Adjoints

Definition 2.2.1   Let $\mathcal{H}, \mathcal{K}$ be a Hilbert space. Then $u : \mathcal{H}\rightarrow \mathcal{K}\rightarrow \mathbb{F}$ is a sesquilinear form. if $u$ is linear in $\mathcal{H}$ and conjugate linear in $\mathcal{K}$, i.e.,

$$u(\alpha x + \beta y, z ) = \alpha u(x,z) + \beta u(y,z)$$

$$u(x,\alpha y + \beta z) = \bar \alpha u(x,y) + \bar \beta u(x,z).$$

We say a sesquilinear form is bounded if there is $M > 0$ such that $\vert u(h,k)\vert \leq M \Vert h\Vert \Vert k\Vert$ for all $h \in \mathcal{H}, k \in \mathcal{K}$.
Define $\Vert u\Vert$ to be the least such $M$.

Lemma 2.2.2   There is a bijective isometric correspondence between $B(\mathcal{H},\mathcal{K})$ and the bounded sesquilinear forms on $\mathcal{H}\times \mathcal{K}$, given by $T \in B(\mathcal{H}, \mathcal{K})$ maps to $u_T$ where $u_T(x,y) = \left\langle{ x, Ty }\right\rangle$.

Proof. First, $u_T$ is a bounded sesqulinear form; sesquilinear follows from linearity of $T$ and $\left\langle{ \cdot, \cdot }\right\rangle$ is a sesquilinear form.
To see $u_T$ is bounded,

$$\Vert u_T\Vert = \sup \{ \vert u_T(x,y)\vert : \Vert x\Vert \leq 1, \Vert y\Vert \leq 1 \}$$

$$= \sup\{ \vert\left\langle{ x, Ty }\right\rangle \vert :\Vert x\Vert \leq 1, \Vert y\Vert \leq 1 \}$$

$$\leq \sup \{ \Vert x\Vert \Vert Ty\Vert : \Vert x\Vert \leq 1, \Vert y\Vert \leq 1 \}$$

$$\leq \sup \{ \Vert x\Vert \cdot \Vert T\Vert \cdot \Vert y\Vert \Vert x\Vert \leq 1, \Vert y\Vert \leq 1 \} = \Vert T\Vert$$

Next, to see $\Vert u_T\Vert = \Vert T\Vert$,

$$\Vert Tx\Vert^2 = \left\langle{ Tx, Tx }\right\rangle = u_T(Tx,x) \leq \Vert u_T\Vert \cdot \Vert Tx\Vert \cdot \Vert x\Vert$$

Thus, $\Vert Tx\Vert \leq \Vert u_T\Vert\cdot \Vert x\Vert$ for all $x$. So, $\Vert T\Vert \leq \Vert u_T\Vert$.

Finally, for $u$ a sesquilinear form, we must find $T$ such that $u = u_T$. For $y \in \mathcal{H}$, define the function $L_y$ on $\mathcal{K}$ by

$$L_y(x) = u(x,y)$$

Since $u$ is bounded, $L_y$ is bounded, and clearly $L_y$ is linear.

By the Riesz Representation Theorem, there is a vector $T_y \in \mathcal{K}$ such that $L_y(x) = \left\langle{ x, Ty }\right\rangle$. This defines a map $T : \mathcal{H}\rightarrow \mathcal{K}$ by $y \mapsto Ty$.

1. $T$ is linear. For, $y,z \in \mathcal{H}$, $x \in sK$,
   $$\left\langle{ x, T(y + z) }\right\rangle = L_{y+z}(x) = u(x,y+z) = u(x,y) + u(x,z)$$
   $$= \left\langle{ x, Ty }\right\rangle + \left\langle{ x, Tz }\right\rangle = \left\langle{ x, Ty + Tz }\right\rangle$$
   Similarly, $T(\alpha y) = \alpha T(y)$ (conjugate twice).
   

2. Exercise: $\Vert T\Vert \leq \Vert u_T\Vert$.
   

   Thus, $T \in B(\mathcal{H}, \mathcal{K})$ and so $u = u_T$.
   

$\qedsymbol$

Theorem 2.2.3   For each $T \in B(\mathcal{H}, \mathcal{K})$, there is a unique operator $T^* \in B(\mathcal{K},\mathcal{H})$ such that $\left\langle{ Tx, y }\right\rangle = \left\langle{ x, T^* y }\right\rangle$ for all $x \in \mathcal{H}$ and $y \in \mathcal{K}$.

