I have been a lot of time trying to understand a key step on a paper about spectral analysis but I have no clue how to prove it (and the authors only said "by standard analysis"). Let me state the question (this is how I interpret actually, I tried to clean it since I prefer to avoid all the details and definitions of the paper). Consider the following 1D-Schrödinger operator $$
L:=-\partial\_x^2+c\_1-c\_2\Phi
$$
where $c\_1,c\_2>0$ and $\Phi$ is a positive Schwartz function. 


Now, I can prove that due to the specific structure of the operator and $\Phi$, $L$ is a **nonnegative** operator (zero is its **first** eigenvalue, which is simple). Moreover, the eigenfunction associated to its first eigenvalue (zero), let's say $\zeta$, is a **positive** Schwartz function (it's some power of $\Phi$ actually). Here is where everything become a bit dark for me. Is it true that if I consider a **nonnegative** function $g\in L^2\setminus\{0\}$, then, due to the spectral information of $L$ above, there exist $\lambda>0$ (depending on my election of $g$) such that for all $f\in H^1(\mathbb{R})$ it holds: $$
\langle Lf,f\rangle\geq \lambda\int f^2-\dfrac{1}{\lambda}\left(\int fg\right)^2?
$$
I am quite surprised that it seems that I can "choose" this $g$ (as soon as restricting myself to nonnegative functions not identically zero). Does anyone has any idea on how to prove it? Or maybe some recommended references?


**PS:** Notice that since the eigenfunction associated to the zero eigenvalue is positive, and we are also 
 assuming $g$ positive, then $g$ cannot be orthogonal to $\zeta$. Thus, you cannot choose $g\perp \zeta$ and then try to choose $f=\zeta$ to make the left-hand side equals zero (while the right-hand side would remain strictly positive).



Nonnegativity implies $\langle Lf,f\rangle\geq \int f^2-(\int fg)^2$ for $g\geq 0$
Under your assumptions the essential spectrum is $[c\_1, \infty[$, hence the part of the spectrum in $[0,c\_1[$ is discrete. Let $\mu>0$ be the second eigenvalue (the first is 0), if it exists, or $c\_1$. Then $(Lh,h) \ge \mu (h,h)$ if $h$ is orthogonal to $\zeta$. Next assume by contradiction that $(Lf\_n, f\_n) +(f\_n,g)^2 \le n^{-1}\|f\_n\|^2$ and $\|f\_n\|=1$ and split $f\_n=c\_n \zeta+h\_n$ with $(\zeta, h\_n)=0$. Then $(Lf\_n,f\_n)=(Lh\_n,h\_n) \to 0$, hence$\|h\_n\| \to 0$ and $|c\_n| \to 1$. Next $(f\_n,g)=(c\_n \zeta+h\_n,g) \to 0$, which is impossible since $|c\_n| \to 1$ and $(\zeta,g)>0$.


