Let $\Omega$ be an open set of $\mathbb{R}^d$ and $K \in L^2(\Omega\times \Omega)$ such that for almost all $x,y \in \Omega$ :
- $K(x,y)=K(y,x)$
- $K(x,y)>0$
One can show that under these hypothesis, the map
$$A : L^2(\Omega) \longrightarrow L^2(\Omega)$$
defined for all $u \in L^2(\Omega)$ by
$$Au : \Omega \longrightarrow \mathbb{R}, x \mapsto \int_{\Omega} K(x,y)u(y)dy$$
is a linear, continuous, and compact map of $L^2(\Omega)$ that can be diagonilized i.e., there exist $(\lambda_k)_k \subset \mathbb{R}$ converging to 0 and a Hilbert basis $(u_k)_{k \in \mathbb{N}}$ of $L^2(\Omega)$ such that for all $k \in \mathbb{N}$ :
$$Au_k=\lambda_k u_k$$
On the other hand, the number $\hat{\lambda} = \max_{k \in \mathbb{N}} \lambda _k $ exist and is a strictly positive eigenvalue of A. And, moreover, each eigenvector $\hat{u}$ associated to $\hat{\lambda}$ realizes :
$$\sup_{u \in L^2(\Omega), \|u\|_{L^2}=1} <Au,u> = \sup_{(\alpha_k)_k : \sum\limits_{k\ge 0} \alpha_k^2 = 1}\sum\limits_{k \ge 0} \alpha_k ^2 \lambda_k $$
I would like to show some inequalities on the positive and negative part of an eigenvector $\hat{u}$ associated to $\hat{\lambda}$. They are defined respectively by :
$$\hat{u}_{+}=max(\hat{u},0)$$ and $$\hat{u}_{-}=max(-\hat{u},0)$$
in such a manner that $\hat{u}=\hat{u}_{+}-\hat{u}_{-}$
I want to show that $$<A\hat{u}_{+},\hat{u}_{+}>\le \hat{\lambda}<\hat{u}_{+},\hat{u}_{+}>$$ and after this, that : $$<A\hat{u}_{+},\hat{u}_{-}> \le 0$$
Can anyone help me with the first inequality ?