Abstract multiplicity results for (p, q)-Laplace equations with two parameters

Abstract

We investigate the existence and multiplicity of abstract weak solutions of the equation $-\Delta _p u -\Delta _q u=\alpha |u|^{p-2}u+\beta |u|^{q-2}u$ in a bounded domain under zero Dirichlet boundary conditions, assuming $1<q<p$ and $\alpha ,\beta \in {\mathbb {R}}$. We determine three generally different ranges of parameters $\alpha$ and $\beta$ for which the problem possesses a given number of distinct pairs of solutions with a prescribed sign of energy. As auxiliary results, which are also of independent interest, we provide alternative characterizations of variational eigenvalues of the q-Laplacian using smaller and larger constraint sets than in the standard minimax definition.

Appendices

Characterizations of $\lambda _k(q)$

In this section, we discuss alternative characterizations of variational eigenvalues $\lambda _k(q)$ using smaller and larger constraint sets than in the standard minimax definition (1.1). The results of this section require less restrictive assumptions on a bounded domain than that additionally imposed in Sect. 1. Because of this, we will provide explicit assumptions on $\Omega$ in each statement.

Let $q, r>1$. Recall the definition (1.1) of $\lambda _k(q)$:

$$\begin{aligned} \lambda _k(q) = \inf \left\{ \max _{u\in A} \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q}:~ A\in \Sigma _k(q) \right\} , \end{aligned}$$

and consider a related object $\lambda _k(q;r)$ defined as

$$\begin{aligned} \lambda _k(q;r) = \inf \left\{ \sup _{u\in A} \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q}:~ A\in {\Sigma }_k(r) ~\text {and}~ A \subset W^{1,q} \right\} , \end{aligned}$$

(A.1)

where $\Sigma _k(r)$ is given by (1.2), i.e.,

$$\begin{aligned} \Sigma _k(r) =\left\{ \, A\subset W_0^{1,r} \setminus \{0\}:~ A\ \mathrm{is\ symmetric,\ compact\ in}\ W^{1,r}_0,\ \textrm{and}\ \gamma (A;W_0^{1,r})\ge k\,\right\} . \end{aligned}$$

Throughout this section, we use the extended notation $\gamma (A;X)$ for the Krasnoselskii genus to emphasize its dependence on a topological vector space X with respect to which the continuity of an odd mapping h in the definition (1.3) of the genus is understood.

Clearly, $\lambda _k(q;q)=\lambda _k(q)$ for any $q>1$, while $\lambda _k(q;r)$ is defined using different constraints than $\lambda _k(q)$ whenever $q \ne r$. The constraint $A \subset W^{1,q}$ in (A.1) is trivial in the case $q<r$ since $W_0^{1,r}\subset W_0^{1,q}$, but it is required when $q>r$, in order for the Rayleigh quotient $\Vert \nabla u\Vert _q^q/\Vert u\Vert _q^q$ to be defined over $A\in {\Sigma }_k(r)$.

We are interested in the relation between $\lambda _k(q;r)$ and $\lambda _k(q)$. Let us start with the following observation on the Krasnoselskii genus.

Lemma A.1

Let X, Y be topological vector spaces. Let $A \subset X {\setminus } \{0\}$ be symmetric and closed. Let $\pi : A \rightarrow Y$ be odd and continuous. Let $\pi (A) \subset Y \setminus \{0\}$ be (symmetric and) closed. Then $\gamma (A;X) \le \gamma (\pi (A);Y)$. If, in addition, the inverse mapping $\pi ^{-1}: \pi (A) \rightarrow A$ is continuous (and hence $\pi (A)$ and A are homeomorphic), then $\gamma (A;X) = \gamma (\pi (A);Y)$.

Proof

If $\gamma (\pi (A);Y) = +\infty$, then the inequality $\gamma (A;X) \le \gamma (\pi (A);Y)$ is trivial. Assume that $k=\gamma (\pi (A);Y) < +\infty$. By the definition (1.3) of $\gamma (\pi (A);Y)$, there exists an odd mapping $h \in C(\pi (A);{\mathbb {R}}^k {\setminus }\{0\})$. Since $\pi$ is odd and continuous, we see that $h \circ \pi$ is odd and $h \circ \pi \in C(A;{\mathbb {R}}^k {\setminus }\{0\})$. Therefore, $h \circ \pi$ is admissible for the definition (1.3) of $\gamma (A;X)$, which implies that $\gamma (A;X) \le \gamma (\pi (A);Y)$. If, in addition, $\pi ^{-1}: \pi (A) \rightarrow A$ is continuous, then the same arguments give $\gamma (\pi (A);Y) \le \gamma (A;X)$, which yields the equality between genuses. $\square$

Remark A.2

Evidently, if the set A in Lemma A.1 is compact in X, then $\pi (A)$ is compact in Y. We also refer to [31, Proposition 7.5: $2^\circ$] or [33, Chapter II, Proposition 5.4 ($4^\circ$)] for the statement of Lemma A.1 in the case when X and Y coincide.

In view of the continuity of the canonical embedding $i: W_0^{1,r}\rightarrow W_0^{1,q}$ for $1<q<r$ defined as $i(u)=u$, Lemma A.1 implies that

$$\begin{aligned} k \le \gamma (A;W_0^{1,r}) \le \gamma (A;W_0^{1,q}) \quad \text {for any}~ A \in \Sigma _k(r), \end{aligned}$$

which yields $A \in \Sigma _k(q)$. A similar relation holds in the case $q>r>1$. This leads to the following remark.