Proof. $u(x,y) = \left\langle{ Tx, y }\right\rangle$ is a bounded sesquilinear form on $\mathcal{H}\times \mathcal{K}$. Apply Lemma. $\qedsymbol$

Properties

1. For $U \in B(\mathcal{H},\mathcal{K})$, the $U$ is an isomorphism if and only if $U$ is invertible and $U^{-1} = U^*$.
   Proof. For the forward direction, $U$ is clearly invertible..
   For $x \in \mathcal{H}, y \in \mathcal{K}$,

   $$\left\langle{ x, U^* y }\right\rangle = \left\langle{ Ux, y }\rig... ...e{ Ux, U ( U^{-1} y) }\right\rangle = \left\langle{ x, U^{-1} y }\right\rangle$$

   Since this is true for all $x$,

   $$U^* y = U^{-1} y$$

   for all $y$, so $U^* = U^{-1}$.
   For the other direction, note that for $x,y \in \mathcal{H}$,

   $$\left\langle{ Ux, Uy }\right\rangle = \left\langle{ U^* U x, y }\... ...= \left\langle{ U^{-1} U x y }\right\rangle = \left\langle{ x,y }\right\rangle$$

   Since $U$ is invertible, it is onto and so $U$ is a surjective isometry. $\qedsymbol$

2. $(\alpha A + B)^* = \bar \alpha A^* + B^*$
3. $(AB)^* = B^* A^*$
4. $(A^*)^* = A$
5. If $A$ is invertible, $(A^*)^{-1} = (A^{-1})^*$.
   Proof.

   $$\left\langle{ x,y }\right\rangle = \left\langle{ A A^{-1} x, y }\... ...{-1} x, A^*y }\right\rangle = \left\langle{ x, (A^{-1})^* A^* y }\right\rangle$$

   So $I = (A^{-1})^* A^*$. Similarly, $I = A^*(A^{-1})^*$. Thus, $A^*$ is invertible and $(A^*)^{-1} = (A^{-1})^*$. $\qedsymbol$

6. $\Vert A \Vert = \Vert A^*\Vert = \Vert A^* A\Vert^\frac12$
   Proof.

   $$A^* A \leq \Vert A^*\Vert \Vert A\Vert$$

   is clear from the definition.

   $$\Vert Ax\Vert^2 = \left\langle{ Ax, Ax }\right\rangle = \left\langle{ A^*Ax, x }\right\rangle \leq \Vert A^* A\Vert \Vert x\Vert^2$$

   Therefore,

   $$\Vert Ax \Vert \leq \Vert A^*A\Vert^\frac12 \Vert x\Vert$$

   So $\Vert A\Vert \leq \Vert A^* A\Vert^\frac12$.
   By (a) and (b), $\Vert A\Vert^2 \leq \Vert A^* A\Vert \leq \Vert A\Vert \Vert A^* \Vert$, so $\Vert A\Vert \leq \Vert A^*\Vert$. In particular, $\Vert A^*\Vert \leq \Vert(A^*)^*\Vert = \Vert A\Vert$. $\qedsymbol$

Examples:

1. $\dim \mathcal{H}= n < \infty$. For $A$ with matrix $[a_{i,j}]$, $A^*$ has matrix $[b_{i,j}]$ where $b_{i,j} = \overline{a_{ji}}$.
2. For a multiplication operator $M_\phi$,
   $$M_\phi^* = M_{\bar \phi}.$$

3. For an integral operator $K$ with kernel $k$, then $K^*$ is an integral operator with kernel $(x,y) \mapsto \overline{k(y,x)}$.
4. For the unilateral shift $S$ on $\ell^2(\mathbb{N})$, $S^*$ is given by $S^*(\alpha_1, \alpha_2, \ldots) = (\alpha_2, \alpha_3, \ldots )$. $S^*$ is called the backward shift.
   Proof. One way - use matrices.
   Second -

   $$\left\langle{ S^*(\alpha_n), (\beta_n) }\right\rangle = \left\lan... ...\right\rangle = \left\langle{ (\alpha_n), (0, \beta_1, \beta_2) }\right\rangle$$

   $$= \alpha_2 \bar \beta_1 + \alpha_3 \bar \beta_2 + \alpha_4 \bar \beta_3 + \cdots$$

   $$= \left\langle{ (\alpha_2, \alpha_3, \ldots), (\beta_1, \beta_2, \ldots ) }\right\rangle$$

   Since this is true for all possible $(\beta_n) \in \ell^2(\mathbb{N})$, $S^*(\alpha_1, \alpha_2, \ldots) = (\alpha_2, \alpha_3, \ldots )$. $\qedsymbol$

Definition 2.2.4   $A \in \mathcal{B}(\mathcal{H})$ is hermitian (or self-adjoint) if $A = A^*$. $A \in \mathcal{B}(\mathcal{H})$ is normal if $A A^* = A^* A$. Recall $A$ is unitary if $A^* = A^{-1}$.

