Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

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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 Aself-adjoint operator∈element ofℒbounded linear operators⁡(HHilbert space) \( A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) positive if 〈Aself-adjoint operator⁡(xvector), xvector〉≥greater than or equal to0zero \( \mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\geq 0 \) for all xvector∈element ofHHilbert space \( x\in H \) and write it as Aself-adjoint operator≥greater than or equal to0zero \( A\geq 0 \). Note: Positive implies self-adjoint.

Theorem II.62

The following are equivalent:

(a)
Aself-adjoint operator≥greater than or equal to0zero \( A\geq 0 \).
(b)
Aself-adjoint operator=equalsAself-adjoint operator* \( A= A^{*} \) and σ⁡(Aself-adjoint operator)⊆subset[interval0zero, +⁢∞infinity)interval \( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left[0, {+}\infty\right)\mathclose{} \).
(c)
Aself-adjoint operator=equalsBpositive operator2two \( A= {B}^{2} \) for some Bpositive operator≥greater than or equal to0zero \( B\geq 0 \).
(d)
Aself-adjoint operator=equalsTlinear map*⁢timesTlinear map \( A= T^{*}T \) for some Tlinear map∈element ofℒbounded linear operators⁡(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \).

Proof.

  1. ((a) ⇒ (b)) Aself-adjoint operator=equalsAself-adjoint operator* \( A= A^{*} \) follows immediately from Proposition II.54, so σ⁡(Aself-adjoint operator)⊆subsetRreal numbers \( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathbb{R} \). For rreal number>greater than0zero \( r\gt 0 \) we have −⁢rreal number∉not an element ofσ⁡(Aself-adjoint operator) \( {-}r\notin \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \) because |modulus〈(rreal number+plusAself-adjoint operator)⁡(xvector), xvector〉|modulus=equalsrreal number⁢times〈xvector, xvector〉+plus〈Aself-adjoint operator⁡(xvector), xvector〉≥greater than or equal torreal number⁢times‖xvector‖2two \( \mathopen{}\left\lvert{}\mathopen{}\left\langle{}\mathopen{}\left(r+A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\right\rvert\mathclose{}= r\mathopen{}\left\langle{}x, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\geq r{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2} \) and so rreal number+plusAself-adjoint operator∈element of ℒbounded linear operators⁡(HHilbert space) −⁢1invertible elements \( r+A\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1} \) by Lemma II.53.
  2. ((b) ⇒ (c)) Notice that the square root function belongs to Cspace of continuous functions⁡(σ⁡(Aself-adjoint operator)) \( \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \). Use functional calculus. Let Bpositive operator=equalsAself-adjoint operator=equalsAself-adjoint operator1one2two \( B= \sqrt{A}= {A}^{\frac{1}{2}} \). Notice that 〈Bpositive operator⁡(xvector), xvector〉=equals〈Aself-adjoint operator1one4four⁡(xvector), Aself-adjoint operator1one4four⁡(xvector)〉 \( \mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}{A}^{\frac{1}{4}}\mathopen{}\left( x\right)\mathclose{}, {A}^{\frac{1}{4}}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{} \), so Bpositive operator≥greater than or equal to0zero \( B\geq 0 \).
  3. ((c) ⇒ (d)) is obvious.
  4. ((d) ⇒ (a)) 〈Tlinear map*⁡(Tlinear map⁡(xvector)), xvector〉=equals〈Tlinear map⁡(xvector), Tlinear map⁡(xvector)〉=equals‖Tlinear map⁡(xvector)‖2two≥greater than or equal to0zero \( \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}\geq 0 \).

Theorem II.63

For any self-adjoint operator Aself-adjoint operator\( A \) and any function ffunction∈element ofC⁡(σ⁡(Aself-adjoint operator)) \( f\in C\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \), we have σ⁡(ffunction⁡(Aself-adjoint operator))=equals{setffunction⁡(treal number)|such that treal number∈element ofσ⁡(Aself-adjoint operator) }set=equalsffunction⁡(σ⁡(Aself-adjoint operator)) \( \mathop{\sigma}\mathopen{}\left( f\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, f\mathopen{}\left( t\right)\mathclose{}\,\middle\vert\, , t\in \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}, \,\right\}\mathclose{}= f\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \).