Remark A.3

The following assertions hold:

• (i) If $1<q<r$, then $\Sigma _k(q) \supset \Sigma _k(r)$ and hence $\lambda _k(q) \le \lambda _k(q;r)$.

• (ii) If $q>r>1$, then $\Sigma _k(q) \subset \Sigma _k(r)$ and hence $\lambda _k(q) \ge \lambda _k(q;r)$.

1.1 The case $q<r$

First, we show that $\lambda _k(q) = \lambda _k(q;r)$ whenever $q<r$, which therefore provides an alternative characterization of $\lambda _k(q)$. In the case $k=1$, this claim is simple thanks to the density of $C_0^\infty (\Omega )$ in both $W_0^{1,r}$ and $W_0^{1,q}$, and our aim is to develop this approach for higher indices k.

Lemma A.4

Let $\Omega \subset {\mathbb {R}}^N$ be a bounded domain, $N \ge 1$. Let $1<q<r$. Then ${\lambda }_k(q) = \lambda _k(q;r)$ for all $k \in {\mathbb {N}}$.

Proof

We know from Remark A.3 (i) that ${\lambda }_k(q)\le \lambda _k(q;r)$ for all k. Suppose, by contradiction, that ${\lambda }_k(q) < \lambda _k(q;r)$ for some k. Consequently, there exists $A \in \Sigma _k(q)$ such that

$$\begin{aligned} \lambda _k(q) \le \max _{u\in A} \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q} < \lambda _k(q;r). \end{aligned}$$

(A.2)

To prove the lemma, we show the existence of an element of $\Sigma _k(r)$ sufficiently close to A in the norm of $W^{1,q}$, which will give a contradiction to (A.2).

Let us take any $\varepsilon \in (0,\text {dist}_q(A,0)/2)$, where

$$\begin{aligned} \text {dist}_q(A,0) = \min _{u \in A}\Vert \nabla u\Vert _q > 0 \end{aligned}$$

since A is compact in $W_0^{1,q}$ and does not contain 0. Let $B_\varepsilon ^q(u)$ be an open ball in $W_0^{1,q}$ of radius $\varepsilon >0$ centered at u. Thanks to the compactness of A in $W_0^{1,q}$, we can extract a finite symmetric subcover $\{B_\varepsilon ^q(u_n) \cup B_\varepsilon ^q(-u_n)\}_{n=1}^K$ from a cover $\cup _{u \in A}B_\varepsilon ^q(u)$, where $K= K(\varepsilon ) \ge 1$.

For every $n \in \{1,\dots ,K\}$ we choose some $v_n \in C_0^\infty (\Omega ) \cap B_\varepsilon ^q(u_n)$. In particular, we have $\Vert \nabla (u - v_n)\Vert _q < 2\varepsilon$ for any $u \in B_\varepsilon ^q(u_n)$. Consider a closed finite dimensional linear subspace of $W_0^{1,q}$ spanned by $\{v_n\}_{n=1}^K$:

$$\begin{aligned} V = V(\varepsilon ) = \text {span} \{v_1,\dots ,v_K\}, \end{aligned}$$

and define the metric projection $P_\varepsilon : W_0^{1,q} \mapsto V$ in the standard way:

$$\begin{aligned} \Vert \nabla (u-P_\varepsilon (u))\Vert _q = \inf _{v \in V}\Vert \nabla (u-v)\Vert _q, \quad u \in W_0^{1,q}. \end{aligned}$$

Our aim is to prove that $P_\varepsilon (A) \in \Sigma _k(r)$, after a possible decrease in the value of $\varepsilon \in (0,\text {dist}_q(A,0)/2)$. The operator $P_\varepsilon$ is well defined in the sense that $P_\varepsilon (u)$ exists and unique for any $u \in W_0^{1,q}$, see, e.g., [32, Corollary 3.4 $(1^\circ )$, $(3^\circ )$, p. 111]. Moreover, $P_\varepsilon$ is a continuous mapping, see, e.g., [32, Theorem 5.4, p. 251]. It is also clear that $P_\varepsilon$ is odd, i.e., $P_\varepsilon (-u)=-P_\varepsilon (u)$. As a consequence of the last two facts, we have $\gamma (P_\varepsilon (A);W_0^{1,q}) \ge k$, see, e.g., Lemma A.1. The continuity of $P_\varepsilon$ and the compactness of A in $W_0^{1,q}$ imply that $P_\varepsilon (A)$ is compact in $W_0^{1,q}$. Moreover, $0 \not \in P_\varepsilon (A)$ thanks to the upper bound on $\varepsilon$. It is clear from the regularity of the basis elements of V that V is a closed finite dimensional linear subspace of $W_0^{1,r}$, and hence $P_\varepsilon (A) \subset W_0^{1,r}$. Since in the finite dimensional space V all norms are equivalent, we conclude that $P_\varepsilon (A)$ is compact in $W_0^{1,r}$. Applying Lemma A.1 with the inclusion mapping $i:V \subset W_0^{1,q}\rightarrow W_0^{1,r}$ defined as $i(u)=u$, we get $\gamma (P_\varepsilon (A);W_0^{1,r}) = \gamma (P_\varepsilon (A);W_0^{1,q}) \ge k$, which finishes the proof that $P_\varepsilon (A) \in \Sigma _k(r)$.

