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 $$
V(u)=\dfrac{1}{2}\int_\mathbb{R} u^2dx \quad \hbox{and}\quad E(u)=\int_\mathbb{R}(\tfrac{1}{2}u\partial_x^2u-\tfrac{1}{2}u^2-\tfrac{1}{3}u^3)dx.
$$