Proof. Notice that if ffunction\( f \) does not vanish on σ⁡(Aself-adjoint operator) \( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \), then ffunction⁡(Aself-adjoint operator) \( f\mathopen{}\left( A\right)\mathclose{} \) is invertible; the inverse is of course (1oneffunction)⁡(Aself-adjoint operator) \( \mathopen{}\left(\frac{1}{f}\right)\mathclose{}\mathopen{}\left( A\right)\mathclose{} \). In the other direction, let's show that if ffunction⁡(treal number0zero)=equals0zero \( f\mathopen{}\left( {t}_{0}\right)\mathclose{}= 0 \) for some treal number0zero∈element ofσ⁡(Aself-adjoint operator) \( {t}_{0}\in \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \), then ffunction⁡(Aself-adjoint operator) \( f\mathopen{}\left( A\right)\mathclose{} \) cannot be invertible. We obtain positive εpositive real number1one\( {ε}_{1} \), εpositive real number2two\( {ε}_{2} \), … such that |modulusffunction⁡(treal number)|modulus<less than1oneninteger \( \mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\lt \frac{1}{n} \) for treal number∈element of(intervaltreal number0zero-minusεpositive real numberninteger, treal number0zero+plusεpositive real numberninteger)interval∩intersectionσ⁡(Aself-adjoint operator) \( t\in \mathopen{}\left({t}_{0}-{ε}_{n}, {t}_{0}+{ε}_{n}\right)\mathclose{}\cap \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \). Restricting tent functions with tent poles at treal number0zero \( {t}_{0} \) and stakes at treal number0zero±plus or minusεpositive real number0zero \( {t}_{0}\pm {ε}_{0} \) to σ⁡(Aself-adjoint operator) \( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \), we obtain functions ggroup element1one\( {g}_{1} \), ggroup element2two\( {g}_{2} \), … in Cspace of continuous functions⁡(σ⁡(Aself-adjoint operator)) \( \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \) such that 0zero≤less than or equal toggroup elementninteger≤less than or equal to1one \( 0\leq {g}_{n}\leq 1 \), ggroup elementninteger⁡(treal number0zero)=equals1one \( {g}_{n}\mathopen{}\left( {t}_{0}\right)\mathclose{}= 1 \), and ggroup elementninteger≡equivalent0zero \( {g}_{n}\equiv 0 \) off (intervaltreal number0zero-minusεpositive real numberninteger, treal number0zero+plusεpositive real numberninteger)interval \( \mathopen{}\left({t}_{0}-{ε}_{n}, {t}_{0}+{ε}_{n}\right)\mathclose{} \). Then |modulusffunction⁢timesggroup elementninteger|modulus≤less than or equal to1oneninteger \( \mathopen{}\left\lvert{}f{g}_{n}\right\rvert\mathclose{}\leq \frac{1}{n} \), while the uniform norm of ggroup elementninteger \( {g}_{n} \) is 1one\( 1 \) for all ninteger\( n \). The functional calculus is isometric between the uniform norm and the operator norm, so ‖ffunction⁡(Aself-adjoint operator)⁢timesggroup elementninteger⁡(Aself-adjoint operator)‖≤less than or equal to1oneninteger \( \mathopen{}\left\lVert{}f\mathopen{}\left( A\right)\mathclose{}{g}_{n}\mathopen{}\left( A\right)\mathclose{}\right\rVert\mathclose{}\leq \frac{1}{n} \), while ‖ggroup elementninteger⁡(Aself-adjoint operator)‖=equals1one \( \mathopen{}\left\lVert{}{g}_{n}\mathopen{}\left( A\right)\mathclose{}\right\rVert\mathclose{}= 1 \). This is not compatible with ffunction⁡(Aself-adjoint operator) \( f\mathopen{}\left( A\right)\mathclose{} \) having a bounded inverse.

We have shown that 0zero∈element offfunction⁡(σ⁡(Aself-adjoint operator)) \( 0\in f\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \) if and only if 0zero∈element ofσ⁡(ffunction⁡(Aself-adjoint operator)) \( 0\in \mathop{\sigma}\mathopen{}\left( f\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \). To finish the proof, replace ffunction\( f \) by λcomplex number-minusffunction \( λ-f \) for arbitrary complex λcomplex number\( λ \).