Finally, let us explicitly obtain an upper bound for the Rayleigh quotient for $\lambda _k(q;r)$ to get a contradiction to (A.2). For any $u \in A$ there exists $n \in \{1,\dots ,K\}$ such that $u \in B_\varepsilon ^q(u_n)$. By the triangle inequality and the definition of $P_\varepsilon (u)$, we have

$$\begin{aligned} \left| \Vert \nabla u\Vert _q - \Vert \nabla P_\varepsilon (u)\Vert _q \right| \le \Vert \nabla (u-P_\varepsilon (u))\Vert _q = \inf _{v \in V}\Vert \nabla (u-v)\Vert _q \le \Vert \nabla (u-v_n)\Vert _q < 2\varepsilon , \nonumber \\ \end{aligned}$$

(A.3)

and hence

$$\begin{aligned} \Vert \nabla P_\varepsilon (u)\Vert _q \le \Vert \nabla u\Vert _q + 2\varepsilon \quad \text {for any}~ u \in A. \end{aligned}$$

(A.4)

The inequality (A.3) and the Poincaré inequality give

$$\begin{aligned} \left| \Vert u\Vert _q - \Vert P_\varepsilon (u)\Vert _q\right| \le \Vert u-P_\varepsilon (u)\Vert _q \le \lambda _1^{-1/q}(q) \, \Vert \nabla (u-P_\varepsilon (u))\Vert _q \le 2 \lambda _1^{-1/q}(q) \varepsilon \nonumber \\ \end{aligned}$$

(A.5)

for any $u \in A$, which yields

$$\begin{aligned} 1 \le \frac{\Vert P_\varepsilon (u)\Vert _q}{\Vert u\Vert _q} + \frac{2 \lambda _1^{-1/q}(q)\varepsilon }{\Vert u\Vert _q}. \end{aligned}$$

(A.6)

Combining (A.4) and (A.6), we get

$$\begin{aligned} \Vert \nabla P_\varepsilon (u)\Vert _q \le \Vert \nabla u\Vert _q +2\varepsilon \le \Vert \nabla u\Vert _q \frac{\Vert P_\varepsilon (u)\Vert _q}{\Vert u\Vert _q} + \Vert \nabla u\Vert _q \frac{2 \lambda _1^{-1/q}(q)\varepsilon }{\Vert u\Vert _q} +2\varepsilon . \end{aligned}$$

Recalling that $0 \not \in P_\varepsilon (A)$ and dividing by $\Vert P_\varepsilon (u)\Vert _q$, we arrive at

$$\begin{aligned} \frac{\Vert \nabla P_\varepsilon (u)\Vert _q}{\Vert P_\varepsilon (u)\Vert _q} \le \frac{\Vert \nabla u\Vert _q}{\Vert u\Vert _q} + \frac{\Vert \nabla u\Vert _q}{\Vert u\Vert _q} \cdot \frac{2 \lambda _1^{-1/q}(q)\varepsilon }{\Vert P_\varepsilon (u)\Vert _q} + \frac{2\varepsilon }{\Vert P_\varepsilon (u)\Vert _q}. \end{aligned}$$

(A.7)

Moreover, we also deduce from (A.5) that

$$\begin{aligned} \Vert P_\varepsilon (u)\Vert _q \ge \Vert u\Vert _q - 2 \lambda _1^{-1/q}(q) \varepsilon \ge \min _{v \in A} \Vert v\Vert _q - 2 \lambda _1^{-1/q}(q) \varepsilon . \end{aligned}$$

(A.8)

Combining (A.7) and (A.8) and taking $\varepsilon >0$ smaller if necessary, we conclude that

$$\begin{aligned} \lambda _k^{1/q}(q;r) \le \max _{u \in A}\frac{\Vert \nabla P_\varepsilon (u)\Vert _q}{\Vert P_\varepsilon (u)\Vert _q} \le \max _{u \in A} \frac{\Vert \nabla u\Vert _q}{\Vert u\Vert _q} + O(\varepsilon ) < \lambda _k^{1/q}(q;r), \end{aligned}$$

where the last inequality is given by our contradictory assumption (A.2). A contradiction. $\square$

Remark A.5

Lemma A.4 is needed for the proofs of Theorems 1.2 and 1.4 (see, more explicitly, the proof of Lemma 2.10), cf. [41].

1.2 The case $q>r$

Let us now discuss the relation between $\lambda _k(q;r)$ and ${\lambda }_k(q)$ in the case $q<r$, which is converse to that considered above. Although $\lambda _k(q;r) \le {\lambda }_k(q)$ for all k (see Remark A.3 (ii)), the precise relation between $\lambda _k(q;r)$ and ${\lambda }_k(q)$ heavily depends on the regularity of $\Omega$ and it is closely connected to the problem of continuity of the mapping $q \mapsto \lambda _k(q)$ from the left, see [26, Section 7], [13, 19, 28], and references therein. We provide two opposite results in this direction.

Lemma A.6

Let $N \ge 2$ and $1<r<q$. Let $\Omega \subset {\mathbb {R}}^N$ be a bounded domain such that

$$\begin{aligned} W_0^{1,q_-}:= W^{1,q} \cap \bigcap _{1<s<q} W_0^{1,s} \ne W_0^{1,q}. \end{aligned}$$

(A.9)

Then $\lambda _k(q;r) < {\lambda }_k(q)$ for any $k \in {\mathbb {N}}$.

