Let $x\in N_{R}\cup N_{r}$ be arbitrary. A direct computation gives

Next, from (2.4), (2.5) and (2.1),

Notice that (2.1) applied since both $\bar{J}\left(\mu Jx\right),\ \bar{J}\left(\mu Jx-Dx\right)$ are bounded independently on $x\in N_{R}\cup N_{r}$ as a consequence of (2.2), (2.3) and of the fact that $J,\bar{J},F^{\prime}$ and $G^{\prime}$ map bounded sets into bounded sets. ∎

In case (a), the conclusion follows from (2.6), the continuity of $G,$ and the strict inequalities $r<G\left(x\right)<R.$ Assume now that condition (b) holds. From the definition of the Fréchet derivative of $G,$ for each $\varepsilon>0,$ there exists $\delta_{\varepsilon}>0$ such that for each $t\in(0,\delta_{\varepsilon})$ we have

Hence

Since $\langle G^{\prime}(x),z\rangle>0,$ we may take $\varepsilon:=\langle G^{\prime}(x),z\rangle$ to obtain

Hence, for $t$ sufficiently small such that $r\leq R-2t\langle G^{\prime}(x),z\rangle$ and $t\in(0,\delta_{\varepsilon}),$ we have $\ r\leq G(x-tz)\leq R.$ This together with (2.6) shows that $y\in K_{r,R}$ for $t>0$ small enough. Finally, if (c) holds, then (2.7) gives

In this case we may take $\varepsilon:=-\langle G^{\prime}(x),z\rangle$ and obtain

Hence, for $t$ sufficiently small such that $r-2t\langle G^{\prime}(x),z\rangle\leq R$ and $t\in(0,\delta_{\varepsilon}),$ we have $\ r\leq G(x-tz)\leq R,$ that is the desired conclusion. ∎

(i) First note that using (2.8), the representation

and the convexity of $K$ yield that $x-t\left[x-\bar{J}\left(Jx-F^{\prime}\left(x\right)\right)\right]\in K\$ for every $t\in\left(0,1\right).$ Then the conclusion of (i) follows from Lemma 2.2.

(ii) According to Lemma 2.2 (b), we first check that $\langle G^{\prime}(x),z\rangle>0,$ where $z:=\varepsilon x+E\left(x\right).$ Indeed, using (2.4) and (2.3), we can see that

Next we need to check (2.6). One has

where $\mu=\mu\left(x\right)$ is as in (2.9), assumed to be nonzero. Clearly, for small $t,$ $1-t\mu\left|\mu\right|^{\frac{2-p}{p-1}}-t\varepsilon+t\lambda>0,$ and thus (2.10) together with (2.9) shows that $y\in K$ for all small enough $t>0.$ The conclusion now follows from Lemma 2.2 (b). The case $\mu=0$ is investigated similarly.

(iii) We proceed as at the case (ii) and find that

Hence, if $z:=-\varepsilon x+E\left(x\right),$ then $\langle G^{\prime}(x),z\rangle<0.$ Furthermore

and we obtain as above that $y\in K$ for all small enough $t>0.$ ∎

We shall apply Ekeland’s variational principle for $M:=K_{r,R}$ (we use here that $K$ is closed and $G$ is continuous, hence $K_{r,R}$ is a closed subset of the Banach space $X$) endowed with the metric $d(x,y):=\|x-y\|,$ for the function $F$ (which from (A3) is $C^{1}$ and bounded from below), and for $\varepsilon:=\frac{1}{n}$ $\left(n\in\mathbb{N}\setminus\left\{0\right\}\right).$ It follows that there exists a sequence $(x_{n})$ in $K_{r,R}$ such that

and

Clearly (2.15) implies (2.11).

