# Minimizer of square root operator norm

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.

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, $$\sqrt{A}u_0=\sqrt{\lambda_0}u_0$$.

• 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 :) – lasik43 Dec 19 '18 at 13:34
• @bavor42 : The elementary solution would be nice to see. :) – DisintegratingByParts Dec 19 '18 at 14:05
• I posted it! :) – lasik43 Dec 20 '18 at 12:53

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.

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