Note that \(\Vert f_{\xi }\Vert _{L^2(\mathbb R^d)} = 1\) for any \(\xi \in \mathbb R^d\). We write

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+&= \sum _{k}(\lambda - \lambda _k)_+ \Vert \varphi _k\Vert _{L^2(\Omega )}^2 = \frac{1}{(2\pi )^{d}} \sum _{k}(\lambda - \lambda _k)_+ \Vert \widehat{\varphi _k}\Vert _{L^2(\mathbb R^{d})}^2 \\&=\frac{|\Omega |}{(2\pi )^{d}} \sum _{k}(\lambda - \lambda _k)_+ \int _{\mathbb R^{d}} |\langle f_\xi ,\varphi _k \rangle |^2\,d\xi \\&=\frac{|\Omega |}{(2\pi )^{d}} \int _{\mathbb R^{d}} \sum _{k}(\lambda - \lambda _k)_+ |\langle f_\xi ,\varphi _k \rangle |^2\,d\xi . \end{aligned}$$

Since \(\sum \limits _{k} |\langle f_\xi ,\varphi _k \rangle |^2 = \Vert f_{\xi }\Vert _{L^2(\mathbb R^d)}^2 = 1\) for \(\xi \in \mathbb R^d\), Jensen’s inequality gives

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+&\ge \frac{|\Omega |}{(2\pi )^{d}} \int _{\mathbb R^{d}} \Bigl ( \lambda \sum _{k}|\langle f_\xi ,\varphi _k \rangle |^2 - \sum _k \lambda _k |\langle f_\xi ,\varphi _k \rangle |^2\Bigr )_+\,d\xi \nonumber \\&=\frac{|\Omega |}{(2\pi )^{d}} \int _{\mathbb R^{d}} \Bigl ( \lambda - \sum _k \lambda _k |\langle f_\xi ,\varphi _k \rangle |^2\Bigr )_+\,d\xi \nonumber \\&=\frac{|\Omega |}{(2\pi )^{d}} \int _{\mathbb R^{d}} \Bigl ( \lambda - ( \mathcal H f_\xi , f_\xi ) \Bigr )_+\,d\xi . \end{aligned}$$

Here, since

$$\begin{aligned} \sqrt{|\Omega |}\, \widehat{f_\xi }(\eta -\xi )= \int _{\Omega }e^{-i(\eta -\xi ) x}e^{-ix \xi }\,dx = \int _{\Omega }e^{-i \eta x} \,dx = \widehat{1_\Omega }(\eta ) \end{aligned}$$

for \(\eta , \xi \in {\mathbb R}^d\), we have

$$\begin{aligned}&|\Omega |(2\pi )^d \left( \mathcal H f_{\xi }, f_{\xi } \right) = |\Omega | \int _{\mathbb R^d} \log |\eta | |\widehat{f_\xi }(\eta )|^2d \eta = |\Omega | \int _{\mathbb R^d} \log |\eta -\xi | |\widehat{f_\xi }(\eta -\xi )|^2d \eta \nonumber \\&\quad =\int _{\mathbb R^d} \log |\eta -\xi | |\widehat{1_\Omega }(\eta )|^2\,d\eta \le \int _{\mathbb R^d} \left[ \log |\xi | +\log (1+|\eta |/|\xi |)\right] |\widehat{1_\Omega }(\eta )|^2\,d\eta \nonumber \\&\quad \le \log |\xi | \int _{\mathbb R^d} |\widehat{1_\Omega }(\eta )|^2\,d\eta + \max \left\{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau }\right\} \int _{\mathbb R^d}(1+|\eta |)^{\tau } (\log (1+|\eta |)|\widehat{1_\Omega }(\eta )|^2\,d\eta \nonumber \\&\quad = |\Omega |(2\pi )^d \Bigl (\log |\xi | + \max \left\{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau }\right\} C_{\Omega ,\tau } \Bigr )\qquad \text {for}\quad \xi \in {\mathbb R}^d, \end{aligned}$$

where \(C_{\Omega ,\tau }\) is defined in (7.1). Here we used Lemma 7.2. Combining (7.5) and (7.6), we get

$$\begin{aligned} \sum _{k}(\lambda - \lambda _k)_+ \ge \frac{|\Omega |}{(2\pi )^{d}} \int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \left\{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau }\right\} C_{\Omega ,\tau }\Bigr )_+d\xi . \end{aligned}$$

