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Let $\Omega$ be a non-empty bounded open subset of $\mathbb{R}^N$, $\lambda\in \mathbb{R}$ be an eigenvalue of $-\Delta$ on the Sobolev space $H^1_0(\Omega)$ and $f\in L^\infty(\Omega\times\mathbb{R})$ such that

  • $\forall x\in \Omega, t\mapsto f(x,t)\in C(\mathbb{R});$
  • $\forall M>0, \exists r>0, \forall |s|>r, \forall x\in \Omega, \int_0^s f(x,t)\operatorname{d}t\ge M;$
  • $\forall \varepsilon>0, \exists r>0, \forall|s|>r, \forall x\in\Omega,\left| \frac{1}{s}\int_0^s f(x,t)\operatorname{d}t\right|\le\varepsilon.$

Define: $$I:H^1_0(\Omega)\to\mathbb{R}, u\mapsto\frac{1}{2}\|u\|^2_{H^1_0}-\frac {\lambda}{2}\|u\|_2^2-\int_\Omega\int_0^{u(x)}f(x,t)\operatorname{d}t\operatorname{d}x.$$ Then $I\in C^1(H^1_0(\Omega),\mathbb{R})$ and $$\forall u,v\in H^1_0(\Omega), \operatorname{d}I(u)(v)=\int_\Omega \nabla u(x)\cdot \nabla v(x) \operatorname{d}x-\lambda\int_\Omega u(x)v(x) \operatorname{d}x-\int_\Omega f(x,u(x))v(x)\operatorname{d}x$$

Is it true that $I$ satisfies the Palais-Smale condition? I.e. is it true that for all $(u_n)_{n\in\mathbb{N}}\subset H^1_0(\Omega)$ such that $(I(u_n))_{n\in\mathbb{N}}$ is bounded and $\|\operatorname{d}I(u_n)\|\to0, n\rightarrow\infty$ there exists a subsequence $(u_{n_k})_{k\in\mathbb{N}}$ that converges in $H^1_0(\Omega)$ to some $\bar u \in H^1_0(\Omega)$?

In my lecture notes on calculus of variations, it is claimed that every sequence $(u_n)_{n\in\mathbb{N}}$ that satisfies the previous condition is actually bounded in $H^1_0(\Omega)$ and so, by the fact that there exists a subsequence that weakly converges in $H^1_0(\Omega)$ to some $\bar u \in H^1_0(\Omega)$, it easily follows (from the fact that the differential of $I$ can be expressed as a sum of a homeomorphism and a compact operator) that this subsequence also converges to $\bar u$ in $H^1_0(\Omega)$.

My problem is proving the fact that a sequence $(u_n)_{n\in\mathbb{N}}$ as before is actually bounded in $H^1_0(\Omega)$. In particular, what I have proved is the following. First, decompose $H^1_0(\Omega)$ as the orthogonal sum: $$H^1_0(\Omega)=E_-\oplus E_0\oplus E_+$$ where $E_-$ is the vector space generated by the eigenfunctions relative to eigenvalues less than $\lambda$, $E_0$ is the eigenspace relative to $\lambda$ and $E_+$ is the closure of the vector space generated by the the eigenfunctions relative to eigenvalues greater than $\lambda$. Define $P_-$ as the orthogonal projection of $H^1_0(\Omega)$ onto $E_-$, define $P_0$ as the orthogonal projection of $H^1_0(\Omega)$ onto $E_0$ and define $P_+$ as the orthogonal projection of $H^1_0(\Omega)$ onto $E_+$.

Then, using the relations: $$\operatorname{d}I(u_n)(P_-u_n)\ge -C\|P_-u_n\|_{H^1_0}$$ and $$\operatorname{d}I(u_n)(P_+u_n)\le C\|P_+u_n\|_{H^1_0}$$ and the estimates of $\|\cdot\|_2^2$ from below with respect to $\|\cdot\|_{H^1_0}^2$ on $E_-$ and of $\|\cdot\|_2^2$ from above with respect to $\|\cdot\|_{H^1_0}^2$ on $E_+$, we obtain that $(P_-u_n)_{n\in\mathbb{N}}$ and $(P_+u_n)_{n\in\mathbb{N}}$ are bounded in $H^1_0(\Omega)$.

It remains to show that $(P_0 u_n)_{n\in\mathbb{N}}$ is bounded in $H^1_0(\Omega)$.

Thanks to the previous estimates, what I proved is that for some constant $B,C>0$ we have that: $$C\ge |I(u_n)|=\left|\frac{1}{2}\|u_n\|^2_{H^1_0}-\frac{\lambda}{2}\|u_n\|_2^2-\int_\Omega\int_0^{u_n(x)}f(x,t)\operatorname{d}t\operatorname{d}x\right|\ge \left|\int_\Omega\int_0^{u_n(x)}f(x,t)\operatorname{d}t\operatorname{d}x\right|-B$$ and so the sequence: $$\left(\int_\Omega\int_0^{u_n(x)}f(x,t)\operatorname{d}t\operatorname{d}x\right)_{n\in\mathbb{N}}$$ is bounded in $\mathbb{R}$.

Now, I suspect that I have to use the hypothesis $$\forall M>0, \exists r>0, \forall |s|>r, \forall x\in \Omega, \int_0^s f(x,t)\operatorname{d}t\ge M$$ with the boundedness of the sequences $$\left(\int_\Omega\int_0^{u_n(x)}f(x,t)\operatorname{d}t\operatorname{d}x\right)_{n\in\mathbb{N}}, \left(P_-u_n\right)_{n\in\mathbb{N}}, \left(P_+u_n\right)_{n\in\mathbb{N}}$$ to conclude that actually $(P_0u_n)_{n\in\mathbb{N}}$ (or directly the sequence $(u_n)_{n\in\mathbb{N}}$) is bounded in $H^1_0(\Omega)$, but I can't see how...

Any suggestion?

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