Proof

It is not hard to observe that $W_0^{1,q_-}$ is a closed vector subspace of $W^{1,q}$ satisfying $W_0^{1,q} \subset W_0^{1,q_-} \subset W_0^{1,r}$ for any $r<q$, see, e.g., [13, Proposition 2.1]. Define

$$\begin{aligned} {\underline{\lambda }}_k(q) = \inf \left\{ \max _{u\in A} \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q}:~ A\in {\Sigma }_k(q_-) \right\} , \end{aligned}$$

where

$$\begin{aligned} \Sigma _k(q_-) =\left\{ \, A\subset W_0^{1,q_-} \setminus \{0\}:~ A\ \mathrm{is\ symmetric,\ compact\ in}\ W^{1,q},\ \textrm{and}\ \gamma (A;W^{1,q})\ge k\,\right\} . \end{aligned}$$

Using the continuous embedding $i_1: W_0^{1,q}\rightarrow W^{1,q}$ defined as $i_1(u)=u$, we apply Lemma A.1 to deduce that $\Sigma _k(q) \subset \Sigma _k(q_-)$. In the same way, Lemma A.1 with the continuous embedding $i_2: W^{1,q} \rightarrow W^{1,r}$ defined as $i_2(u)=u$ yields $\gamma (A;W^{1,r}) \ge \gamma (A;W^{1,q}) \ge k$ for any $A \in \Sigma _k(q_-)$. Moreover, any such A is symmetric and compact in $W_0^{1,r}$. Since the topologies induced by the norms of $W^{1,r}$ and $W_0^{1,r}$ are equivalent in $W_0^{1,r}$, we conclude from Lemma A.1 that $\gamma (A;W^{1,r})=\gamma (A;W_0^{1,r})$ for any $A \in \Sigma _k(q_-)$. Therefore, $\Sigma _k(q_-) \subset \Sigma _k(r)$. Consequently,

$$\begin{aligned} \lambda _k(q;r) \le {\underline{\lambda }}_k(q) \le \lambda _k(q). \end{aligned}$$

(A.10)

It follows from [13, Theorem 6.1 and Corollary 6.2] that ${\underline{\lambda }}_k(q) = \lambda _k(q)$ for $k \ge 2$ if and only if ${\underline{\lambda }}_1(q) = \lambda _1(q)$. However, our assumption (A.9) is equivalent to ${\underline{\lambda }}_1(q) < \lambda _1(q)$, see [13, Theorem 4.1 (c), (d)]. Thus, we conclude from (A.10) that $\lambda _k(q;r) \le {\underline{\lambda }}_k(q) < \lambda _k(q)$ for any $k \in {\mathbb {N}}$. $\square$

Remark A.7

A domain $\Omega$ satisfying the assumption (A.9), or, equivalently, $\lim _{s \nearrow q} \lambda _1(s) < \lambda _1(q)$, was first constructed in [26, Section 7]. We refer to [13, 19] for further developments and for characterizations of the family of such domains.

Lemma A.8

Let $N \ge 1$ and $1<r<q$. Let $\Omega \subset {\mathbb {R}}^N$ be a bounded domain such that

$$\begin{aligned} W^{1,q} \cap W_0^{1,r}= W_0^{1,q}. \end{aligned}$$

(A.11)

Then ${\lambda }_k(q) = \lambda _k(q;r)$ for all $k \in {\mathbb {N}}$.

Proof

Let us take any $\varepsilon >0$ and any $A_\varepsilon \in \Sigma _k(r)$ such that $A_\varepsilon \subset W^{1,q}$ and

$$\begin{aligned} \sup _{u\in A_\varepsilon } \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q} \le \lambda _k(q;r) + \varepsilon . \end{aligned}$$

(A.12)

It is known from [8] (see, more precisely, [8, Eq. (2.3) and Corollary 3.6]) that $\lambda _k(q)$ can be equivalently defined as

$$\begin{aligned} \lambda _k(q) = \min \left\{ \sup _{u\in A} \frac{\Vert \nabla u\Vert _q^q}{\Vert u\Vert _q^q}:~ A\in {\mathcal {G}}_k(q) \right\} , \end{aligned}$$

(A.13)

where

$$\begin{aligned}&{\mathcal {G}}_k(q) \\&=\left\{ \, A\subset W_0^{1,q} \setminus \{0\}:~ A\ \mathrm{is\ symmetric,\ closed \ and \ bounded\ in}\ W^{1,q}_0,\ \textrm{and}\ \gamma (A; L^q)\ge k\,\right\} . \end{aligned}$$

Our aim is to show that $A_\varepsilon \in {\mathcal {G}}_k(q)$, i.e., $A_\varepsilon$ is admissible for the definition (A.13) of $\lambda _k(q)$. If it is true, then $\lambda _k(q) \le \lambda _k(q;r)+\varepsilon$ in view of (A.12). Since $\varepsilon >0$ is chosen arbitrarily, we get $\lambda _k(q) \le \lambda _k(q;r)$. On the other hand, we have $\lambda _k(q;r) \le {\lambda }_k(q)$ by Remark A.3 (ii), which gives the desired conclusion of the lemma. (If we suppose that for any $\varepsilon >0$ there exists $A_\varepsilon$ which is compact in $W_0^{1,q}$, then the above argument applies to the original definition (1.1) of $\lambda _k(q)$ instead of (A.13), which would give the result in a simpler way.)

