H_k=\mbox{span}\left\{e_{1},\ldots,e_k\right\}$$ and $$\mathbb P_k=
\left\{u\in X_{0}:\,\,\left\langle u,e_{j}\right\rangle_{X_{0}}=0\,\,\,\,\mbox{for any}\,\, j=1,\ldots,k \right\},$$ where $$\left\langle u,v\right\rangle_{X_{0}}:= \int_{{\mathbb R}^n\times {\mathbb R}^n} \big(u(x)-u(y)\big)\big(\varphi(x)-\varphi(y)\big)K(x-y)\,dx\,dy$$ makes $X_0$ a Hilbert space, see [@svmountain Lemma 7]. In this way, the variational characterization of the eigenvalues (see [@svlinking Proposition 9]) implies that $$\label{poincarevincolata}
\int_{{\mathbb R}^n\times {\mathbb R}^n}|u(x)-u(y)|^2 K(x-y)\,dx\,dy\geq \lambda_{k+1} \int_\Omega |u(x)|^2\,dx \ \mbox{ for all $u\in \mathbb P_k$},$$ while, using the orthogonality properties of the eigenvalues, a standard Fourier decomposition gives $$\label{antipoincare}
\int_{{\mathbb R}^n\times {\mathbb R}^n}|u(x)-u(y)|^2 K(x-y)\,dx\,dy\leq \lambda_k \int_\Omega |u(x)|^2\,dx \ \mbox{ for all $u\in \mathbb H_k$}.$$

Gradient in $X_0$ {#secgrad}
-----------------

Being $X_0$ a Hilbert space and $\mathcal J_\lambda$ of class $C^1$, the gradient $\nabla \mathcal J_\lambda$ of $\mathcal J_\lambda$ is immediately defined as $$\label{gradJ}
\begin{aligned}
\langle \nabla \mathcal J_\lambda(u),v\rangle_{X_0} :&=\langle \mathcal J'_\lambda(u), v\rangle\\
& =\langle u,v\rangle_{X_0}-\lambda \int_\Omega u(x)v(x)\,dx-\int_\Omega f(x,u(x))v(x)\,dx
\end{aligned}$$ for any $u,v\in X_0$.

Let $\nu \in [1, 2^*]$ and $\nu'$ be its conjugate, that is $1/\nu+1/\nu'=1$. Introducing the operator $\mathcal L_K^{-1}:L^{\nu'}(\Omega)\to X_0$, defined as $\mathcal L_K^{-1}g=v$ if and only if $v\in X_0$ solves $$\begin{cases}
\mathcal L_Kv=g(x) & \mbox{ in } \Omega,\\
v=0 & \mbox{ in } {\mathbb R}^n\setminus \Omega,
\end{cases}$$ it is readily seen that $$\label{grad}
\nabla \mathcal J_\lambda(u)=u-\mathcal L_K^{-1}(\lambda u+f(x,u))$$ for all $u\in X_0$. Indeed, setting 