$Lu = -u'' + c u$ where c is some constant
The question is when it's positive definite in square integrable on $[0; 1]$ with $u(0)=u(1)=0$
$(Lu, u) = \int^1_0 u Lu dx = -u u''+c u^2 dx = \int^1_0 (u')^2 + cu^2 dx$>0$
How do I proceed?
I know the answer is $c>-\pi^2$ but I have no idea how to prove it.
Let $L$ be the operator operator $Lu = -u''$ on the domain $\mathcal{D}(L)$ of twice absolutely continuous functions $u$ on $[0,1]$ for which $u(0)=u(1)=0$. Then $L$ a selfadjoint linear operator with a complete orthogonal basis of eigenfunctions $\{ \sin(n\pi x) \}_{n=1}^{\infty}$. The spectrum of $L$ consists of the corresponding eigenvalues $\pi^{2}n^{2}$ for $n=1,2,3,\cdots$. Therefore, $$ (Lu,u) \ge \pi^{2}(u,u),\;\;\; u \in \mathcal{D}(L), $$ with equality for some $u \ne 0$ iff $u$ is a constant multiple of $\sin(\pi x)$. Therefore, $L+cI$ is positive definite iff $c > -\pi^{2}$. It is positive, but not positive definite if $c = -\pi^{2}$ because $(L-\pi^{2}I)\sin(\pi x)=0$.