Proposition II.64

For Tlinear map∈element ofℒbounded linear operators⁡(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) we have Kerkernel⁡(Tlinear map)=equals Tlinear map*⁡(HHilbert space) ⊥ \( \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}= { T^{*}\mathopen{}\left( H\right)\mathclose{} }^{\perp} \) and hence Tlinear map*⁡(HHilbert space) ¯=equals Kerkernel⁡(Tlinear map) ⊥ \( \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} }= { \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} }^{\perp} \).

Proof. xvector∈element ofKerkernel⁡(Tlinear map) \( x\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \) if and only if 〈Tlinear map⁡(xvector), yvector〉=equals0zero \( \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= 0 \) for all yvector\( y \) if and only if 〈xvector, Tlinear map*⁡(yvector)〉=equals0zero \( \mathopen{}\left\langle{}x, T^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= 0 \) for all yvector\( y \), i.e., xvector\( x \) is orthogonal to Tlinear map*⁡(HHilbert space) \( T^{*}\mathopen{}\left( H\right)\mathclose{} \).

In particular, HHilbert space=equalsKerkernel⁡(Tlinear map)⊕Tlinear map*⁡(HHilbert space)¯ \( H= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}\oplus \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}} \). So, for instance, two bounded operators that agree on Kerkernel⁡(Tlinear map)\( \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \) and on Tlinear map*⁡(HHilbert space) \( T^{*}\mathopen{}\left( H\right)\mathclose{} \) must be the same operator.

Proposition II.65

If Apositive operator\( A \) and Bpositive operator\( B \) are positive operators with Bpositive operator2two=equalsApositive operator \( {B}^{2}= A \) then Bpositive operator=equalsApositive operator1one2two \( B= {A}^{\frac{1}{2}} \).

Proof. Consider Tlinear map=equalsBpositive operator+plusApositive operator1one2two \( T= B+{A}^{\frac{1}{2}} \) (where Apositive operator1one2two \( {A}^{\frac{1}{2}} \) is the square root from the functional calculus). We want to show that Bpositive operator-minusApositive operator1one2two=equals0zero \( B-{A}^{\frac{1}{2}}= 0 \). Bpositive operator\( B \) commutes with Apositive operator\( A \) because Apositive operator=equalsBpositive operator2two \( A= {B}^{2} \) and Bpositive operator\( B \) commutes with Bpositive operator2two\( {B}^{2} \), so Bpositive operator\( B \) commutes with any polynomial in Apositive operator\( A \) and thus with any continuous function of Apositive operator\( A \). Thus (Bpositive operator-minusApositive operator1one2two)⁢times(Bpositive operator+plusApositive operator1one2two)=equalsBpositive operator2two-minusApositive operator1one2two⁢timesBpositive operator+plusBpositive operator⁢timesApositive operator1one2two-minusApositive operator=equalsBpositive operator⁢timesApositive operator1one2two-minusApositive operator1one2two⁢timesBpositive operator=equals0zero \( \mathopen{}\left(B-{A}^{\frac{1}{2}}\right)\mathclose{}\mathopen{}\left(B+{A}^{\frac{1}{2}}\right)\mathclose{}= {B}^{2}-{A}^{\frac{1}{2}}B+B{A}^{\frac{1}{2}}-A= B{A}^{\frac{1}{2}}-{A}^{\frac{1}{2}}B= 0 \).

