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I have come across a linear bounded operator $B:K\to \mathcal{L}(U,Z)$ where $K$, $U$, and $Z$ are separable Hilbert spaces. I need a reference (any source) to find out about:

  • The adjoint of such an operator, something that enables me to write, roughly speaking, $$\langle z,(Bk)u\rangle_Z= \langle (B^*u)z,k\rangle_K$$

  • A way to expand an element $B(k)\in \mathcal{L}(U,Z)$ based on eigen functions of $U$ and $Z$ (if separable).

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  • $\begingroup$ This question deserves a more specific title but I can't come up with one at the moment. $\endgroup$
    – j.c.
    Commented Mar 27, 2018 at 23:28

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For fixed $u \in U$, the map that takes $(z,k)$ to $\langle z,(Bk)u\rangle_Z$ is a bounded sesquilinear form on $Z\times K$. (Bounded because $|\langle z,(Bk)u\rangle| \leq \|z\|\|(Bk)u\| \leq \|B\|\|u\|\|z\|\|k\|$.) Therefore there exists a bounded linear operator $B^*u$ from $Z$ to $K$ satisfying $\langle (B^*u)z,k\rangle_K = \langle z,(Bk)u\rangle_Z$ for all $(z,k) \in Z\times K$. That's a standard consequence of the Riesz representation theorem. I can't really make sense of what is meant by "eigenfunction of $U$" in the second question.

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  • $\begingroup$ I meant a basis for U, and Z as well. $\endgroup$
    – Saj_Eda
    Commented Jan 26, 2018 at 21:02
  • $\begingroup$ Well, if $(e_i)$ is an orthonormal basis for $U$ and $(f_j)$ is an orthonormal basis for $Z$ then the matrix entries for $Bk$ would be the values $\langle (Bk)e_i, f_j\rangle$. $\endgroup$
    – Nik Weaver
    Commented Jan 26, 2018 at 22:21
  • $\begingroup$ Is it possible to find out what $B^*$ yields acting on particular elements $z=(z_i)$, $k=(k_i)$, $u=(u_i)$ of $\ell ^2$? $\endgroup$
    – Saj_Eda
    Commented Jan 26, 2018 at 23:50

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