Therefore, let us justify that $A_\varepsilon \in {\mathcal {G}}_k(q)$. By the assumption (A.11), we have $A_\varepsilon \subset W_0^{1,q}$. Since $r<q$ and $A_\varepsilon \subset W_0^{1,r}{\setminus } \{0\}$ is compact in $W_0^{1,r}$, we get $A_\varepsilon \subset W_0^{1,q}{\setminus } \{0\}$. The set $A_\varepsilon$ is bounded in $W_0^{1,q}$. Indeed, we have from (A.12) that

$$\begin{aligned} \Vert \nabla v\Vert _q^q \le (\lambda _k(q;r) + \varepsilon ) \Vert v\Vert _q^q \quad \text {for any}~ v \in A_\varepsilon , \end{aligned}$$

and hence [35, Lemma 9] and the Poincaré inequality give the existence of a constant $C>0$ such that

$$\begin{aligned} \Vert \nabla v\Vert _q \le C \Vert v\Vert _r \le C \lambda _1^{-1/r}(r) \Vert \nabla v\Vert _r \le C \lambda _1^{-1/r}(r) \, \max _{u \in A_\varepsilon } \Vert \nabla u\Vert _r <+\infty \quad \text {for any}~ v \in A_\varepsilon . \nonumber \\ \end{aligned}$$

(A.14)

Moreover, $A_\varepsilon$ is closed in $W_0^{1,q}$. To show this, let us take any sequence $\{u_n\} \subset A_\varepsilon$ which strongly converges in $W_0^{1,q}$ to a function $u_0 \in W_0^{1,q}$. By the compactness of $A_\varepsilon$ in $W_0^{1,r}$ and the assumption $r<q$, we see that $u_0 \in W_0^{1,r}{\setminus } \{0\}$ and $u_n \rightarrow u_0$ strongly in $W_0^{1,r}$, and so $u_0 \in A_\varepsilon$, which gives the desired closedness of $A_\varepsilon$ in $W_0^{1,q}$.

It remains to prove that $\gamma (A_\varepsilon ; L^q)\ge k$. Consider the mapping $i: A_\varepsilon \subset W_0^{1,r}\rightarrow L^q$ defined as $i(u)=u$. This mapping is continuous. (This claim is trivial when q does not exceed the critical Sobolev exponent $r^*$.) Indeed, let us take any sequence $\{u_n\} \subset A_\varepsilon$ which converges to some $u_0 \in A_\varepsilon$ strongly in $W_0^{1,r}$. Since $A_\varepsilon$ is bounded in $W_0^{1,q}$, we see that any subsequence of $\{u_n\}$ has a subsubsequence which converges to $u_0$ weakly in $W_0^{1,q}$ and hence strongly in $L^q$. Therefore, the whole sequence $\{u_n\}$ also converges to $u_0$ strongly in $L^q$, and hence i is continuous. Since i is odd, we apply Lemma A.1 to get $\gamma (A_\varepsilon ; L^q)\ge \gamma (A_\varepsilon ; W_0^{1,r}) \ge k$.

Thus, we conclude that $A_\varepsilon \in {\mathcal {G}}_k(q)$, which finishes the proof. $\square$

Remark A.9

We refer to [19, Sections 3 and 4] for several assumptions equivalent to/sufficient for (A.11). In particular, if $\Omega$ is Lipschitz (in the case $N \ge 2$), then (A.11) holds, see [19, Remarks (ii), p. 252]. Moreover, it is clear that (A.11) is satisfied when $N=1$.

Relation between $\lambda _k(q)$ and $\beta _*$

The aim of this section is to show that, for an appropriate domain $\Omega \subset {\mathbb {R}}^N$ with $N \ge 2$, the assumptions (ii), (iii), (iv) of Theorem 1.2 can be satisfied for $k \ge 2$. Throughout this section, in order to represent the dependence of quantities on a domain $\Omega$, we will use the expanded notation $\beta _*(\Omega )$, $\beta _*(\alpha ;\Omega )$, $\lambda _k(q;\Omega )$ for $\beta _*$, $\beta _*(\alpha )$, $\lambda _k(q)$, respectively. In particular, recall the definition of $\beta _*(\Omega )$, see (1.6):

$$\begin{aligned} \beta _*(\Omega ) = \frac{\Vert \nabla \varphi _p\Vert _q^q}{\Vert \varphi _p\Vert _q^q}. \end{aligned}$$

Lemma B.1

Let $1<q<p$, $N \ge 2$, and $k \ge 2$. Then there exists a bounded $C^2$-smooth domain $\Omega \subset {\mathbb {R}}^N$ such that $\lambda _k(q;\Omega ) < \beta _*(\Omega )$.

Proof

Let us construct a beads-type domain $\Omega _\varepsilon$ as follows. Take k points in ${\mathbb {R}}^N$ lying on the $x_1$-axis:

$$\begin{aligned} z_j = (j, 0, \dots , 0), \quad j = 1,\dots ,k. \end{aligned}$$

Denote by $\Omega _0$ the union over j of open balls $B_r(z_j)$ of radius $r \in (0,1/2)$ centered at $z_j$. Thanks to the choice of r, these balls are disjoint. Let us now consider a domain $\Omega _\varepsilon$ as the union of $\Omega _0$ with a set of $k-1$ thin cylindrical type channels $T_{j,\varepsilon }$ of maximal width $\varepsilon >0$ subsequently connecting the balls $B_r(z_j)$ and $B_r(z_{j+1})$ in a $C^2$-smooth way, so that $\Omega _\varepsilon$ is of class $C^2$ and $\Omega _{\varepsilon _1} \supset \Omega _{\varepsilon _2}$ provided $\varepsilon _1> \varepsilon _2>0$, see Fig. 3. For instance, one can take

$$\begin{aligned} T_{j,\varepsilon } = \{ (x_1,\dots ,x_N) \in {\mathbb {R}}^N:~ j+r-\varepsilon< x_1< j+1-r+\varepsilon ,~ x_2^2+\dots +x_{N}^2 < g_\varepsilon ^2(x_1) \}, \end{aligned}$$

where $\{g_\varepsilon \}$ is an appropriate family of smooth positive functions in the interval $[1,k+1]$ satisfying $\max _{s \in [1,k+1]}g_\varepsilon (s) \rightarrow 0$ as $\varepsilon \rightarrow 0$, and $g_{\varepsilon _1} > g_{\varepsilon _2}$ provided $\varepsilon _1> \varepsilon _2>0$, so that $T_{j,\varepsilon _1} \supset T_{j,\varepsilon _2}$.

