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).

  • $\begingroup$ 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? $\endgroup$ – Jochen Glueck Apr 24 '20 at 19:09
  • $\begingroup$ @JochenGlueck Yes, in the whole space, sorry for the misunderstanding $\endgroup$ – Sharik Apr 24 '20 at 19:19
  • 1
    $\begingroup$ It would be nice if you linked the paper from which the doubt arose. $\endgroup$ – Adrián González-Pérez Apr 25 '20 at 10:21

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 $(\zeta,g)>0$.


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