0$ a.e. on $ [a,b]$. Then $\lambda _n(W)\geq \lambda _n(w)$ if $\lambda _n(W)<0$ and $ \lambda _n(w)<0;$ but $\lambda _n(W)\leq \lambda _n(w)$ if $\lambda _n(W)>0$ and $\lambda _n(w)>0$. Furthermore, if strict inequality holds in the hypothesis on a set of positive Lebesgue measure, then strict inequality holds in the conclusion. \end{enumerate} \end{theorem} \begin{proof} We give the proof for (1), the proofs of (2) and (3) are similar. Define a function $f:\mathbb{R}$ $\to \mathbb{R}$ by \[ f(t)=\lambda _n(s(t)),\;s(t)=q+t(Q-q),\quad t\in [ 0,1]. \] Then $s(t)\in L ((a,b),\mathbb{R})$ for each $t\in [ 0,1]$. From the chain rule in Banach space and formula \eqref{5.8} for $\lambda _n'(q)$ we have \[ f'(t)=\lambda _n'((s(t))\,s'(t)=\int_{a}^{b}|u^2(r,s(t))|\,(Q(r)-q(r))\,dr\geq 0,\;t\in [ 0,1]. \] Hence $f$ is nondecreasing on $[0,1]$ and $f(1)=\lambda _n(Q)\geq \lambda _n(q)=f(0)$. 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