Fig. 3

A possible shape of $\Omega _\varepsilon$ for $k=2$ and some $\varepsilon >0$

Evidently, we have $\lambda _1(q;B_r(z_i)) = \lambda _1(q;B_r(z_j))$ for any $i,j \in \{1,\dots ,k\}$, and when there is no ambiguity, we denote, for short,

$$\begin{aligned} \lambda _1(q;B_r) = \lambda _1(q;B_r(z_j)). \end{aligned}$$

Our aim is to prove that $\lambda _k(q; \Omega _\varepsilon ) < \beta _*(\Omega _\varepsilon )$ for any sufficiently small $\varepsilon >0$. First, let us show that

$$\begin{aligned} \lambda _1(q;\Omega _\varepsilon ) \le \lambda _2(q;\Omega _\varepsilon ) \le \dots \le \lambda _k(q;\Omega _\varepsilon ) \le \lambda _1(q;B_r). \end{aligned}$$

(B.1)

For this purpose, denote by $\varphi _{q,j} \in W_0^{1,q}(B_r(z_j)) \cap C^1(\overline{B_r(z_j)})$ the first eigenfunction corresponding to $\lambda _1(q;B_r(z_j))$. Extending $\varphi _{q,j}$ by zero outside of $B_r(z_j)$, we have $\varphi _{q,j} \in W_0^{1,q}(\Omega _\varepsilon )$. Let us consider the following subset of $W_0^{1,q}(\Omega _\varepsilon )$:

$$\begin{aligned} A = \{a_1 \varphi _{q,1} + \ldots + a_k \varphi _{q,k}:~ a_1,\dots ,a_k \in {\mathbb {R}},~ |a_1|^q + \ldots + |a_k|^q = 1\}. \end{aligned}$$

It is clear that A is symmetric and compact. Moreover, the mapping

$$\begin{aligned} a_1 \varphi _{q,1} + \ldots + a_k \varphi _{q,k} \mapsto (a_1,\dots ,a_k) \end{aligned}$$

is continuous and odd, and hence $\gamma (A;W_0^{1,q}) = k$, see [31, Proposition 7.7]. Therefore, A is an admissible set for the definition (1.1) of $\lambda _k(q;\Omega _\varepsilon )$. Since $\varphi _{q,i}$ and $\varphi _{q,j}$ have disjoint supports and equal norms when $i \ne j$, and

$$\begin{aligned} \int _{\Omega _\varepsilon }|\nabla \varphi _{q,j}|^q \,dx = \lambda _1(q;B_r) \int _{\Omega _\varepsilon }|\varphi _{q,j}|^q \,dx, \quad j =1,\dots ,k, \end{aligned}$$

we have

$$\begin{aligned}&\lambda _k(q;\Omega _\varepsilon ) \\&\quad \le \max _{u \in A} \frac{\int _{\Omega _\varepsilon }|\nabla u|^q \,dx}{\int _{\Omega _\varepsilon }|u|^q \,dx} = \max \left\{ \frac{\int _{\Omega _\varepsilon }|\nabla (a_1 \varphi _{q,1} + \ldots + a_k \varphi _{q,k})|^q \,dx}{\int _{\Omega _\varepsilon }|a_1 \varphi _{q,1} + \ldots + a_k \varphi _{q,k}|^q \,dx} : |a_1|^q + \ldots + |a_k|^q = 1 \right\} \\&\quad = \max \left\{ \frac{|a_1|^q \int _{\Omega _\varepsilon }|\nabla \varphi _{q,1}|^q \,dx + \ldots + |a_k|^q \int _{\Omega _\varepsilon }|\nabla \varphi _{q,k}|^q \,dx}{|a_1|^q \int _{\Omega _\varepsilon }|\varphi _{q,1}|^q \,dx + \ldots + |a_k|^q \int _{\Omega _\varepsilon }|\varphi _{q,k}|^q \,dx} : |a_1|^q + \ldots + |a_k|^q = 1 \right\} = \lambda _1(q;B_r), \end{aligned}$$

which establishes the chain of inequalities (B.1). Exactly the same construction but in the space $W_0^{1,p}(\Omega _0)$ instead of $W_0^{1,q}(\Omega _\varepsilon )$ shows that

$$\begin{aligned} \lambda _1(p;\Omega _0) = \lambda _2(p;\Omega _0) = \dots = \lambda _k(p;\Omega _0) = \lambda _1(p;B_r). \end{aligned}$$

(B.2)

As it can be seen from (B.1), in order to prove the lemma, it is sufficient to justify that

$$\begin{aligned} \lambda _1(q;B_r) < \beta _*(\Omega _\varepsilon ) \quad \text {for any sufficiently small}~ \varepsilon >0. \end{aligned}$$

(B.3)

