# Does this functional satisfies the Palais-Smale condition?

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?