Sturm-Liouville Show negative eigenvalue Let $q(x)$ be a continuous function such that $\int_0^1 q = 0$ but $q$ is not identically zero. I want to show that $Lu = -u'' +q(x)u$ with BC's $u'(0)=u'(1)=0$ has a strictly negative eigenvalue. 
The given hint is that I should show that $\int_0^1 uLu$ can be negative.
My approach: Assume $u$ is non-zero, then dividing by $u$ we have $$-u''+qu=\lambda u$$
$$\int_0^1 -\frac{u''}{u}=\lambda$$
Integration by parts,
$$-\int_0^1 \frac{u'^2}{u^2}=\lambda$$ So for non-constant $u$ we should have a negative eigenvalue. I don't think this is legitimate though because I am assuming $u\ne0$ and I am not using the hint given to me. So I would like to find a better approach.
 A: Let $\mathcal{D}(L)$ consist of all twice continuously differentiable functions on $[0,1]$ for which $u'(0)=u'(1)=0$. Because $q$ is continuous on $[0,1]$, then $L$ has a complete orthonormal basis of real eigenfunctions $\{ e_{n} \}_{n=1}^{\infty}\subset \mathcal{D}(L)$. Furthermore, because the conditions are separated (i.e., each condition involves only values of $u$ at one endpoint), the eigenspaces are one-dimensional. Furthermore, the eigenvalues of $L$ are bounded below. To see this, let $M$ be a bounded for $q$ on $[0,1]$, and notice that, for any real function $u \in \mathcal{D}(L)$,
$$
     (Lu,u) = \int_{0}^{1}\{-u''u + qu^{2}\} dx = \int_{0}^{1}\{u'^{2}+qu^{2}\}dx \ge \int_{0}^{1}qu^{2}dx \ge -M\int_{0}^{1}u^{2}\,dx.
$$
Therefore, if $Lu=\lambda u$ for some non-zero $u \in \mathcal{D}(L)$ and real $\lambda$, then $\lambda(u,u)\ge -M(u,u)$, or $\lambda \ge -M$. So we may assume $Le_{n}=\lambda_{n}e_{n}$ where $-M \le \lambda_{1} < \lambda_{2} < \lambda_{3} < \cdots$. Furthermore, $(Le_{1},e_{1}) \le (Lu,u)$ for all $u \in \mathcal{D}(L)$ with $\|u\|^{2}=\int_{0}^{1}u^{2}\,dx =1$, and equality holds for some real $u$ iff $u=\alpha e_{1}$ for $\alpha = \pm 1$.
The function $u_{1}$ which is identically $1$ is in $\mathcal{D}(L)$ and $Lu_{1}=qu_{1}=q$ is not the zero function (by assumption). Furthermore $(Lu_{1},u_{1})=0$ because $\int_{0}^{1}qdx=0$ (by assumption). So we know that $\lambda_{1} \le 0$. In fact, $\lambda_{1} < 0$ because $\lambda_{1}=0$ implies that $(Lu,u)$ is minimized by the unit vector $u_{1}$, which then forces $Lu_{1}=\lambda_{1}u_{1}=0$, a contradiction. So $\lambda_{1} < 0$, as desired.