Hereinafter, we assume that $j \in \{1,\dots ,k\}$ is fixed and we denote $B_r:= B_r(z_j)$, for brevity. We will prove (B.3) by noting that $\lambda _1(q;B_r) < \beta _*(B_r)$ (see Sect. 1.1) and showing that

$$\begin{aligned} \beta _*(B_r) = \beta _*(\Omega _0) \le \beta _*(\Omega _\varepsilon ) - o(1). \end{aligned}$$

Let us recall from Sect. 1.1 that $\beta _*(\Omega _\varepsilon )$ for any $\varepsilon \ge 0$ can be characterized as

$$\begin{aligned} \beta _*(\Omega _\varepsilon )&= \beta _*(\lambda _1(p;\Omega _\varepsilon ); \Omega _\varepsilon ) \nonumber \\&= \inf \Bigg \{ \frac{\int _{\Omega _\varepsilon }|\nabla u|^q \,dx}{\int _{\Omega _\varepsilon }|u|^q \,dx}: ~ u \in W_0^{1,p}(\Omega _\varepsilon ) \setminus \{0\},\int _{\Omega _\varepsilon }|\nabla u|^p \,dx - \lambda _1(p;\Omega _\varepsilon ) \int _{\Omega _\varepsilon }|u|^p \,dx \le 0 \Bigg \}. \end{aligned}$$

(B.4)

In particular, taking the first eigenfunction $\varphi _{p,j}$ corresponding to $\lambda _1(p;B_r)$ as a test function for (B.4) with $\varepsilon =0$ and using (B.2), it is not hard to deduce that $\beta _*(\Omega _0) = \beta _*(B_r)$.

Recall from Sect. 1.1 that the mapping $\alpha \mapsto \beta _*(\alpha ;\Omega _\varepsilon )$ does not increase. Therefore, noting that $\lambda _1(p;B_r) > \lambda _1(p;\Omega _\varepsilon )$ since $B_r \subsetneq \Omega _\varepsilon$, we get

$$\begin{aligned} \beta _*(\lambda _1(p;B_r); \Omega _\varepsilon ) \le \beta _*(\lambda _1(p;\Omega _\varepsilon ); \Omega _\varepsilon ) \equiv \beta _*(\Omega _\varepsilon ). \end{aligned}$$

(B.5)

Let us investigate the behavior of $\beta _*(\lambda _1(p;B_r); \Omega _\varepsilon )$ with respect to $\varepsilon$. The first eigenfunction $\varphi _{p,j}$ corresponding to $\lambda _1(p;B_r)$, being extended by zero outside of $B_r$ so that $\varphi _{p,j} \in W_0^{1,p}(\Omega _\varepsilon )$, is an admissible test function for the definition (1.5) of $\beta _*(\lambda _1(p;B_r); \Omega _\varepsilon )$, and hence

$$\begin{aligned} \beta _*(\lambda _1(p;B_r); \Omega _\varepsilon ) \le \frac{\int _{\Omega _\varepsilon }|\nabla \varphi _{p,j}|^q \,dx}{\int _{\Omega _\varepsilon }|\varphi _{p,j}|^q \,dx} = \frac{\int _{B_r} |\nabla \varphi _{p,j}|^q \,dx}{\int _{B_r} |\varphi _{p,j}|^q \,dx} = \beta _*(B_r) = \beta _*(\Omega _0). \end{aligned}$$

We want to show that

$$\begin{aligned} \lim _{\varepsilon \searrow 0} \beta _*(\lambda _1(p;B_r); \Omega _\varepsilon ) = \beta _*(\Omega _0). \end{aligned}$$

(B.6)

Suppose, by contradiction, that

$$\begin{aligned} \liminf _{\varepsilon \searrow 0} \beta _*(\lambda _1(p;B_r); \Omega _\varepsilon ) < \beta _*(\Omega _0). \end{aligned}$$

Consequently, there exists a sequence $\{u_\varepsilon \}$ such that $u_\varepsilon \in W_0^{1,p}(\Omega _\varepsilon )$, $\int _{\Omega _\varepsilon }|u_\varepsilon |^q \,dx = 1$,

$$\begin{aligned} \int _{\Omega _\varepsilon }|\nabla u_\varepsilon |^p \,dx - \lambda _1(p;B_r) \int _{\Omega _\varepsilon }|u_\varepsilon |^p \,dx \le 0, \end{aligned}$$

(B.7)

and

$$\begin{aligned} \liminf _{\varepsilon \searrow 0} \int _{\Omega _\varepsilon }|\nabla u_\varepsilon |^q \,dx < \beta _*(\Omega _0). \end{aligned}$$

(B.8)

By the construction, we have $\Omega _{\varepsilon _2} \subset \Omega _{\varepsilon _1}$ provided $0<\varepsilon _2 < \varepsilon _1$, which yields $W_0^{1,p}(\Omega _{\varepsilon _2}) \subset W_0^{1,p}(\Omega _{\varepsilon _1})$. Therefore, extending $u_\varepsilon$ by zero outside of $\Omega _\varepsilon$, we get $u_\varepsilon \in W_0^{1,p}(\Omega _{\varepsilon _1})$ for a fixed $\varepsilon _1>0$ and all $\varepsilon \in (0,\varepsilon _1)$. In view of (B.8), $\{u_\varepsilon \}$ is bounded in $W_0^{1,p}(\Omega _{\varepsilon _1})$ and hence there exists $u_0 \in W_0^{1,p}(\Omega _{\varepsilon _1})$ such that $u_\varepsilon \rightarrow u_0$ weakly in $W_0^{1,p}(\Omega _{\varepsilon _1})$ and $W_0^{1,q}(\Omega _{\varepsilon _1})$, strongly in $L^p(\Omega _{\varepsilon _1})$ and $L^q(\Omega _{\varepsilon _1})$, and a.e. in ${\mathbb {R}}^N$, up to a subsequence.

