$m_3(\cdot, \cdot)\in\mathcal X_3$, and $m_3(t, x)\not \equiv m_3(t)$ for $t\in \R$ and $x\in \R^N$. The eigenvalue problem has exactly one positive principal spectrum point, denoted by $\lambda_3^p$, if and only if \[P\] P (m\_3)&gt;0 \_[D\_3]{} \_0\^Tm\_3(t, x)dtdx&lt;0. More precisely, we show the following.

-   There is a $\lambda_3^p>0$ such that $\mu_3^n(\lambda_3^p)=0$ if and only if (P) holds.

-   If $\lambda_3^{p, 1},\lambda_3^{p, 2}>0$ are such that $\mu_3^n(\lambda_3^{p, 1})=\mu_3^n(\lambda_3^{p, 2})=0$, then $\lambda_3^{p, 1}=\lambda_3^{p, 2}$.

\[PSP-peri-coro\]

-   If $m_3(t, x)=m_3(t)$, then $$\mu_3^n(\lambda_3)=\lambda_3 \widehat m_3.$$ It then follows that, if $\int_0^T m_3(t)dt\not =0$, then there is no positive principal spectrum point of , and if $\int_0^T m_3(t)dt=0$, then any $\lambda_3>0$ is a positive principal spectrum point of .

-   If $m_3(t, x)\equiv m_3(x)$, consider $$\label{autonomous-periodic-ori}
    \int_{\R^N}\kappa(y-x)[u(y)-u(x)]dy+\lambda_3m_3(x)u(x)=0.$$ There is exactly one positive principal spectrum point $\lambda_3^p$ of if and only if $$m_3(x_0)>0\quad \text{for some}\,\, x_0\in D_3,\quad {\rm and}\,\,\int_{D_3}  m_3(x)dx<0.$$

Next, we state the main result on the upper bounds of principal spectrum points of , , and , and sufficient conditions for the principal spectrum points of , , and to be principal eigenvalues.

Consider the eigenvalue problem with indefinite weight function $\hat m_i$, $$\label{average-weight-pev}
K_iu-B_iu+\lambda_i \widehat m_i u=0,$$ and the regular eigenvalue problem $$\label{average-pev}