Take xvector∈element ofKerkernel⁡(Tlinear map*)=equalsKerkernel⁡(Tlinear map) \( x\in \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \). 0zero=equals〈(Bpositive operator+plusApositive operator1one2two)⁡(xvector), xvector〉=equals〈Bpositive operator⁡(xvector), xvector〉+plus〈Apositive operator1one2two⁡(xvector), xvector〉 \( 0= \mathopen{}\left\langle{}\mathopen{}\left(B+{A}^{\frac{1}{2}}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{} \) implies 〈Bpositive operator⁡(xvector), xvector〉=equals0zero=equals〈Apositive operator1one2two⁡(xvector), xvector〉 \( \mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= 0= \mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{} \) so 〈Bpositive operator1one2two⁡(Bpositive operator1one2two⁡(xvector)), xvector〉=equals0zero \( \mathopen{}\left\langle{}{B}^{\frac{1}{2}}\mathopen{}\left( {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= 0 \) implies 〈Bpositive operator1one2two⁡(xvector), Bpositive operator1one2two⁡(xvector)〉=equals ‖Bpositive operator1one2two⁡(xvector)‖ 2two \( \mathopen{}\left\langle{}{B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}{B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \) so Bpositive operator1one2two⁡(xvector)=equals0zero \( {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}= 0 \) so Bpositive operator⁡(xvector)=equalsBpositive operator1one2two⁡(Bpositive operator1one2two⁡(xvector))=equals0zero \( B\mathopen{}\left( x\right)\mathclose{}= {B}^{\frac{1}{2}}\mathopen{}\left( {B}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= 0 \). Likewise Apositive operator1one2two⁡(xvector)=equals0zero \( {A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}= 0 \). Bpositive operator\( B \) and Apositive operator1one2two \( {A}^{\frac{1}{2}} \) agree on Tlinear map⁡(HHilbert space)¯ \( \overline{T\mathopen{}\left( H\right)\mathclose{}} \) and Kerkernel⁡(Tlinear map*) \( \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{} \), 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 HHilbert space\( H \) and KHilbert space\( K \) be Hilbert spaces

  1. An operator U∈element ofℒbounded linear operators⁡(HHilbert spaceKHilbert space) \( U\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{} \) is called unitary if U*⁢timesU=equalsIHHilbert spaceidentity operator on HHilbert space \( U^{*}U= \mathrm{I}_{H} \) and U⁢timesU*=equalsIKHilbert spaceidentity operator on KHilbert space \( U U^{*}= \mathrm{I}_{K} \).
  2. An operator P∈element ofℒbounded linear operators⁡(HHilbert space) \( P\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) is called a projection if P=equalsP2two=equalsP* \( P= {P}^{2}= P^{*} \).
  3. An operator Vvector space∈element ofℒbounded linear operators⁡(HHilbert spaceKHilbert space) \( V\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{} \) is called a partial isometry if Vvector space*⁢timesVvector space \( V^{*}V \) is a projection. When Vvector space*⁢timesVvector space=equalsIHHilbert spaceidentity operator on HHilbert space \( V^{*}V= \mathrm{I}_{H} \), we call Vvector space\( V \) an isometry.

Remark II.67

The following assertions are readily verified.

  1. U∈element ofℒbounded linear operators⁡(HHilbert spaceKHilbert space) \( U\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{} \) is unitary if and only if U⁡(HHilbert space)=equalsKHilbert space \( U\mathopen{}\left( H\right)\mathclose{}= K \) and 〈U⁡(ξvector), U⁡(ηvector)〉=equals〈ξvector, ηvector〉 \( \mathopen{}\left\langle{}U\mathopen{}\left( ξ\right)\mathclose{}, U\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, η\right\rangle\mathclose{} \) for all ξvector\( ξ \) and ηvector\( η \) in HHilbert space\( H \), if and only if U⁡(HHilbert space)=equalsKHilbert space \( U\mathopen{}\left( H\right)\mathclose{}= K \) and ‖U⁡(ξvector)‖=equals‖ξvector‖ \( \mathopen{}\left\lVert{}U\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{} \) for all ξvector∈element ofHHilbert space\( ξ\in H \).
  2. The orthogonal projection on a closed subspace of HHilbert space\( H \) is a projection operator. Conversely, if P∈element ofℒbounded linear operators⁡(HHilbert space) \( P\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) is a projection operator, then P⁡(HHilbert space) \( P\mathopen{}\left( H\right)\mathclose{} \) is closed, and P\( P \) is the orthogonal projection on P⁡(HHilbert space) \( P\mathopen{}\left( H\right)\mathclose{} \).
  3. Vvector space∈element ofℒbounded linear operators⁡(HHilbert spaceKHilbert space) \( V\in \mathcal{L}\mathopen{}\left( H, K\right)\mathclose{} \) is a partial isometry if and only if Vvector space*⁢timesVvector space⁢timesVvector space*=equalsVvector space \( V^{*}V V^{*}= V \) if and only if Vvector space*\( V^{*} \) is a partial isometry. A partial isometry Vvector space\( V \) maps Vvector space*⁡(Vvector space⁡(HHilbert space))=equalsVvector space*⁡(KHilbert space) \( V^{*}\mathopen{}\left( V\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{}= V^{*}\mathopen{}\left( K\right)\mathclose{} \) isometrically onto Vvector space⁡(Vvector space*⁡(KHilbert space))=equalsVvector space⁡(HHilbert space) \( V\mathopen{}\left( V^{*}\mathopen{}\left( K\right)\mathclose{}\right)\mathclose{}= V\mathopen{}\left( H\right)\mathclose{} \).

