# Nonnegativity implies $\langle Lf,f\rangle\geq \int f^2-(\int fg)^2$ for $g\geq 0$

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).

• Since you are mentioning Schwartz functions and functions $f \in H^1(\mathbb{R})$, do I get this right that your problem is posed on the whole real line rather than on a bounded interval? – Jochen Glueck Apr 24 '20 at 19:09
• @JochenGlueck Yes, in the whole space, sorry for the misunderstanding – Sharik Apr 24 '20 at 19:19
• It would be nice if you linked the paper from which the doubt arose. – Adrián González-Pérez Apr 25 '20 at 10:21

## 1 Answer

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 $$(\zeta,g)>0$$.