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Let $A:D(A) \to \mathcal H$ be a positive self-adjoint operator and $\sqrt{A}$ defined by via the spectral theorem on $D(\sqrt{A}) = Q(A)$ where $Q(A)$ is the quadratic form domain. Let $$E=\inf\{\lVert\sqrt{A}u\rVert^2 : u \in D(\sqrt{A}), \lVert u \rVert = 1\}.$$ Assume there exists a minimizer $u_0 \in D(\sqrt{A})$ for $E$. Prove that $u_0\in D(A)$ and $Au_0 = E u_0$.

First I tried to show $u_0 \in D(A)$. Since $u_0 \in D(\sqrt{A}) = Q(A)$, it suffices to show $$\sup_{y \in Q(A)\\ \lVert y \rVert \leq 1}\lvert \langle u_0, Ay \rangle \rvert < \infty.$$ But I don't know how to proceed - We can write $$\langle u_0, A y \rangle = \langle \sqrt{A} u_0, \sqrt{A} y\rangle $$ but from here I don't know which estimate I can use. Any help appreciated, also any hint on the second part.

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Let $A=\int_0^{\infty} \lambda dP(\lambda)$ be the spectral decomposition of $A$. Then $u\in\mathcal{D}(\sqrt{A})$ iff $$ \|\sqrt{A}u\|^2= \int_{0}^{\infty}\lambda d\|P(\lambda)u\|^2 < \infty. $$

Suppose $\lambda_0 = \inf \{ \|\sqrt{\lambda}u\| : u\in\mathcal{D}(\sqrt{A}),\;\; \|u\|=1 \}$. If $u_0$ is a minimizer, meaning that $\|u_0\|=1$, $u_0\in\mathcal{D}(\sqrt{A})$, and $\|\sqrt{A}u_0\|=\lambda_0$, then it's not hard to see that $\mu(S)=\|P(S)u_0\|^2$ is a probability measure that must be concentrated at $\{\lambda_0\}$. Otherwise, the probability measure $\mu$, which is concentrated on $[\lambda_0,\infty)$, could not satisfy the following: $$ \lambda_0=\int_{\lambda_0}^{\infty}\lambda d\mu(\lambda). $$

Therefore, $P\{\lambda_0\}u=u$ must hold and, hence, $\sqrt{A}u_0=\sqrt{\lambda_0}u_0$.

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  • $\begingroup$ Very nice approach. In the meantime I found an elementary solution using the fact that $t\mapsto \frac{\langle (u_0 + tv), A(u_0 + tv)\rangle}{\lVert u_0 +tv \rVert}$ is minimized at $t=0$ but yours is way more elegant :) $\endgroup$ – lasik43 Dec 19 '18 at 13:34
  • $\begingroup$ @bavor42 : The elementary solution would be nice to see. :) $\endgroup$ – DisintegratingByParts Dec 19 '18 at 14:05
  • $\begingroup$ I posted it! :) $\endgroup$ – lasik43 Dec 20 '18 at 12:53
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Here's an elementary solution:

Let $v\in D(\sqrt{A})$. Notice that the function $f: t \mapsto \frac{\lVert \sqrt{A}(u_0 + tv) \rVert^2 }{\lVert u_0 + tv \rVert^2}$ gets minimized at $t=0$ by assumption, so $f'(0) = 0$. Now expanding everything yields $$f'(t) = \frac{(2 \operatorname{Re } \langle \sqrt{A}u_0, \sqrt{A}v \rangle + 2t\lVert \sqrt{A} v \rVert^2)\lVert u_0 +tv \rVert^2 - (2 \operatorname{Re } \langle u_0, v\rangle + 2t \Vert v \rVert^2)\lVert \sqrt{A} (u_0 + tv) \rVert^2}{\lVert u_0 + tv \rVert^4}$$ and evaluating at $t = 0$ gives $$0 = 2 \operatorname{Re } \langle \sqrt{A} u_0 , \sqrt{A} v \rangle - 2\operatorname{Re } \langle u_0, v \rangle \lVert \sqrt{A} u_0 \rVert ^2,$$ or equivalently $$\operatorname{Re} \langle \sqrt{A}u_0, \sqrt{A}v\rangle = E \operatorname{Re }\langle u_0, v \rangle.$$ Doing the same with $iv$ instead of $v$ we get the same result for the imaginary part and together $$\langle \sqrt{A} u_0, \sqrt{A} v \rangle = \langle Eu_0, v \rangle.$$ Since $A$ is self-adjoint, $D(A) = D(A^*)$, so it suffices to prove $u_0 \in D(A^*).$ Indeed, we have $$\sup_{\substack{v\in D(A) \\ \lVert v \rVert = 1}} \lvert \langle u_0, Av \rangle \rvert \leq \sup_{\substack{v\in D(\sqrt{A}) \\ \lVert v \rVert = 1}} \lvert \langle u_0, Av\rangle \rvert = \sup_{\substack{v\in D(\sqrt{A}) \\ \lVert v \rVert = 1}} \lvert \langle \sqrt{A}u_0, \sqrt{A}v\rangle \rvert = \sup_{\substack{v\in D(\sqrt{A}) \\ \lVert v \rVert = 1}} \lvert \langle Eu_0, v \rangle \rvert = E < \infty,$$ implying $u_0 \in D(A^*) = D(A)$. Then we have for each $v\in D(A)$ $$\langle Eu_0 , v \rangle = \langle \sqrt{A} u_0 , \sqrt{A} v \rangle = \langle A u_0 , v \rangle$$ and since $D(A)$ is dense $Eu_0 = Au_0$, as desired.

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    $\begingroup$ Very nice. +1 from me. It's good to have a solution that does not require the Spectral Theorem. $\endgroup$ – DisintegratingByParts Dec 20 '18 at 16:25

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