Note: for Tlinear map∈element ofℒbounded linear operators⁡(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) we will define |modulusTlinear map|modulus=equals (Tlinear map*⁢timesTlinear map) 1one2two \( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}= {\mathopen{}\left( T^{*}T\right)\mathclose{}}^{\frac{1}{2}} \).

Proposition II.68 (Polar Decomposition)

For Tlinear map∈element ofℒbounded linear operators⁡(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) and for all xvector∈element ofHHilbert space \( x\in H \),

  1. ‖|modulusTlinear map|modulus⁡(xvector)‖=equals‖Tlinear map⁡(xvector)‖ \( \mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} \)
  2. Kerkernel⁡(|modulusTlinear map|modulus)=equalsKerkernel⁡(Tlinear map) \( \operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \) and |modulusTlinear map|modulus⁡(HHilbert space) ¯=equals Tlinear map*⁡(HHilbert space) ¯ \( \overline{ \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{} }= \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} } \)
  3. There exists Vvector space∈element ofℒbounded linear operators⁡(HHilbert space) \( V\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) such that Vvector space*⁢timesVvector space \( V^{*}V \) is orthogonal projection on Tlinear map*⁡(HHilbert space) ¯ \( \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{} } \), Vvector space⁢timesVvector space* \( V V^{*} \) is projection on Tlinear map⁡(HHilbert space) ¯ \( \overline{ T\mathopen{}\left( H\right)\mathclose{} } \), and Tlinear map=equalsVvector space⁢times|modulusTlinear map|modulus \( T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{} \).

Proof.

