This is a problem from Haim Brezis' functional analysis book (Exercise 8.41). I solved parts of it, but am stuck on some parts/want confirmation on the method. The problem is as follows:
Let $q \in C([0,1])$ and consider the bilinear form \begin{equation} a(u,v) = \int_0^1 (u'v' + quv), \, u, v \in H_0^1(0,1) \end{equation} on $H_0^1(0,1)$. Equip the space $H_0^1(0,1)$ with the scalar product $a(u,v)$, denoted $(u,v)_H$, and the norm $\lvert u \rvert_H = (u,u)^{1/2}_H$. Assume that $a$ is not coercive and more precisely that there exists some $v_1 \in H_0^1$ such that $v_1 \neq 0$ and $a(v_1, v_1) \leq 0$. Prove that there is no solution to \begin{equation} \begin{cases} -u'' + qu = u^3 \text{ on } (0,1) \\ u > 0 \text{ on } (0,1), \text{ and } u(0) = u(1) = 0 \end{cases} \end{equation}
The hint is to check that the first eigenvalue of $Au = -u'' + qu$ is nonpositive, and multiply the above equation by the corresponding eigenfunctions. How do I find this nonpositive eigenvalue? My first thought was to use a theorem about compact self-adjoint operators, but $A$ seems to be only self-adjoint and not compact. Nonetheless, using the fact that $a(v_1,v_1) \leq 0$, I can deduce that there is some nonpositive element in the spectrum of $A$. How can I fix this to get a nonpositive eigenvalue?
Now, assuming that I have a nonpositive eigenvalue $\lambda_1$ with corresponding eigenfunction $\phi_1$, I get \begin{align} \int_0^1 -u'' \phi_1 + qu \phi_1 &= \int_0^1 u^3 \phi_1 \\ \int_0^1 -u \phi_1'' + qu \phi_1 &= \int_0^1 u^3 \phi_1 \\ \int_0^1 \lambda_1 \phi_1 u &= \int_0^1 u^3 \phi_1 \\ \int_0^1 u\phi_1 (u^2 - \lambda_1) &= 0. \end{align}
Since $u > 0$, $\lambda_1 \leq 0$, and we can choose $\phi_1 > 0$, this contradicts the LHS $= 0$.
Any help or suggestions would be extremely appreciated with finding the eigenvalue.