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Let

  • $d\in\mathbb N$
  • $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
  • $\Omega\subseteq\mathbb R^d$ be open
  • $H:=L^2(\Omega,\mathbb R^d)$
  • $U$ be a separable $\mathbb R$-Hilbert space
  • $Q:U\to H$ be a Hilbert-Schmidt Operator

It's easy to see that for $\lambda$-almost all $x\in\Omega$ $$\tilde Q(x)u:=(Qu)(x)\;\;\;\text{for }u\in U$$ is a Hilbert-Schmidt Operator from $U$ to $\mathbb R^d$, since $$\int_\Omega\left\|\tilde Q(x)\right\|_{\operatorname{HS}(U,\:\mathbb R^d)}^2\;{\rm d}\lambda(x)=\left\|Q\right\|_{\operatorname{HS}(U,\:H)}^2<\infty\;.\tag 1$$

How can we determine the adjoint ${\tilde Q(x)}^\ast$ of $\tilde Q(x)$? In particular, how can we calculate $\tilde Q(x){\tilde Q(x)}^\ast e_i$ for the $i$th standard basis vector $e_i$ of $\mathbb R^d$?

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  • $\begingroup$ it means nothing to me. make things simple, nobody cares $Q$ is HS, it won't change anything that it is compact $\endgroup$ – reuns Jul 28 '16 at 14:21
  • $\begingroup$ @user1952009 The question is short. How can I further simplify it? Maybe it helps you, if you know my motivation: I've hoped that the special form of $\tilde Q$ would help me in my other question. $\endgroup$ – 0xbadf00d Jul 28 '16 at 14:24
  • $\begingroup$ what is $(Qu)(x)$ and $\tilde{Q}(x)u$ ? you know that $h \mapsto h(x)$ isn't a bounded operator ? but if you replace $x$ by ... then $h \mapsto h(x)$ becomes bounded and $\tilde{Q}$ is simply $Q^*$ $\endgroup$ – reuns Jul 28 '16 at 14:26
  • $\begingroup$ so no I don't want to help you if you make no efforts : make your questions clear $\endgroup$ – reuns Jul 28 '16 at 14:27
  • $\begingroup$ @user1952009 $Q$ is a given HS-operator from $U$ to $H=L^2(\Omega,\mathbb R^d)$ and hence $Qu$ is a function in $H$. $\endgroup$ – 0xbadf00d Jul 28 '16 at 14:38
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Just curious... Your eq (2) seems to be \begin{align*} \left\langle u,\widetilde{Q}\left(x\right)^{*}y\right\rangle _{U} & =\left\langle \widetilde{Q}\left(x\right)u,y\right\rangle _{\mathbb{R}^{d}}, \end{align*} for $x,y\in\mathbb{R}^{d}$, and $u\in U$.

Fix $x,y\in\mathbb{R}^{d}$, and choose an ONB $\left\{ \varphi_{j}\right\} $ in $U$. (I'm going to assume $\left\langle \cdot,\cdot\right\rangle $ is linear in the second variable.) Then \begin{align*} \widetilde{Q}\left(x\right)^{*}y & =\sum_{j}\left\langle \varphi_{j},\widetilde{Q}\left(x\right)^{*}y\right\rangle _{U}\varphi_{j}=\sum_{j}\left\langle \widetilde{Q}\left(x\right)\varphi_{j},y\right\rangle _{\mathbb{R}^{d}}\varphi_{j}. \end{align*} So this gives \begin{align*} \widetilde{Q}\left(x\right)^{*} & =\sum_{j}\left|\varphi_{j}\left\rangle \right\langle \widetilde{Q}\left(x\right)\varphi_{j}\right|=\sum_{j}\left|\varphi_{j}\left\rangle \right\langle \left(Q\varphi_{j}\right)\left(x\right)\right|. \end{align*}

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  • $\begingroup$ Clearly, there was a typo in $(2)$ and the equation should be as you wrote. $\endgroup$ – 0xbadf00d Jul 31 '16 at 9:54
  • $\begingroup$ I don't understand the notation in your last equation. $\endgroup$ – 0xbadf00d Jul 31 '16 at 10:06
  • $\begingroup$ The last line is a notation for rank-1 operators. So $\left|\varphi_{j}\left\rangle \right\langle \left(Q\varphi_{j}\right)\left(x\right)\right|:\mathbb{R}^{d}\longrightarrow U$, by \begin{align*} \mathbb{R}^{d} & \in y\longmapsto\left\langle \left(Q\varphi_{j}\right)\left(x\right),y\right\rangle \varphi_{j}\in U. \end{align*} $\endgroup$ – James Jul 31 '16 at 10:35

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