  1. ‖Tlinear map⁡(xvector)‖2two=equals〈Tlinear map⁡(xvector), Tlinear map⁡(xvector)〉=equals〈Tlinear map*⁡(Tlinear map⁡(xvector)), xvector〉=equals〈|modulusTlinear map|modulus⁡(|modulusTlinear map|modulus⁡(xvector)), xvector〉=equals〈|modulusTlinear map|modulus⁡(xvector), |modulusTlinear map|modulus⁡(xvector)〉=equals‖|modulusTlinear map|modulus⁡(xvector)‖2two \( {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \).
  2. Kerkernel⁡(Tlinear map)=equalsKerkernel⁡(|modulusTlinear map|modulus) \( \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \) is immediate. Then Tlinear map*⁡(HHilbert space)¯=equalsKerkernel⁡(Tlinear map)⊥=equals(Kerkernel⁡(|modulusTlinear map|modulus*))⊥=equals|modulusTlinear map|modulus⁡(HHilbert space)¯ \( \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}}= {\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}}^{\perp}= {\mathopen{}\left(\operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}^{*}\right)\mathclose{}\right)\mathclose{}}^{\perp}= \overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}} \).
  3. We have HHilbert space=equals|modulusTlinear map|modulus⁡(HHilbert space)¯⊕Kerkernel⁡(Tlinear map) \( H= \overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}\oplus \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \). Notice that for xvector∈element ofHHilbert space \( x\in H \) and wvector∈element ofKerkernel⁡(Tlinear map) \( w\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \), ‖|modulusTlinear map|modulus⁡(xvector)+pluswvector‖ 2two =equals‖|modulusTlinear map|modulus⁡(xvector)‖2two+plus‖wvector‖2two=equals‖Tlinear map⁡(xvector)‖2two+plus‖wvector‖2two≥greater than or equal to‖Tlinear map⁡(xvector)‖2two . \[ {\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \text{.} \] Extending by continuity from |modulusTlinear map|modulus⁡(HHilbert space)+plusKerkernel⁡(Tlinear map) \( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}+\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \) to |modulusTlinear map|modulus⁡(HHilbert space)¯+plusKerkernel⁡(Tlinear map) \( \overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}+\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \), we obtain a map Vvector space∈element ofℒbounded linear operators⁡(HHilbert space) \( V\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) such that Vvector space⁡(|modulusTlinear map|modulus⁡(xvector)+pluswvector)=equalsTlinear map⁡(xvector) \( V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{} \) for xvector∈element ofHHilbert space \( x\in H \) and wvector∈element ofKerkernel⁡(Tlinear map) \( w\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \). We now figure out what Vvector space*\( V^{*} \) is using HHilbert space=equalsTlinear map⁡(HHilbert space)¯⊕Kerkernel⁡(Tlinear map*) \( H= \overline{T\mathopen{}\left( H\right)\mathclose{}}\oplus \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{} \). For wvector∈element ofKerkernel⁡(Tlinear map*) \( w\in \operatorname{Ker}\mathopen{}\left( T^{*}\right)\mathclose{} \), wvector′∈element ofKerkernel⁡(Tlinear map) \( w'\in \operatorname{Ker}\mathopen{}\left( T\right)\mathclose{} \), and xvector\( x \) and xvector′∈element ofHHilbert space \( x'\in H \), 〈Vvector space*⁡(Tlinear map⁡(xvector)+pluswvector), |modulusTlinear map|modulus⁡(xvector′)+pluswvector′〉=equals〈Tlinear map⁡(xvector)+pluswvector, Vvector space⁡(|modulusTlinear map|modulus⁡(xvector′)+pluswvector′)〉=equals〈Tlinear map⁡(xvector)+pluswvector, Tlinear map⁡(xvector′)〉=equals〈Tlinear map⁡(xvector), Tlinear map⁡(xvector′)〉=equals〈Tlinear map*⁡(Tlinear map⁡(xvector)), xvector′〉=equals〈|modulusTlinear map|modulus⁡(|modulusTlinear map|modulus⁡(xvector)), xvector′〉=equals〈|modulusTlinear map|modulus⁡(xvector), |modulusTlinear map|modulus⁡(xvector′)〉=equals〈|modulusTlinear map|modulus⁡(xvector), |modulusTlinear map|modulus⁡(xvector′)+pluswvector〉 . \[ \mathopen{}\left\langle{} V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+ w'\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}+w, V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+ w'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}+w, T\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x'\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x'\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}, \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x'\right)\mathclose{}+w\right\rangle\mathclose{} \text{.} \] So Vvector space*⁡(Tlinear map⁡(xvector)+pluswvector)=equals|modulusTlinear map|modulus⁡(xvector) \( V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}= \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{} \), and hence Vvector space⁡(Vvector space*⁡(Tlinear map⁡(xvector)+pluswvector))=equalsTlinear map⁡(xvector) \( V\mathopen{}\left( V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{} \), so Vvector space⁡(Vvector space*) \( V\mathopen{}\left( V^{*}\right)\mathclose{} \) is the projection on Tlinear map⁡(HHilbert space)¯ \( \overline{T\mathopen{}\left( H\right)\mathclose{}} \). Furthermore, Vvector space*⁡(Vvector space⁡(|modulusTlinear map|modulus⁡(xvector)+pluswvector))=equalsVvector space*⁡(Tlinear map⁡(xvector))=equals|modulusTlinear map|modulus⁡(xvector) \( V^{*}\mathopen{}\left( V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{}+w\right)\mathclose{}\right)\mathclose{}= V^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( x\right)\mathclose{} \), which makes Vvector space*⁡(Vvector space) \( V^{*}\mathopen{}\left( V\right)\mathclose{} \) the projection on Tlinear map*⁡(HHilbert space)¯ \( \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}} \).

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