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Let $H=(H,(\cdot, \cdot))$ be a Hilbert space and $L:D(L) \subset H \longrightarrow H$ a linear operator densely defined. If $L$ is self-adjoint operator, then $L$ is coercive, that is, there exists $C>0$ such that $$(L(x),x)\geq C ||x||^2,\: \forall \: x \in D(L)?$$

I don't know if that's true. I couldn't prove it or set a counterexample.

Guilherme
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  • Did you mean to have a square on $|x|$ in your inequality? – Disintegrating By Parts Mar 31 '20 at 16:48
  • Yes, $||x||^2$. I fixed. – Guilherme Mar 31 '20 at 17:10
  • What about $L x = -x$? – PhoemueX Mar 31 '20 at 19:38
  • @PhoemueX Does not occur. – Guilherme Mar 31 '20 at 19:49
  • Well, this satisfies the assumptions that you put in your question. If you have some further properties you should state these. Otherwise we can only try to guess what you really want to ask. – PhoemueX Mar 31 '20 at 19:51
  • @PhoemueX by the hour, I don't know properties anymore. The ones I know are already stated, there are no guesswork and my question is very well put. – Guilherme Mar 31 '20 at 19:58
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    If you choose $D(L)=H$ and $L x := -x$ for all $x \in H$, then $L$ satisfies your assumptions (densely defined, linear, self adjoint operator), but not your conclusion. So either what you want to prove is wrong, or one of us is misunderstanding something. – PhoemueX Mar 31 '20 at 20:08
  • @PhoemueX But $D(L)$ is not necessarily equal to $H$. That's why I didn't put that hypothesis, because it doesn't necessarily happen. You're the one who's assuming that. I say the same for the other assumption you made. – Guilherme Mar 31 '20 at 20:13
  • @PhoemueX And I don't want to prove anything, I don't even know if this is true, under the hypotheses stated there. – Guilherme Mar 31 '20 at 20:14
  • It is not necessarily true that $D(L)=H$, but it might be. I have given you an example which satisfies all your assumptions, but not the desired conclusion. Therefore, you know that your desired conclusion is in general not true under your assumptions. – PhoemueX Apr 01 '20 at 06:00

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You inequality would imply that $$ \langle (L-(C+\epsilon)I)f,f\rangle \ge \epsilon \|f\|^2,\;\; f\in H,\;\; \epsilon > 0. $$ And that would force $(L-CI+\epsilon I)$ to be invertible for every $\epsilon > 0$, assuming that $L=L^*$. So, if your conjecture were true, then every self-adjoint operator would have a spectrum that is bounded below.

Disintegrating By Parts
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  • Suppose for instance $L$ has two eigenvalues $\lambda_1, \lambda_2 \in \mathbb{R}$, with $v_1,v_1 \in D(L)$ eigenvectors associated. Is true that, for all $w \in D(L)$ such that $(w,v_1)=(w,v_2)=0$ there exist $C>0$ so that $$(L(w),w)\geq C ||w||^2?$$ Is a consequence of the Spectral Theorem? – Guilherme Jul 29 '21 at 22:52
  • In the same way as here. – Guilherme Jul 29 '21 at 22:57
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What about $D(L) = H$ and $L(x) = 0$ for all $x \in H$?

gerw
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