Let us redefine the spectral parameter \(\lambda = \log \mu \) again. Then we find

$$\begin{aligned}&\int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \left\{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau } \right\} C_{\Omega ,\tau } \Bigr )_+d\xi \nonumber \\&\quad = \left| \mathbb S^{d-1} \right| \, \int _0^\infty \left( \log \frac{\mu }{r} - \max \left\{ \frac{1}{r}, \frac{1}{r^\tau }\right\} C_{\Omega ,\tau } \right) _+ \, r^{d-1}dr \nonumber \\&\quad = \mu ^d\, \left| \mathbb S^{d-1} \right| \, \int _0^\infty \left( -\log r - \max \left\{ \frac{1}{\mu ^{1-\tau } r}, \frac{1}{r^\tau }\right\} \frac{C_{\Omega ,\tau }}{\mu ^\tau } \right) _+ \, r^{d-1}dr \nonumber \\&\quad \ge \mu ^d\, \left| \mathbb S^{d-1} \right| \, \int _{\frac{1}{\mu }}^\infty \left( -\log r - \frac{1}{r^\tau }\frac{C_{\Omega ,\tau }}{\mu ^\tau } \right) _+ \, r^{d-1}dr. \end{aligned}$$

For the last inequality, we used the fact that \(\frac{1}{\mu ^{1-\tau } r} \le \frac{1}{r^\tau }\) for \(r \ge \frac{1}{\mu }\).

Next we note that the function \(r \mapsto f_\mu (r) = -\log r - \frac{1}{r^\tau }\frac{C_{\Omega ,\tau }}{\mu ^\tau }\) satisfies

$$\begin{aligned} f_\mu (r)<0 \quad \text {for}\quad r \ge 1\qquad \text {and}\qquad \lim _{r \rightarrow 0^+}f_\mu (r)= -\infty . \end{aligned}$$

Moreover, this function has two zeros \(r_1(\mu ), r_2(\mu )\) with \(0<r_1(\mu )< \frac{1}{\mu }< r_2(\mu )<1\) and

$$\begin{aligned} f_\mu (r)\ge 0 \qquad \text {if and only if}\quad r_1(\mu ) \le r \le r_2(\mu ). \end{aligned}$$

To see this, we write

$$\begin{aligned} f_\mu (r)= \frac{1}{r^\tau }g(r^\tau ) \qquad \text {with}\qquad g(s)=-\frac{s}{\tau } \log s - \frac{C_{\Omega ,\tau }}{\mu ^\tau } \end{aligned}$$

and note that g is strictly concave since \(s \mapsto g'(s)= -\frac{1}{\tau } - \log s\) is strictly decreasing. Consequently, g has at most two zeros, and the same is true for f. Combining this with (7.9) and the fact that

$$\begin{aligned} f(1/\mu ) = \log \mu - C_{\Omega ,\tau } > 0 \end{aligned}$$

since \(\lambda \ge 2 C_{\Omega ,\tau } >C_{\Omega ,\tau }\) by assumption, the claim above follows. From (7.8), we thus obtain the lower bound

$$\begin{aligned}&\int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \bigl \{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau } \bigl \} C_{\Omega ,\tau } \Bigr )_+d\xi \nonumber \\&\quad \ge \mu ^d\, \left| \mathbb S^{d-1} \right| \, \int _{\frac{1}{\mu }}^{r_2(\mu )} \left( -\log r - \frac{1}{r^\tau }\frac{C_{\Omega ,\tau }}{\mu ^\tau } \right) _+ \, r^{d-1}dr. \end{aligned}$$

Next, we claim that

$$\begin{aligned} r_2(\mu ) \ge r_3(\mu ):= e^{\frac{1}{2\tau }\bigl (\sqrt{1- \frac{4\tau C_{\Omega ,\tau }}{\mu ^\tau }}\;-1\bigr )}. \end{aligned}$$

Here we note that \(\frac{4\tau C_{\Omega ,\tau }}{\mu ^\tau }=\frac{4\tau C_{\Omega ,\tau }}{e^{\tau \lambda }} <1\) since \(\lambda \ge 2 C_{\Omega ,\tau }\) by assumption. To see (7.11), we write

$$\begin{aligned} r_3(\mu )= e^{- c_\mu \frac{C_{\Omega ,\tau }}{\mu ^\tau }}\qquad \text {with}\qquad c_\mu = \frac{\mu ^\tau }{2 \tau C_{\Omega ,\tau }}\Bigl (1 - \sqrt{1- \frac{4\tau C_{\Omega ,\tau }}{\mu ^\tau }}\Bigr ), \end{aligned}$$

noting that

$$\begin{aligned} \frac{\tau C_{\Omega ,\tau }}{\mu ^\tau } c_\mu ^2 -c_\mu +1= 0 \end{aligned}$$

and therefore

$$\begin{aligned}&f(r_3(\mu )) = f(e^{-\frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }})=\frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau } - \frac{1}{e^{-\tau \frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }}} \frac{C_{\Omega ,\tau }}{\mu ^\tau }\\&\quad =\frac{C_{\Omega ,\tau }}{\mu ^\tau e^{-\tau \frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }}} \Bigl ( c_\mu e^{-\tau \frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }} - 1\Bigr ) h\ge \frac{C_{\Omega ,\tau }}{\mu ^\tau e^{-\tau \frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }}}\Bigl ( c_\mu \bigl (1 - \tau \frac{c_\mu C_{\Omega ,\tau }}{\mu ^\tau }\bigr )-1\Bigr ) = 0. \end{aligned}$$

