Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

## G. Positive Operators

Definition II.61

Call $$$A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ positive if $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\geq 0$$$ for all $$$x\in H$$$ and write it as $$$A\geq 0$$$. Note: Positive implies self-adjoint.

Theorem II.62

The following are equivalent:

(a)
$$$A\geq 0$$$.
(b)
$$$A= A^{*}$$$ and $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left[0, {+}\infty\right)\mathclose{}$$$.
(c)
$$$A= {B}^{2}$$$ for some $$$B\geq 0$$$.
(d)
$$$A= T^{*}T$$$ for some $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$.

Proof.

1. ((a) ⇒ (b)) $$$A= A^{*}$$$ follows immediately from Proposition II.54, so $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathbb{R}$$$. For $$$r\gt 0$$$ we have $$${-}r\notin \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ because $$$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}\mathopen{}\left(r+A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\right\rvert\mathclose{}= r\mathopen{}\left\langle{}x, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\geq r{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}$$$ and so $$$r+A\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1}$$$ by Lemma II.53.
2. ((b) ⇒ (c)) Notice that the square root function belongs to $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$. Use functional calculus. Let $$$B= \sqrt{A}= {A}^{\frac{1}{2}}$$$. Notice that $$$\mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}{A}^{\frac{1}{4}}\mathopen{}\left( x\right)\mathclose{}, {A}^{\frac{1}{4}}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}$$$, so $$$B\geq 0$$$.
3. ((c) ⇒ (d)) is obvious.
4. ((d) ⇒ (a)) $$$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}\geq 0$$$.

Theorem II.63

For any self-adjoint operator $$$A$$$ and any function $$$f\in C\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$, we have $$$\mathop{\sigma}\mathopen{}\left( f\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, f\mathopen{}\left( t\right)\mathclose{}\,\middle\vert\, , t\in \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}, \,\right\}\mathclose{}= f\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$.

Proof. Notice that if $$$f$$$ does not vanish on $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$, then $$$f\mathopen{}\left( A\right)\mathclose{}$$$ is invertible; the inverse is of course $$$\mathopen{}\left(\frac{1}{f}\right)\mathclose{}\mathopen{}\left( A\right)\mathclose{}$$$. In the other direction, let's show that if $$$f\mathopen{}\left( {t}_{0}\right)\mathclose{}= 0$$$ for some $$${t}_{0}\in \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$, then $$$f\mathopen{}\left( A\right)\mathclose{}$$$ cannot be invertible. We obtain positive $$${ε}_{1}$$$, $$${ε}_{2}$$$, … such that $$$\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\lt \frac{1}{n}$$$ for $$$t\in \mathopen{}\left({t}_{0}-{ε}_{n}, {t}_{0}+{ε}_{n}\right)\mathclose{}\cap \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$. Restricting tent functions with tent poles at $$${t}_{0}$$$ and stakes at $$${t}_{0}\pm {ε}_{0}$$$ to $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$, we obtain functions $$${g}_{1}$$$, $$${g}_{2}$$$, … in $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$ such that $$$0\leq {g}_{n}\leq 1$$$, $$${g}_{n}\mathopen{}\left( {t}_{0}\right)\mathclose{}= 1$$$, and $$${g}_{n}\equiv 0$$$ off $$$\mathopen{}\left({t}_{0}-{ε}_{n}, {t}_{0}+{ε}_{n}\right)\mathclose{}$$$. Then $$$\mathopen{}\left\lvert{}f{g}_{n}\right\rvert\mathclose{}\leq \frac{1}{n}$$$, while the uniform norm of $$${g}_{n}$$$ is $$$1$$$ for all $$$n$$$. The functional calculus is isometric between the uniform norm and the operator norm, so $$$\mathopen{}\left\lVert{}f\mathopen{}\left( A\right)\mathclose{}{g}_{n}\mathopen{}\left( A\right)\mathclose{}\right\rVert\mathclose{}\leq \frac{1}{n}$$$, while $$$\mathopen{}\left\lVert{}{g}_{n}\mathopen{}\left( A\right)\mathclose{}\right\rVert\mathclose{}= 1$$$. This is not compatible with $$$f\mathopen{}\left( A\right)\mathclose{}$$$ having a bounded inverse.

We have shown that $$$0\in f\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$ if and only if $$$0\in \mathop{\sigma}\mathopen{}\left( f\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$. To finish the proof, replace $$$f$$$ by $$$λ-f$$$ for arbitrary complex $$$λ$$$.

Proposition II.64

For $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ we have $$$\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}= { T^{*}\mathopen{}\left( H\right)\mathclose{} }^{\perp}$$$ and hence $$$\overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} }= { \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} }^{\perp}$$$.

Proof. $$$x\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$ if and only if $$$\mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= 0$$$ for all $$$y$$$ if and only if $$$\mathopen{}\left\langle{}x, T^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= 0$$$ for all $$$y$$$, i.e., $$$x$$$ is orthogonal to $$$T^{*}\mathopen{}\left( H\right)\mathclose{}$$$.

In particular, $$$H= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}\oplus \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}}$$$. So, for instance, two bounded operators that agree on $$$\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$ and on $$$T^{*}\mathopen{}\left( H\right)\mathclose{}$$$ must be the same operator.

Proposition II.65

If $$$A$$$ and $$$B$$$ are positive operators with $$${B}^{2}= A$$$ then $$$B= {A}^{\frac{1}{2}}$$$.