Examples

1. Unitaries and hermitians are normal
2. $M_\phi$ is always normal . $M_\phi$ is hermitian if and only if $\phi$ is real-valued a.e. $M_\phi$ is unitary iff $\vert\phi\vert = 1$ a.e.
3. An integral operator is hermitian if and only if $k(x,y) = \overline{k(y,x)}$ almost everywhere on $\mu \times \mu$.
4. $S$ is not normal.

Properties

1. For $A \in \mathcal{B}(\mathcal{H})$, the following are equivalent:

   (a)

   $A$ is an isometry

   (b)

   $A^* A = I$

   (c)

   $\left\langle{ Ag,Ah }\right\rangle = \left\langle{ g,h }\right\rangle$ for all $g,h \in \mathcal{H}$

2. For $A \in \mathcal{B}(\mathcal{H})$, the following are equivalent:

   (a)

   $A$ is unitary.

   (b)

   $A^* A = A A^* = I$

   (c)

   $A$ is a normal isometry.

3. For all $A \in \mathcal{B}(\mathcal{H})$, the following are equivalent:

   (a)

   $A$ is normal

   (b)

   $\Vert Ah\Vert = \Vert A^* h\Vert$ for all $h$

   (c)

   If $\mathcal{H}$ is a $\mathbb{C}$-Hilbert space, $Re A$ and $Im A$ commute.

   To prove 3, we need:
   Proposition 2.2.5   For $A \in \mathcal{B}(\mathcal{H})$, $A = A^*$, $\Vert A\Vert = \sup \{ \vert\left\langle{ Ah, h }\right\rangle : h \in \mathcal{H}\}$ provided $\mathcal{H}$ is a $\mathbb{C}$-Hilbert space.
   Proof. Let $M = \sup\{ \vert\left\langle{ Ah, h }\right\rangle \vert : \Vert h\Vert = 1$. Clearly, $\vert\left\langle{ Ah, h }\right\rangle \vert \leq \Vert Ah\Vert\Vert h\Vert \leq \Vert A\Vert\Vert h\Vert^2$, so $M \leq \Vert A\Vert$.

   $$\left\langle{ A(h + g), (h + g) }\right\rangle = \left\langle{ Ah... ...ngle + \left\langle{ Ag, h }\right\rangle + \left\langle{ Ag, g }\right\rangle$$

   $$= \left\langle{ Ah, h }\right\rangle + \left\langle{ Ah, g }\righ... ...e + \left\langle{ g, A^* h }\right\rangle + \left\langle{ Ag, g }\right\rangle$$

   $$= \left\langle{ Ah, h }\right\rangle + 2 Re \left\langle{ Ah, g }\right\rangle + \left\langle{ Ag, g }\right\rangle$$

   Similarly,

   $$\left\langle{ A(h-g), (h - g) }\right\rangle = \left\langle{ Ah, ... ... - 2 Re\left\langle{ Ah, g }\right\rangle + \left\langle{ Ag, g }\right\rangle$$

   Thus,

   $$4 Re \left\langle{ Ah, g }\right\rangle = \left\langle{ A(h+g), (h+g) }\right\rangle - \left\langle{ A(h-g), h-g }\right\rangle$$

   As $\vert\left\langle{ Af, f }\right\rangle \vert \leq M \Vert f\Vert^2$,

   $$4 Re \left\langle{ Ah, g }\right\rangle \leq M (\Vert h+g\Vert^2 + \Vert h-g\Vert^2 ) = 2M( \Vert h\Vert^2 + \Vert g\Vert^2 )$$

   If $\Vert h\Vert, \Vert g\Vert$ are $1$, then $4 Re\left\langle{ Ah, g }\right\rangle \leq 4 M$. If $\left\langle{ Ah, g }\right\rangle = e^{i \theta} \vert\left\langle{ Ah, g }\right\rangle \vert$, then we can replace $h$ with $e^{-i \theta} h$ to get $\vert\left\langle{ Ah, g }\right\rangle \vert \leq M$. Letting $g = \frac{Ah}{\Vert Ah\Vert}$, we get $\Vert Ah\Vert \leq M$ for all $h \in \mathcal{H}$ with $\Vert h\Vert = 1$. Thus, $\Vert A\Vert \leq M$.
   Corollary, if $A = A^*$ and $\left\langle{ Ah, h }\right\rangle = 0$ for all $h \in \mathcal{H}$, then $A = 0$. $\qedsymbol$
   To prove 3, first notice that
   $$\Vert Ah\Vert^2 - \Vert A^*h\Vert^2 = \left\langle{ Ah, Ah }\righ... ...ft\langle{ A^* A h, h }\right\rangle - \left\langle{ A A^* h, h }\right\rangle$$
   $$= \left\langle{ (A^* A - A A^*)h, h }\right\rangle$$
   Thus $\Vert Ah\Vert = \Vert A^* h\Vert$ for all $h$ if and only if $A^* A = A^* A$. So, (a) and (b) are equivalent. For (a) implies (c), multiply $Re A \cdot Im A - IM A \cdot Re A$ For (c) implies (b), if $A = B + iC$,
   $$A^* A = B^2 - i CB + i BC + C^2$$
   $$A A^* = B^2 + i CB - i BC + C^2$$
   Thus, $A^* A - A A^* = 0$ if and only if $CB = BC$.

