Lecture Notes in Functional Analysis
by William L. Paschke
edition 0.9
image/svg+xml
A. Tensor Notation
For x vector \( x \) and y vector \( y \) in H Hilbert space \( H \) , define
x vector ⊗ y vector
: maps H Hilbert space → to H Hilbert space
\(
x\otimes y
: H \to H
\)
by
( x vector ⊗ y vector ) ( ξ function ) = equals 〈 ξ function , y vector 〉 times x vector
\(
\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}= \mathopen{}\left\langle{}ξ, y\right\rangle\mathclose{}x
\) . Then
‖ ( x vector ⊗ y vector ) ( ξ function ) ‖ ≤ less than or equal to ‖ ξ function ‖ times ‖ y vector ‖ times ‖ x vector ‖
\(
\mathopen{}\left\lVert{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}
\)
and so
‖ ( x vector ⊗ y vector ) ( y vector ) ‖ = equals ‖ y vector ‖ 2 two times ‖ x vector ‖
\(
\mathopen{}\left\lVert{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}
\) . Thus
x vector ⊗ y vector ∈ element of ℒ bounded linear operators ( H Hilbert space )
\(
x\otimes y\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}
\) ,
and
‖ x vector ⊗ y vector ‖ = equals ‖ x vector ‖ times ‖ y vector ‖
\(
\mathopen{}\left\lVert{}x\otimes y\right\rVert\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}
\) .
Finally,
〈 ( x vector ⊗ y vector ) ( ξ function ) , ν 〉 = equals 〈 〈 y vector , ξ function 〉 times x vector , ν 〉 = equals 〈 y vector , ξ function 〉 times 〈 x vector , y vector 〉
,
\[
\mathopen{}\left\langle{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, ν\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{}x, ν\right\rangle\mathclose{}= \mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{}\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}
\text{,}
\] and
〈 ξ function , ( y vector ⊗ x vector ) ( ν ) 〉 = equals 〈 ξ function , 〈 ν , x vector 〉 times y vector 〉 = equals 〈 x vector , y vector 〉 times 〈 y vector , ξ function 〉
.
\[
\mathopen{}\left\langle{}ξ, \mathopen{}\left(y\otimes x\right)\mathclose{}\mathopen{}\left( ν\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, \mathopen{}\left\langle{}ν, x\right\rangle\mathclose{}y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{}
\text{.}
\] Thus
( x vector ⊗ y vector ) * = equals y vector ⊗ x vector
\(
\mathopen{}\left(x\otimes y\right)\mathclose{}^{*}= y\otimes x
\) .
For
T linear map ∈ element of ℒ bounded linear operators ( H Hilbert space )
\(
T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}
\) ,
we have
T linear map times ( x vector ⊗ y vector ) = equals T linear map ( x vector ) ⊗ y vector
\(
T\mathopen{}\left(x\otimes y\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}\otimes y
\)
and
( ( x vector ⊗ y vector ) times T linear map ) * = equals T linear map * ( y vector ⊗ x vector ) = equals T linear map * ( y vector ) ⊗ x vector = equals ( x vector ⊗ T linear map * ( y vector ) ) *
\(
\mathopen{}\left(\mathopen{}\left(x\otimes y\right)\mathclose{}T\right)\mathclose{}^{*}= T^{*}\mathopen{}\left( y\otimes x\right)\mathclose{}= T^{*}\mathopen{}\left( y\right)\mathclose{}\otimes x= \mathopen{}\left(x\otimes T^{*}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}^{*}
\) .
So,
( x vector ⊗ y vector ) ( T linear map ) = equals x vector ⊗ T linear map * ( y vector )
\(
\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( T\right)\mathclose{}= x\otimes T^{*}\mathopen{}\left( y\right)\mathclose{}
\) .