Proof. Consider $$$T= B+{A}^{\frac{1}{2}}$$$ (where $$${A}^{\frac{1}{2}}$$$ is the square root from the functional calculus). We want to show that $$$B-{A}^{\frac{1}{2}}= 0$$$. $$$B$$$ commutes with $$$A$$$ because $$$A= {B}^{2}$$$ and $$$B$$$ commutes with $$${B}^{2}$$$, so $$$B$$$ commutes with any polynomial in $$$A$$$ and thus with any continuous function of $$$A$$$. Thus $$$\mathopen{}\left(B-{A}^{\frac{1}{2}}\right)\mathclose{}\mathopen{}\left(B+{A}^{\frac{1}{2}}\right)\mathclose{}= {B}^{2}-{A}^{\frac{1}{2}}B+B{A}^{\frac{1}{2}}-A= B{A}^{\frac{1}{2}}-{A}^{\frac{1}{2}}B= 0$$$.

Take $$$x\in \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$. $$$0= \mathopen{}\left\langle{}\mathopen{}\left(B+{A}^{\frac{1}{2}}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}$$$ implies $$$\mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= 0= \mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}$$$ so $$$\mathopen{}\left\langle{}{B}^{\frac{1}{2}}\mathopen{}\left( {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= 0$$$ implies $$$\mathopen{}\left\langle{}{B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}{B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$$ so $$${B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}= 0$$$ so $$$B\mathopen{}\left( x\right)\mathclose{}= {B}^{\frac{1}{2}}\mathopen{}\left( {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= 0$$$. Likewise $$${A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}= 0$$$. $$$B$$$ and $$${A}^{\frac{1}{2}}$$$ agree on $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$ and $$$\operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{}$$$, so they are the same by the comment above.

The following definitions are part of the essential vocabulary of our subject.

Definition II.66

Let $$$H$$$ and $$$K$$$ be Hilbert spaces

1. An operator $$$U\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{}$$$ is called unitary if $$$U^{*}U= \mathrm{I}_{H}$$$ and $$$U U^{*}= \mathrm{I}_{K}$$$.
2. An operator $$$P\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is called a projection if $$$P= {P}^{2}= P^{*}$$$.
3. An operator $$$V\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{}$$$ is called a partial isometry if $$$V^{*}V$$$ is a projection. When $$$V^{*}V= \mathrm{I}_{H}$$$, we call $$$V$$$ an isometry.

Remark II.67

The following assertions are readily verified.

1. $$$U\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{}$$$ is unitary if and only if $$$U\mathopen{}\left( H\right)\mathclose{}= K$$$ and $$$\mathopen{}\left\langle{}U\mathopen{}\left( ξ\right)\mathclose{}, U\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, η\right\rangle\mathclose{}$$$ for all $$$ξ$$$ and $$$η$$$ in $$$H$$$, if and only if $$$U\mathopen{}\left( H\right)\mathclose{}= K$$$ and $$$\mathopen{}\left\lVert{}U\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}$$$ for all $$$ξ\in H$$$.
2. The orthogonal projection on a closed subspace of $$$H$$$ is a projection operator. Conversely, if $$$P\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is a projection operator, then $$$P\mathopen{}\left( H\right)\mathclose{}$$$ is closed, and $$$P$$$ is the orthogonal projection on $$$P\mathopen{}\left( H\right)\mathclose{}$$$.
3. $$$V\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{}$$$ is a partial isometry if and only if $$$V^{*}V V^{*}= V$$$ if and only if $$$V^{*}$$$ is a partial isometry. A partial isometry $$$V$$$ maps $$$V^{*}\mathopen{}\left( V\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{}= V^{*}\mathopen{}\left( K\right)\mathclose{}$$$ isometrically onto $$$V\mathopen{}\left( V^{*}\mathopen{}\left( K\right)\mathclose{}\right)\mathclose{}= V\mathopen{}\left( H\right)\mathclose{}$$$.

Note: for $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ we will define $$$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}= {\mathopen{}\left( T^{*}T\right)\mathclose{}}^{\frac{1}{2}}$$$.

Proposition II.68 (Polar Decomposition)

For $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and for all $$$x\in H$$$,

1. $$$\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}$$$
2. $$$\operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$ and $$$\overline{ \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{} }= \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} }$$$
3. There exists $$$V\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ such that $$$V^{*}V$$$ is orthogonal projection on $$$\overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} }$$$, $$$V V^{*}$$$ is projection on $$$\overline{ T\mathopen{}\left( H\right)\mathclose{} }$$$, and $$$T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{}$$$.

Proof.

1. $$${\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$$.
2. $$$\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}$$$ is immediate. Then $$$\overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}}= {\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}}^{\perp}= {\mathopen{}\left(\operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}^{*}\right)\mathclose{}\right)\mathclose{}}^{\perp}= \overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}$$$.
3. We have $$$H= \overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}\oplus \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$. Notice that for $$$x\in H$$$ and $$$w\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$, $${\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \text{.}$$ Extending by continuity from $$$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}+\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$ to $$$\overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}+\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$, we obtain a map $$$V\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ such that $$$V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}$$$ for $$$x\in H$$$ and $$$w\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$. We now figure out what $$$V^{*}$$$ is using $$$H= \overline{T\mathopen{}\left( H\right)\mathclose{}}\oplus \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{}$$$. For $$$w\in \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{}$$$, $$$w'\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}$$$, and $$$x$$$ and $$$x'\in H$$$, $$\mathopen{}\left\langle{} V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+ w'\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}+w, V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+ w'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}+w, T\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x'\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x'\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+w\right\rangle\mathclose{} \text{.}$$ So $$$V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}= \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}$$$, and hence $$$V\mathopen{}\left( V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}$$$, so $$$V\mathopen{}\left( V^{*}\right)\mathclose{}$$$ is the projection on $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$. Furthermore, $$$V^{*}\mathopen{}\left( V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}\right)\mathclose{}= V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}$$$, which makes $$$V^{*}\mathopen{}\left( V\right)\mathclose{}$$$ the projection on $$$\overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}}$$$.