This proves (7.11). As a consequence of the inequality \(\sqrt{1-a} \ge 1-\frac{a}{2} -\frac{a^2}{2}\) for \(0 \le a \le 1\), we also have

$$\begin{aligned} r_3(\mu ) \ge e^{- \bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr )} = : r_4(\mu ). \end{aligned}$$


$$\begin{aligned}&\int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \bigl \{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau } \bigl \} C_{\Omega ,\tau } \Bigr )_+d\xi \\&\quad \ge \mu ^d\, \left| \mathbb S^{d-1} \right| \, \int _{\frac{1}{\mu }}^{r_4(\mu )} \left( -\log r - \frac{1}{r^\tau }\frac{C_{\Omega ,\tau }}{\mu ^\tau } \right) _+ \, r^{d-1}dr\\&\quad = \mu ^d\, \left| \mathbb S^{d-1} \right| \, \Bigl [-\frac{r^d}{d} \log r + \frac{1}{d^2}r^d - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}r^{d-\tau } \Bigr ]_{\frac{1}{\mu }}^{r_4(\mu )}\\&\quad = \mu ^d\, \left| \mathbb S^{d-1} \right| \,\Bigl ( \Bigl [-\frac{r_4(\mu )^d}{d} \log r_4(\mu ) + \frac{1}{d^2}r_4(\mu )^d - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}r_4(\mu )^{d-\tau } \Bigr ]\\&\qquad - \Bigl [\frac{\mu ^{-d}}{d} \log \mu + \frac{1}{d^2}\mu ^{-d} - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}\mu ^{\tau -d} \Bigr ]\Bigr ), \end{aligned}$$

which implies that

$$\begin{aligned}&\int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \bigl \{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau } \bigl \} C_{\Omega ,\tau } \Bigr )_+d\xi \\&\quad \ge \mu ^d\, \left| \mathbb S^{d-1} \right| \,\Bigl (\frac{1}{d^2}r_4(\mu )^d - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}r_4(\mu )^{d-\tau }- \frac{\mu ^{-d}}{d} \log \mu - \frac{1}{d^2}\mu ^{-d} \Bigr )\\&\quad = \mu ^d\, \left| \mathbb S^{d-1} \right| \,\Bigl (\frac{1}{d^2}e^{- d\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr )} - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}e^{- (d-\tau )\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr )}\\&\qquad - \frac{\mu ^{-d}}{d} \log \mu - \frac{1}{d^2}\mu ^{-d} \Bigr ). \end{aligned}$$


$$\begin{aligned} e^{- d\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr )} \ge 1- d\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr ) \end{aligned}$$


$$\begin{aligned} e^{- (d-\tau )\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr )} \le 1, \end{aligned}$$

we conclude that

$$\begin{aligned}&\int _{\mathbb R^{d}} \Bigl (\lambda - \log |\xi | - \max \bigl \{ \frac{1}{|\xi |}, \frac{1}{|\xi |^\tau } \bigl \} C_{\Omega ,\tau } \Bigr )_+d\xi \\&\quad \ge \mu ^d\, \frac{\left| \mathbb S^{d-1} \right| }{d^2} \,\Bigl (1- d\bigl (\frac{C_{\Omega ,\tau }}{\mu ^\tau }+\frac{4\tau C_{\Omega ,\tau }^2}{\mu ^{2\tau }}\bigr ) - \frac{C_{\Omega ,\tau }}{\mu ^\tau (d-\tau )}- \mu ^{-d}(d \log \mu + 1) \Bigr )\\&\quad = \frac{\left| B_d \right| }{d} \,\Bigl (\mu ^d - C_{\Omega ,\tau }(d-\frac{1}{d-\tau })\mu ^{d-\tau } - 4d\tau C_{\Omega ,\tau }^2 \mu ^{d-2\tau } - (d \log \mu + 1) \Bigr )\\&\quad = \frac{\left| B_d \right| }{d} \,\Bigl (e^{d \lambda } - \frac{d(d-\tau )-1}{d-\tau } C_{\Omega ,\tau }e^{(d-\tau )\lambda } - 4d\tau C_{\Omega ,\tau }^2 e^{(d-2\tau )\lambda } - (d \lambda + 1) \Bigr ). \end{aligned}$$

Combining the last estimate with (7.7), we get the asserted lower bound.

Appendix: Note on a bound for Bessel functions

The following elementary bound might be known but seems hard to find in this form.

