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

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  • $\begingroup$ Just to see if I get correctly the question. Does $\chi$ enter the game some way ? The inequality you ask is to bound the $L^2$ bilinear form associated to $\mathcal{L}$, right ? $\endgroup$ – an_ordinary_mathematician Mar 30 at 11:40
  • $\begingroup$ @an_ordinary_mathematician The function $\chi$ seems to play no role on this point (to prove the property I am asking for), just the fact that it is associated to the unique negative direction of $\mathcal{L}$ . I am wondering if from the fact that under both orthogonality conditions we have $\langle \mathcal{L}y,y\rangle>0$, we can deduce the last inequality of my question. $\endgroup$ – Sharik Mar 30 at 12:49
  • $\begingroup$ Your edit doesn't clarify much --- nothing in the expression $E(\phi_c) + cV(\phi_c)$ has been defined ... $\endgroup$ – Nik Weaver Mar 30 at 13:08
  • $\begingroup$ @NikWeaver Yes, I am very sorry for that, I am trying to avoid stating a too long question. Let me re-edit it, I hope it helps. $\endgroup$ – Sharik Mar 30 at 13:10
  • $\begingroup$ That does help, thank you! What about $\phi_c$? $\endgroup$ – Nik Weaver Mar 30 at 14:19
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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 $$(Ly,y)\ge \mu(y,y)\ge \frac{\mu}{2\lambda}\|y\|^2_{H^{s/2}}.$$

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  • $\begingroup$ Are you assuming here $y$ is orthogonal to $\chi$, the eigenfunction of the negative eigenvalue? Or is it enough that $y$ is orthogonal to $\phi_c$ as the OP stated? $\endgroup$ – Willie Wong Mar 31 at 2:19
  • $\begingroup$ I changed the argument so I am no longer assuming y orthogonal to $\chi$. $\endgroup$ – Michael Renardy Mar 31 at 2:46
  • $\begingroup$ @MichaelRenardy Thank you very much for your answer. Just one (maybe silly) question, why does exist this $\mu$ satisfying $(Ly,y)>\mu(y,y)$? $\endgroup$ – Sharik Mar 31 at 7:56
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    $\begingroup$ According to your post, you know $(Ly,y)>0$ for nonzero $y$ satisfying the orthogonality conditions. Now suppose you have a sequence $y_n$ with $(y_n,y_n)=1$ and $(Ly_n,y_n)\to 0$. By taking a subsequence, we may assume $y_n$ converges weakly, let $y$ be the limit. Let $z_n$ be the component of $y_n$ orthogonal to $\chi$ and let $z$ be the weak limit of $z_n$. Since $L$ is positive definite on the subspace orthogonal to $\chi$ and $\phi_c'$, we have $(Lz,z)\le \lim(Lz_n,z_n)$. If the inequality is strict, then $(Ly,y)<0$, a contradiction. If $(Lz_n,z_n)$ converges, we can conclude $(z,z)=1$. $\endgroup$ – Michael Renardy Mar 31 at 11:55

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