In particular,
( x vector 1 one ⊗ y vector 1 one ) ( x vector 2 two ⊗ y vector 2 two ) = equals 〈 x vector 2 two , y vector 1 one 〉 times ( x vector 1 one ⊗ y vector 2 two )
\(
\mathopen{}\left({x}_{1}\otimes {y}_{1}\right)\mathclose{}\mathopen{}\left( {x}_{2}\otimes {y}_{2}\right)\mathclose{}= \mathopen{}\left\langle{}{x}_{2}, {y}_{1}\right\rangle\mathclose{}\mathopen{}\left({x}_{1}\otimes {y}_{2}\right)\mathclose{}
\) .
Notice for
y vector ≠ not equal to 0 zero
\(
y\neq 0
\) ,
Ran range ( x vector ⊗ y vector ) = equals C x vector
\(
\operatorname{Ran}\mathopen{}\left( x\otimes y\right)\mathclose{}= \mathbb{C}x
\) .
Every
T linear map ∈ element of ℱ finite rank operators ( H Hilbert space )
\(
T\in \mathcal{F}\mathopen{}\left( H\right)\mathclose{}
\)
can be written
T linear map = equals ∑ summation i integer = 1 one n integer
x vector i integer ⊗ y vector i integer
\(
T= \sum_{i=1}^{n}{}
{x}_{i}\otimes {y}_{i}
\) .
Indeed, let
{ set e unit vector 1 one … e unit vector n integer } set
\(
\mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{}
\)
be an orthonormal basis for
T linear map ( H Hilbert space )
\(
T\mathopen{}\left( H\right)\mathclose{}
\) .
Then
T linear map ( ξ function ) = equals ∑ summation i integer = 1 one n integer
〈 T linear map ( ξ function ) , e unit vector i integer 〉 times e unit vector i integer
= equals ∑ summation i integer = 1 one n integer
〈 ξ function , T linear map * ( e unit vector i integer ) 〉 times e unit vector i integer
= equals ∑ summation i integer = 1 one n integer
( e unit vector i integer ⊗ T linear map * ( e unit vector i integer ) ) ( ξ function )
\[
T\mathopen{}\left( ξ\right)\mathclose{}= \sum_{i=1}^{n}{}
\mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, {e}_{i}\right\rangle\mathclose{}{e}_{i}
= \sum_{i=1}^{n}{}
\mathopen{}\left\langle{}ξ, T^{*}\mathopen{}\left( {e}_{i}\right)\mathclose{}\right\rangle\mathclose{}{e}_{i}
= \sum_{i=1}^{n}{}
\mathopen{}\left({e}_{i}\otimes T^{*}\mathopen{}\left( {e}_{i}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}
\]
Easily, then
ℱ finite rank operators ( H Hilbert space ) = equals H Hilbert space ⊙ algebraic tensor product H Hilbert space ¯ conjugate Hilbert space
\(
\mathcal{F}\mathopen{}\left( H\right)\mathclose{}= H\odot \overline{H}
\)
where
H Hilbert space ¯ conjugate Hilbert space = equals H Hilbert space
\(
\overline{H}= H
\)
with conjugate scalar multiplication.
Until further notice, H Hilbert space \( H \) is separable and infinite dimensional.
Notice that if
{ set e unit vector 1 one … e unit vector n integer } set
\(
\mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{}
\)
is a finite orthonormal set, then
P = equals ∑ summation i integer = 1 one n integer
e unit vector i integer ⊗ e unit vector i integer
\(
P= \sum_{i=1}^{n}{}
{e}_{i}\otimes {e}_{i}
\)
is the projection of H Hilbert space \( H \) on
span span ( e unit vector 1 one … e unit vector n integer )
\(
\operatorname{span}\mathopen{}\left( {e}_{1}, \dotsc, {e}_{n}\right)\mathclose{}
\) . In particular,
P ( ξ function ) = equals ∑ summation i integer = 1 one n integer
〈 ξ function , e unit vector i integer 〉 times e unit vector i integer
\(
P\mathopen{}\left( ξ\right)\mathclose{}= \sum_{i=1}^{n}{}
\mathopen{}\left\langle{}ξ, {e}_{i}\right\rangle\mathclose{}{e}_{i}
\) .