Lemma 8.1

For \(\nu \ge \sqrt{3}-2\) and \(0 \le x \le 2 \sqrt{2(\nu +2)}\) we have

$$\begin{aligned} |J_\nu (x)| \le \frac{x^\nu }{2^\nu \Gamma (\nu +1)}. \end{aligned}$$


We use the representation

$$\begin{aligned} J_\nu (x)= \Bigl (\frac{x}{2}\Bigr )^{\nu } \sum _{m=0}^\infty \frac{(-1)^m}{m! \Gamma (m+\nu + 1)} \Bigl (\frac{x}{2}\Bigr )^{2m}. \end{aligned}$$

For \(0 \le x \le 2 \sqrt{2(\nu +2)}\) and \(m \ge 1\), we have

$$\begin{aligned} \Bigl (\frac{x}{2}\Bigr )^{2} \le (m+1)(m+\nu +1) = \frac{(m+1) \Gamma (m+\nu + 2)}{\Gamma (m+\nu + 1)} \end{aligned}$$

and therefore

$$\begin{aligned} \frac{\Gamma (\nu +1)}{(m+1)! \Gamma (m+\nu + 2)} \Bigl (\frac{x}{2}\Bigr )^{2(m+1)} \le \frac{\Gamma (\nu +1)}{m! \Gamma (m+\nu + 1)} \Bigl (\frac{x}{2}\Bigr )^{2m}. \end{aligned}$$


$$\begin{aligned} J_\nu (x)&= \frac{x^\nu }{2^\nu \Gamma (\nu +1)}\Bigl [1 + \sum _{m=1}^\infty \frac{(-1)^m \Gamma (\nu +1)}{m! \Gamma (m+\nu + 1)} \Bigl (\frac{x}{2}\Bigr )^{2m}\Bigr ]\le \frac{x^\nu }{2^\nu \Gamma (\nu +1)}. \end{aligned}$$

From (8.1) we also deduce that

$$\begin{aligned} J_\nu (x)&\ge \frac{x^\nu }{2^\nu \Gamma (\nu +1)} \Bigl [1 - \frac{\Gamma (\nu +1)}{\Gamma (\nu + 2)} \Bigl (\frac{x}{2}\Bigr )^{2}+ \frac{\Gamma (\nu +1)}{2\Gamma (\nu + 3)} \Bigl (\frac{x}{2}\Bigr )^{4}- \frac{\Gamma (\nu +1)}{6\Gamma (\nu + 4)} \Bigl (\frac{x}{2}\Bigr )^{6}\Bigr ]\\&= \frac{x^\nu }{2^\nu \Gamma (\nu +1)} \Bigl [1-\frac{1}{\nu + 1}f\bigl (\bigl (\frac{x}{2}\bigr )^{2} \bigr )\Bigr ] \end{aligned}$$

with \(f: {\mathbb R}\rightarrow {\mathbb R}\) given by \(f(t)= t - \frac{t^2}{2(\nu +2)}+ \frac{t^3}{6(\nu +2)(\nu +3)}\). Since

$$\begin{aligned} f'(t)= 1- \frac{t}{\nu +2} + \frac{t^2}{2(\nu +2)(\nu +3)}, \qquad \text {and}\qquad f''(t)= \frac{1}{\nu +2}\bigl (\frac{t}{\nu +3}- 1\bigr ) \end{aligned}$$

we have

$$\begin{aligned} f'(t) \ge f'(\nu +3) = 1- \frac{\nu +3 }{\nu +2} + \frac{\nu +3}{2(\nu +2)} = 1 - \frac{1}{2}\frac{\nu +3 }{\nu +2} \ge 0 \quad \text {for}\quad t \in {\mathbb R}\hbox { if }\nu \ge -1 \end{aligned}$$

and therefore

$$\begin{aligned} f(t) \le f(2(\nu +2))= 2(\nu +2) - \frac{[2(\nu +2)]^2}{2(\nu +2)}+ \frac{[2(\nu +2)]^3}{6(\nu +2)(\nu +3)}= \frac{4(\nu +2)^2}{3(\nu +3)} \end{aligned}$$

for \(t \le 2(\nu +2)\) if \(\nu \ge -1\). Since \(\frac{4(\nu +2)^2}{3(\nu +3)} \le \frac{2}{\nu +1}\) for \(\nu \ge \sqrt{3}-2\), we conclude that

$$\begin{aligned} J_\nu (x) \ge \frac{x^\nu }{2^\nu \Gamma (\nu +1)} \Bigl [1-\frac{1}{\nu + 1}f\bigl (\bigl (\frac{x}{2}\bigr )^{2} \bigr )\Bigr ]\ge - \frac{x^\nu }{2^\nu \Gamma (\nu +1)}. \end{aligned}$$

for \(\nu \ge \sqrt{3}-2\) and \(0 \le x \le 2 \sqrt{2(\nu +2)}\). The claim thus follows. \(\square \)