I am trying to understand the following argument: Let $\mathcal{L}:L^2(\mathbb{R})\to L^2(\mathbb{R})$ be an essentially self-adjoint unbounded linear operator with domain $D(\mathcal{L})=H^s(\mathbb{R})$ for some $s>0$. Let us assume that $\mathcal{L}$ has only one negative eigenvalue (which is simple) with associated eigenfunction $\chi$. Moreover, assume that zero is also a simple eigenvalue with associated eigenfunction $\phi\_c'$ (the derivate of a fixed function $\phi\_c$). Finally, assume that the rest of the spectrum is positive and away from zero. 


Now, I can prove the following lemma: Under some extra hypothesis (I don't think they are relevant for my question), if a function $y\in H^{s/2}(\mathbb{R})$ satisfies $$
\langle y,\phi\_c\rangle=\langle y,\phi\_c'\rangle=0,
$$
then $\langle\mathcal{L}y,y\rangle>0$. Here $\langle\cdot,\cdot\rangle$ denotes the inner product in $L^2$. Now my question is, under these conditions over the spectrum of $\mathcal{L}$, does the previous lemma implies that there exists a constant $C>0$ such that for $y\in H^{s/2}$ satisfying the hypothesis of the lemma we have $\langle \mathcal{L}y,y\rangle\geq C\Vert y\Vert\_{H^{s/2}} ^2$?


**Edit:** To give more context to my question, $\mathcal{L}$ is a self-adjoint differential operator which comes from the linearization of certain PDE around $\phi\_c$, so that we have $\mathcal{L}\phi\_c'=0$. Moreover, the extra hypothesis on my lemma states that if we define $d(c)=E(\phi\_c)+cV(\phi\_c)$, then $d''(c)>0$. Here it might be important to notice that $E'(\phi\_c)+cV'(\phi\_c)=0$ and that $\mathcal{L}=E''(\phi\_c)+cV''(\phi\_c)$. So I think the most important part is that we are assuming that $d''(c)>0$. Of course, this is extremely important to prove the lemma, but I am not sure if we can use it to prove the inequality I am looking for.


**Edit2:** The functionals $E$ and $V$ are defined as follows $$
V(u)=\dfrac{1}{2}\int\_\mathbb{R} u^2dx \quad \hbox{and}\quad E(u)=\int\_\mathbb{R}(\tfrac{1}{2}u\partial\_x^2u-\tfrac{1}{2}u^2-\tfrac{1}{3}u^3)dx.
$$



Coercivity of linear operators
Yes, the inequality is true. Let $\lambda$ be large enough so $(L+\lambda I)$ is positive definite. We have $$((L+\lambda I)y,y)=((L+\lambda I)^{1/2}y,(L+\lambda I)^{1/2}y)\ge C\|y\|^2\_{H^{s/2}},$$
since the domain of $(L+\lambda I)^{1/2}$ is $H^{s/2}$ by the general theory of interpolation spaces. It follows that
$$(Ly,y)\ge C\|y\|^2\_{H^{s/2}}-\lambda (y,y).$$
Now we distinguish two cases. Either $\|y\|^2\_{H^{s/2}}>2\lambda (y,y)$. In that case it follows that
$$(Ly,y)\ge \frac{C}{2} \|y\|^2\_{H^{s/2}}.$$ In the opposite case, let $\mu$ be such that $(Ly,y)\ge \mu (y,y)$. Then it follows that
$$(Ly,y)\ge \mu(y,y)\ge \frac{\mu}{2\lambda}\|y\|^2\_{H^{s/2}}.$$