The spectral theorem for self-adjoint compact operators says if
A self-adjoint operator = equals A self-adjoint operator * ∈ element of 𝒦 compact linear operators ( H Hilbert space )
\(
A= A^{*}\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}
\) ,
then there exist real
λ complex number n integer
\(
{λ}_{n}
\)
with
λ complex number n integer → converges to 0 zero
\(
{λ}_{n} \to 0
\)
and an orthonormal basis
{ set e unit vector 1 one e unit vector 2 two … } set
\(
\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}
\)
for H Hilbert space \( H \) such that
A self-adjoint operator = equals ∑ summation n integer = 1 one ∞ infinity
λ complex number times e unit vector n integer ⊗ e unit vector n integer
\(
A= \sum_{n=1}^{\infty}{}
λ{e}_{n}\otimes {e}_{n}
\)
(norm-convergent series).
In this case
A self-adjoint operator ( e unit vector j integer ) = equals λ complex number j integer times e unit vector j integer
\(
A\mathopen{}\left( {e}_{j}\right)\mathclose{}= {λ}_{j}{e}_{j}
\) .
Example III.18
Let H Hilbert space \( H \) be the space
L 2 Lebesgue space ( [ interval 0 zero , 1 one ] interval )
\(
\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\) .
Let
ξ function 0 zero ( x vector ) = equals 1 one
\(
{ξ}_{0}\mathopen{}\left( x\right)\mathclose{}= 1
\)
and
ξ function 1 one ( x vector ) = equals x vector
\(
{ξ}_{1}\mathopen{}\left( x\right)\mathclose{}= x
\) . Let T linear map \( T \) be the Volterra operator, defined by
( T linear map ( f function ) ) ( t real number ) = equals ∫ integral 0 zero t real number f function ( x vector ) d x vector
\(
\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{0}^{t}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x
\) . Recall
( T linear map * ( f function ) ) ( t real number ) = equals ∫ integral t real number 1 one f function ( x vector ) d x vector
\(
\mathopen{}\left( T^{*}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{t}^{1}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x
\) .
So
( ( T linear map + plus T linear map * ) ( f function ) ) ( t real number ) = equals ∫ integral 0 zero 1 one f function ( x vector ) d x vector = equals 〈 f function , ξ function 0 zero 〉
\(
\mathopen{}\left(\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{0}^{1}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x= \mathopen{}\left\langle{}f, {ξ}_{0}\right\rangle\mathclose{}
\) .
Then
( T linear map + plus T linear map * ) ( f function ) = equals 〈 f function , ξ function 0 zero 〉 times ξ function 0 zero
\(
\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}= \mathopen{}\left\langle{}f, {ξ}_{0}\right\rangle\mathclose{}{ξ}_{0}
\)
and
T linear map + plus T linear map * = equals ξ function 0 zero ⊗ ξ function 0 zero
\(
T+ T^{*}= {ξ}_{0}\otimes {ξ}_{0}
\) .
Define
A self-adjoint operator = equals − T linear map 2 two + plus ξ function 1 one ⊗ ( ξ function 0 zero - minus ξ function 1 one )
\(
A= {-}{T}^{2}+{ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}
\) ,
so
A self-adjoint operator ∈ element of 𝒦 compact linear operators ( H Hilbert space )
\(
A\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}
\) .
We claim that
A self-adjoint operator = equals A self-adjoint operator *
\(
A= A^{*}
\) .
( T linear map * ) 2 two = equals
( ξ function 0 zero ⊗ ξ function 0 zero - minus T linear map )
2 two
= equals
( ξ function 0 zero ⊗ ξ function 0 zero )
2 two
- minus ( ξ function 0 zero ⊗ ξ function 0 zero ) times T linear map - minus T linear map times ( ξ function 0 zero ⊗ ξ function 0 zero ) + plus T linear map 2 two = equals ξ function 0 zero ⊗ ξ function 0 zero - minus ξ function 0 zero ⊗ T linear map * ( ξ function 0 zero ) - minus T linear map ( ξ function 0 zero ) ⊗ ξ function 0 zero + plus T linear map 2 two = equals ξ function 0 zero ⊗ ξ function 0 zero - minus ξ function 0 zero ⊗ ( ξ function 0 zero - minus ξ function 1 one ) - minus ξ function 1 one ⊗ ξ function 0 zero + plus T linear map 2 two = equals T linear map 2 two + plus ξ function 0 zero ⊗ ξ function 1 one - minus ξ function 1 one ⊗ ξ function 0 zero
.
