• $U$ and $H$ be real Hilbert spaces
  • $\iota:U\to H$ be a Hilbert-Schmidt embedding
  • $Q:=\iota\iota^\ast$

Can we find a concrete representation of $Qy$ for some $y\in H$?

By Riesz' representation theorem, $\exists!v\in U$ with $$\langle\iota u,y\rangle_H=\langle u,v\rangle_U\;\;\;\text{for all }u\in U\tag 1\;.$$ By definition of the adjoint, $$\iota^\ast y=v\;.\tag 2$$


  • $U\subseteq H$,
  • $\langle\;\cdot\;,\;\cdot\;\rangle_U$ is the restriction of $\langle\;\cdot\;,\;\cdot\;\rangle_H$ to $U$ and
  • $\iota$ is the inclusion map,

then $(1)$ is equivalent to $$\langle u,y-v\rangle_H=0\;\;\;\text{for all }u\in U\;.\tag 3$$ Since $(3)$ must hold for $u=y-v$, we obtain $v=y$ and hence $$Qy=\iota(\iota^\ast y)=\iota v=v=y\;.\tag 4$$

The questions are:

  1. Did I made any mistake in the special case?
  2. Can we find a concrete representation of $Qy$ in the general case?

$(3)$ must not hold for $u=y-v$ (only if $y\in U$). That was my mistake.

Let $U$ and $H$ be Hilbert spaces and $\iota$ be an embedding of $U$ into $H$. Then, $$\pi x:=u\;\;\;\text{for }x\in H\text{ with }x=\iota u+y\text{ for some }u\in U\text{ and }y\in\left(\iota U\right)^\perp$$ is a well-defined mapping $H\to U$. If $\iota$ is an isometry, then $$\langle\iota u,x\rangle_H=\langle\iota u,\iota\left(\pi x\right)\rangle_H=\langle u,\pi x\rangle_U\;\;\;\text{for all }u\in U\text{ and }x\in H\;,$$ i.e. the adjoint of $\iota$ is given by $$\iota^\ast=\pi\;.$$ So, if $U\subseteq H$ and $\iota$ is the inclusion, $$Q=\pi\;.$$


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