Since $(x_{n})$ belongs to $K_{r,R},$ we distinguish three cases:

Case 1: There exists a subsequence of $(x_{n}),$ still denoted by $(x_{n}),$ in one of the following situations: (i₁) $r<G(x_{n})<R$ for all $n;$ (i₂) $x_{n}\in N_{R}$ and $\langle G^{\prime}(x_{n}),x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\rangle>0$ for all $n;$ (i₃) $x_{n}\in N_{r}$ and $\langle G^{\prime}(x_{n}),x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\rangle<0\$ for all $n.$

Case 2: There exists a subsequence of $(x_{n}),$ still denoted by $(x_{n}),$ such that $x_{n}\in N_{R}$ and $\langle G^{\prime}(x_{n}),x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\rangle\leq 0\$ for all $n.$

Case 3: There exists a subsequence of $(x_{n}),$ still denoted by $(x_{n}),$ such that $x_{n}\in N_{r}$ and $\langle G^{\prime}(x_{n}),x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\rangle\geq 0\$ for all $n.$

Assume Case 1. According to Lemma 2.3 (i), for each $n,$ we have $y:=x_{n}-t\left(x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\right)\in K_{r,R},$ for all $t>0$ sufficiently small. Thus we may apply (2.16) and deduce

Divide by $t$ and let $t$ go to zero to obtain

It follows that

Then from (2.1),

Using these equality in (2.17) we deduce that

Hence $x_{n}-\bar{J}\left[Jx_{n}-F^{\prime}(x_{n})\right]\rightarrow 0$ as $n\rightarrow\infty$ and so, property (a) holds in Case 1.

Assume Case 2. Now Lemma 2.3 (ii) guarantees that for each $n$ and any $\varepsilon>0,\$ $y:=x_{n}-t\left(\varepsilon x_{n}+E\left(x_{n}\right)\right)\in K_{r,R}$ for all $t>0$ sufficiently small. Then (2.16) implies

Letting $\varepsilon\rightarrow 0$ and using Lemma 2.1 we deduce

Let us consider the continuous linear operator

Since $(x_{n})\subset N_{R}$ and the level set $N_{R}$ is bounded, it follows that $(x_{n})$ is a bounded sequence. By the assumption on $G^{\prime}$ it follows that $(G^{\prime}(x_{n}))$ is also bounded. In addition

We have

Hence there exists $\alpha_{R}>0$ (independent on $n$) such that

For $x:=\bar{J}\left(\mu_{n}Jx_{n}\right)-\bar{J}\left(\mu_{n}Jx_{n}-Dx_{n}\right),$ one has $P_{n}x=E(x_{n})$ and thus

Then by (2.18),

Since $\beta>1,$ this yields $\bar{J}\left(\mu_{n}Jx_{n}\right)-\bar{J}\left(\mu_{n}Jx_{n}-Dx_{n}\right)\rightarrow 0\$ as $n\rightarrow\infty,$ that is (2.12) holds.

Finally, in Case 3 we proceed similarly by using Lemma 2.3 (iii) and taking in (2.16) $y:=x_{n}-t\left(-\varepsilon x_{n}+E\left(x_{n}\right)\right).$ The conclusion is the same, namely (2.12) holds.

Assume now that the additional hypotheses of the theorem are satisfied. The (PS) condition guarantees the existence of a subsequence of $\left(x_{n}\right),$ which is still denoted by $\left(x_{n}\right),$ such that $x_{n}\rightarrow x$ as $n\rightarrow\infty,$ for some element $x\in K_{r,R}.$ Clearly, (2.11) gives $F(x)=\inf F(K_{r,R}).$ In case of the property (a), if we denote $y_{n}:=x_{n}-\bar{J}\left(Jx_{n}-F^{\prime}\left(x_{n}\right)\right),$ then $y_{n}\rightarrow 0$ as $n\rightarrow\infty,$ and from

letting $n\rightarrow\infty$ and using the continuity of $F^{\prime}$ and the demicontinuity of $J,$ we obtain $F^{\prime}\left(x\right)=0$ and the proof is finished. Assume that the property (b) holds. Then, if we pass to the limit we obtain

and

where $\mu$ is the limit of some convergent subsequence of $\left(\mu_{n}\right).$ Notice that such a subsequence exists since according to (A4), $\left|\mu_{n}\right|\leq\mu_{\rho},$ where $\rho$ is a bound for the sequence $\left(\left\|x_{n}\right\|\right).$ Next from (2.20)

that is

where

In case that $\eta=0,$ (2.21) shows that $F^{\prime}\left(x\right)=0$ and we are done. Assume $\eta\neq 0.$ From (2.21),

This together with (2.19) gives

Since