On the Adjoint of Linear Relations in Hilbert Spaces

Abstract

Assume that ${\mathfrak {H}}$ and ${\mathfrak {K}}$ are two real or complex Hilbert spaces, A a linear relation from ${\mathfrak {H}}$ to ${\mathfrak {K}}$, and B a linear relation from ${\mathfrak {K}}$ to ${\mathfrak {H}}$, respectively. Necessary and sufficient conditions for B to be equal to the adjoint of A are provided. Several consequences are also presented. More precisely, new characterizations for closed, skew–adjoint, selfadjoint, normal linear relations, and generalized orthogonal projections are obtained.

Appendix

The next result is useful with respect to the multiplication of $2 \times 2$ matrices of linear relations.

Lemma 9.1

Assume that ${\mathfrak {H}}$ and ${\mathfrak {K}}$ are two real ar complex Hilbert spaces and let $A \in {\mathfrak {L}}({\mathfrak {H}},{\mathfrak {K}})$ and $A \in {\mathfrak {L}}({\mathfrak {K}},{\mathfrak {H}})$ The following identity holds true:

$$\begin{aligned} \left( \begin{array}{cc} I &{}\quad -B \\ A &{}\quad I \\ \end{array} \right) \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right) = \left( \begin{array}{cc} I+BA &{}\quad \mathrm{dom\,}B \times \mathrm{mul\,}B \\ \mathrm{dom\,}A \times \mathrm{mul\,}A &{}\quad I + AB \\ \end{array} \right) . \end{aligned}$$

(9.12)

Proof

Assume that $\left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I &{}\quad -B \\ A &{}\quad I \\ \end{array} \right) \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right)$, so that

$$\begin{aligned} \left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} a \\ b \end{array} \right) \right\} \in \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right) , \quad \left\{ \left( \begin{array}{c} a \\ b \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I \,\,&{} -B \\ A \,\,&{} I \\ \end{array} \right) , \end{aligned}$$

for some $a \in {\mathfrak {H}}$ and $b \in {\mathfrak {K}}$. Therefore, $a = x + y^{\prime }$ and $b = -x^{\prime } +y$ for some $\{x,x^{\prime }\} \in A$ and $\{y,y^{\prime }\} \in B$. Also $z = a-b^{\prime }$ and $t = a^{\prime } + b$ for some $\{a,a^{\prime }\} \in A$ and $\{b,b^{\prime }\} \in B$ . Consequently, $z = x + y^{\prime } - b^{\prime }$ and $t = a^{\prime } - x^{\prime } + y$. Furthermore, $\{x,x^{\prime }\} \in A$ implies that $\{x,y-b\} \in A$ and:

$$\begin{aligned} \{y-b, y^{\prime }-b^{\prime }\} = \{y,y^{\prime }\} - \{b,b^{\prime }\} \in B \end{aligned}$$

so that $\{x,y^{\prime }-b^{\prime }\} \in AB$, which further leads to:

$$\begin{aligned} \{x,x+y^{\prime }-b^{\prime }\} = \{x,z\} \in I + AB. \end{aligned}$$

(9.13)

Similar arguments shows that:

$$\begin{aligned} \{y,t\} \in I + BA. \end{aligned}$$

(9.14)

Since also:

$$\begin{aligned} \{y,0\} \in \mathrm{dom\,}B \times \mathrm{mul\,}B, \quad \{x,0\} \in \mathrm{dom\,}A \times \mathrm{mul\,}A, \end{aligned}$$

(9.15)

it follows from (9.13), (9.14) and (9.15) that:

$$\begin{aligned} \left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I + BA &{}\quad \mathrm{dom\,}B \times \mathrm{mul\,}B \\ \mathrm{dom\,}A \times \mathrm{mul\,}A &{}\quad I + AB \\ \end{array} \right) . \end{aligned}$$

Thus:

$$\begin{aligned} \left( \begin{array}{cc} I &{} -B \\ A &{} I \\ \end{array} \right) \left( \begin{array}{cc} I &{} B \\ -A &{} I \\ \end{array} \right) \subset \left( \begin{array}{cc} I+BA &{}\quad \mathrm{dom\,}B \times \mathrm{mul\,}B \\ \mathrm{dom\,}A \times \mathrm{mul\,}A &{}\quad I + AB \\ \end{array} \right) . \end{aligned}$$

(9.16)

Conversely, it will be shown that:

$$\begin{aligned} \left( \begin{array}{cc} I+BA &{} \mathrm{dom\,}B \times \mathrm{mul\,}B \\ \mathrm{dom\,}A \times \mathrm{mul\,}A &{} I + AB \\ \end{array} \right) \subset \left( \begin{array}{cc} I &{}\quad -B \\ A &{}\quad I \\ \end{array} \right) \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right) . \end{aligned}$$

(9.17)

Assume that $\left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I + BA &{} \mathrm{dom\,}B \times \mathrm{mul\,}B \\ \mathrm{dom\,}A \times \mathrm{mul\,}A &{} I + AB \\ \end{array} \right) ,$ so that $z = x^{\prime } + m_{2}$ and $t = y^{\prime } + m_{1}$ for some $\{x,x^{\prime }\} \in I + BA$, $\{y,y^{\prime }\} \in I + AB$, $m_{1} \in \mathrm{mul\,}A$ and $m_{2} \in \mathrm{mul\,}B$. It follows from $\{x,x^{\prime }\} \in I + BA$ that $\{x,x^{\prime }-x\} \in BA$, so that $\{x,a\} \in A$, $\{a,x^{\prime }-x\} \in B$ for some $a \in {\mathfrak {K}}$. Also, it follows from $\{y,y^{\prime }\} \in I+AB$ that $\{y,y^{\prime }-y\} \in AB$, so that $\{y,b\} \in B$, $\{b,y^{\prime }-y\} \in A$ for some $b \in {\mathfrak {H}}$. Then, clearly:

$$\begin{aligned} \left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} x + b \\ -a + y \end{array} \right) \right\} \in \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right) . \end{aligned}$$

(9.18)

Furthermore, the following relations:

$$\begin{aligned} \{x+b,a+y^{\prime }-y+m_{1}\}= & {} \{x,a\} + \{b,y^{\prime }-y\} + \{0,m_{1}\} \in A, \end{aligned}$$

(9.19)

$$\begin{aligned} \{-a+y,b-x^{\prime }+x-m_{2}\}= & {} - \{a,x^{\prime }-x\} + \{y,b\} - \{0,m_{2}\} \in B,\nonumber \\ \end{aligned}$$

(9.20)

imply that:

$$\begin{aligned}&\left\{ \left( \begin{array}{c} x + b \\ -a + y \end{array} \right) , \left( \begin{array}{c} x+b-b+x^{\prime }-x+m_{2} \\ a+y^{\prime }-y+m_{1}-a+y \end{array} \right) \right\} \nonumber \\&\quad = \left\{ \left( \begin{array}{c} x + b \\ -a + y \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I &{}\quad -B \\ A &{}\quad I \\ \end{array} \right) . \end{aligned}$$

(9.21)

A combination of (9.18) and (9.21) shows that:

$$\begin{aligned} \left\{ \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} z \\ t \end{array} \right) \right\} \in \left( \begin{array}{cc} I &{}\quad -B \\ A &{}\quad I \\ \end{array} \right) \left( \begin{array}{cc} I &{}\quad B \\ -A &{}\quad I \\ \end{array} \right) , \end{aligned}$$

so that (9.17) has been proved. Finally, it follows from (9.16) and (9.17) that (9.12) holds true. $\square$