By the construction, we have $u_0 = 0$ a.e. in ${\mathbb {R}}^N {\setminus } \overline{\Omega _0}$. Indeed, if we suppose that $u_0 \ne 0$ in some set ${\mathcal {O}} \subset {\mathbb {R}}^N {\setminus } \overline{\Omega _0}$ with positive measure $|{\mathcal {O}}|>0$, then there exists $\varepsilon _0>0$ and a channel $T_{l,\varepsilon _0}$ (which connects $B_r(z_l)$ with $B_r(z_{l+1})$) of maximal width $\varepsilon _0$ such that $|{\mathcal {O}} \cap T_{l,\varepsilon _0}| > 0$. Since the channel shrinks to a line as $\varepsilon \rightarrow 0$, we see that $|{\mathcal {O}} \cap \left( {\mathbb {R}}^N {\setminus } T_{l,\varepsilon }\right) | > 0$ for any sufficiently small $\varepsilon >0$, which means that $|{\mathcal {O}} \cap \left( {\mathbb {R}}^N {\setminus } \Omega _\varepsilon \right) | > 0$. However, this is impossible since $u_\varepsilon \in W_0^{1,p}(\Omega _\varepsilon )$ and it is extended by zero outside of $\Omega _\varepsilon$, and $u_\varepsilon \rightarrow u_0$ a.e. in ${\mathbb {R}}^N$. Therefore, we have $u_0 \in W_0^{1,p}(\Omega _0)$, see, e.g., [37, Corollary 3.3] for an explicit reference.

The strong convergence in $L^q(\Omega _{\varepsilon _1})$ yields

$$\begin{aligned} 1 = \int _{\Omega _\varepsilon }|u_\varepsilon |^q \,dx = \int _{\Omega _{\varepsilon _1}} |u_\varepsilon |^q \,dx \rightarrow \int _{\Omega _{\varepsilon _1}} |u_0|^q \,dx = \int _{\Omega _{0}} |u_0|^q \,dx ~~\text {as}~ \varepsilon \searrow 0, \end{aligned}$$

(B.9)

and so $u_0 \not \equiv 0$ in $\Omega _{0}$. Applying also the weak convergence in $W_0^{1,p}(\Omega _{\varepsilon _1})$ and the strong convergence in $L^p(\Omega _{\varepsilon _1})$, we obtain

$$\begin{aligned}&\int _{\Omega _{0}} |\nabla u_0|^p \,dx - \lambda _1(p;B_r) \int _{\Omega _{0}} |u_0|^p \,dx \\&\quad = \int _{\Omega _{\varepsilon _1}} |\nabla u_0|^p \,dx - \lambda _1(p;B_r) \int _{\Omega _{\varepsilon _1}} |u_0|^p \,dx \\&\quad \le \liminf _{\varepsilon \searrow 0} \left( \int _{\Omega _{\varepsilon _1}} |\nabla u_\varepsilon |^p \,dx - \lambda _1(p;B_r) \int _{\Omega _{\varepsilon _1}} |u_\varepsilon |^p \,dx \right) \\&\quad = \liminf _{\varepsilon \searrow 0} \left( \int _{\Omega _\varepsilon }|\nabla u_\varepsilon |^p \,dx - \lambda _1(p;B_r) \int _{\Omega _\varepsilon }|u_\varepsilon |^p \,dx \right) \le 0, \end{aligned}$$

where the last inequality follows from (B.7). Thus, we have shown that $u_0$ is an admissible function for the definition (B.4) of $\beta _*(\Omega _0)$, and hence, using (B.8) and (B.9), we arrive at the following contradiction:

$$\begin{aligned} \beta _*(\Omega _0) \le \frac{\int _{\Omega _0}|\nabla u_0|^q \,dx}{\int _{\Omega _0}|u_0|^q \,dx} = \int _{\Omega _0}|\nabla u_0|^q \,dx&\le \liminf _{\varepsilon \searrow 0} \int _{\Omega _{\varepsilon _1}} |\nabla u_\varepsilon |^q \,dx \\&= \liminf _{\varepsilon \searrow 0} \int _{\Omega _\varepsilon }|\nabla u_\varepsilon |^q \,dx < \beta _*(\Omega _0). \end{aligned}$$

Therefore, the convergence (B.6) is established.

Combining now (B.5) and (B.6), we see that

$$\begin{aligned} \beta _*(\Omega _0) + o(1) \le \beta _*(\Omega _\varepsilon ). \end{aligned}$$

That is, we obtain the desired relation

$$\begin{aligned} \lambda _1(q;B_r) < \beta _*(B_r) = \beta _*(\Omega _0) \le \beta _*(\Omega _\varepsilon ) - o(1), \end{aligned}$$

which justifies (B.3). Finally, combining (B.3) and (B.1), we complete the proof of the lemma. $\square$

Remark B.2

Essentially the same proof shows the existence of a smooth bounded domain $\Omega \subset {\mathbb {R}}^N$ such that $\lambda _l(p) < \alpha _*$ for given $N \ge 2$ and $l \ge 2$. We do not provide further details since this inequality can be observed already in the case $N=1$, see [6, Lemma A.3].