Wednesday, Sept. 21, 2005:

Proposition 2.2.6   Let $\mathcal{H}$ be a $\mathbb{C}$-Hilbert space, $A \in \mathcal{B}(\mathcal{H})$. Then, $A = A^*$ if and only if $\left\langle{ Ah,h }\right\rangle \in \mathbb{R}$ for all $h \in \mathcal{H}$.

Proof.

$\Rightarrow$

$\left\langle{ Ah,h }\right\rangle = \left\langle{ h,Ah }\right\rangle = \overline{Ah,h}$.

$\Leftarrow$

We must use the usual trickery of expanding and canceling:

$$\left\langle{ A(h + \alpha g), h+ \alpha g }\right\rangle = \left... ...ht\rangle + \vert\alpha\vert^2 \left\langle{ Ag,g }\right\rangle \in \mathbb{R}$$

Note $\left\langle{ Ah,h }\right\rangle$, $\left\langle{ Ag,g }\right\rangle \in \mathbb{R}$. So,

$$\bar \alpha \left\langle{ Ah,g }\right\rangle + \alpha \left\langle{ Ag,h }\right\rangle \in \mathbb{R}.$$

But,

$$\bar \alpha \left\langle{ Ah, g }\right\rangle + \alpha \left\lan... ...ft\langle{ h, A^*g }\right\rangle + \alpha \left\langle{ g,A^*h }\right\rangle$$

$$= \bar \alpha \overline{\left\langle{ Ag,h }\right\rangle } + \al... ...gle{ A^*g, h }\right\rangle + \bar \alpha \left\langle{ A^*h, g }\right\rangle$$

for all complex numbers $\alpha$. Then, ``equating coefficients",

$$\left\langle{ Ah,g }\right\rangle = \left\langle{ A^*h, g }\right\rangle$$

$$\left\langle{ Ag,h }\right\rangle = \left\langle{ A^*g, h }\right\rangle$$

$\qedsymbol$

Theorem 2.2.7   For $A \in \mathcal{B}(\mathcal{H})$,

$$\ker A = (ran A^*)^\perp$$

Proof. Let $h \in \mathcal{H}$. Then, $h \perp ran A^*$ iff $\left\langle{ h, A^*g }\right\rangle = 0$ for all $g \in \mathcal{H}$ iff $\left\langle{ Ah, g }\right\rangle = 0$ for all $g \in \mathcal{H}$ iff $Ah = 0$. $\qedsymbol$

Corollary 2.2.8  

$$(\ker A)^\perp = \overline{ran A^*}$$

$$(\ker A^*)^\perp = \overline{ran A}$$

$$(\ker A^* = (ran A)^\perp$$

Addition to text:

Definition 2.2.9   $A \in \mathcal{B}(\mathcal{H})$ is positive, written $A \geq 0$ iff $A = A^*$ and $\left\langle{ Ah,h }\right\rangle \geq 0$ for all $h \in \mathcal{H}$. If $\mathcal{H}$ is a $\mathbb{C}$-Hilbert space, $\left\langle{ Ah,h }\right\rangle \geq 0$ and therefore $A = A^*$.

We can extend this to a partial order on self adjoint elements of $\mathcal{B}(\mathcal{H})$ by $A \geq B$ iff $A - B \geq 0$.

Properties:

1. $S,T \geq 0 \implies S + T \geq 0$
2. $\alpha \geq 0, S \geq 0 \implies \alpha S \geq 0$
3. $S,T \geq 0$ does not imply $ST \geq 0$ even if $ST$ is self-adjoint.
4. $T^*T \geq 0$ for all $T \in \mathcal{B}(\mathcal{H})$ Proof:
   $$\left\langle{ T^*T h, h }\right\rangle = \left\langle{ Th,Th }\right\rangle \geq 0$$