\[
{\mathopen{}\left( T^{*}\right)\mathclose{}}^{2}= {\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}-T\right)\mathclose{}}^{2}= {\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}}^{2}-\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}T-T\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}+{T}^{2}= {ξ}_{0}\otimes {ξ}_{0}-{ξ}_{0}\otimes T^{*}\mathopen{}\left( {ξ}_{0}\right)\mathclose{}-T\mathopen{}\left( {ξ}_{0}\right)\mathclose{}\otimes {ξ}_{0}+{T}^{2}= {ξ}_{0}\otimes {ξ}_{0}-{ξ}_{0}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}-{ξ}_{1}\otimes {ξ}_{0}+{T}^{2}= {T}^{2}+{ξ}_{0}\otimes {ξ}_{1}-{ξ}_{1}\otimes {ξ}_{0}
\text{.}
\]
Thus
− A self-adjoint operator * = equals ( T linear map * ) 2 two - minus ( ξ function 0 zero - minus ξ function 1 one ) ⊗ ξ function 1 one = equals ξ function 1 one ⊗ ξ function 1 one - minus ξ function 1 one ⊗ ξ function 0 zero + plus T linear map 2 two = equals ξ function 1 one ⊗ ( ξ function 1 one - minus ξ function 0 zero ) + plus T linear map 2 two = equals − A self-adjoint operator
\(
{-} A^{*}= {\mathopen{}\left( T^{*}\right)\mathclose{}}^{2}-\mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\otimes {ξ}_{1}= {ξ}_{1}\otimes {ξ}_{1}-{ξ}_{1}\otimes {ξ}_{0}+{T}^{2}= {ξ}_{1}\otimes \mathopen{}\left({ξ}_{1}-{ξ}_{0}\right)\mathclose{}+{T}^{2}= {-}A
\) .
Consider eigenvalues
λ complex number ∈ element of R real numbers
\(
λ\in \mathbb{R}
\) ,
A self-adjoint operator ( f function ) = equals λ complex number ( f function )
\(
A\mathopen{}\left( f\right)\mathclose{}= λ\mathopen{}\left( f\right)\mathclose{}
\) ,
f function ≠ not equal to 0 zero
\(
f\neq 0
\) .
There are three cases.
(
λ complex number > greater than 0 zero
\(
λ\gt 0
\) )
Since
T linear map ( H Hilbert space ) ⊆ subset C space of continuous functions ( [ interval 0 zero , 1 one ] interval )
\(
T\mathopen{}\left( H\right)\mathclose{}\subseteq \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\) ,
the fundamental theorem of calculus makes
T linear map 2 two ( f function ) ∈ element of C 1 space of continuously differentiable functions ( [ interval 0 zero , 1 one ] interval )
\(
{T}^{2}\mathopen{}\left( f\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\)
with
( T linear map 2 two ( f function ) ) ′ derivative = equals T linear map ( f function )
\(
\mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = T\mathopen{}\left( f\right)\mathclose{}
\) .
Writing
λ complex number = equals 1 one μ measure 2 two
\(
λ= \frac{1}{{μ}^{2}}
\) ,
we have
f function = equals μ measure 2 two times A self-adjoint operator ( f function )
\(
f= {μ}^{2}A\mathopen{}\left( f\right)\mathclose{}
\) ,
and thus
f function ∈ element of C 1 space of continuously differentiable functions ( [ interval 0 zero , 1 one ] interval )
\(
f\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\) .
It follows that
T linear map ( f function ) ∈ element of C 1 space of continuously differentiable functions ( [ interval 0 zero , 1 one ] interval )
\(
T\mathopen{}\left( f\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\) ,
with
( T linear map ( f function ) ) ′ derivative = equals f function
\(
\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = f
\) .
Further
f function ′ derivative = equals μ measure 2 two times ( A self-adjoint operator ( f function ) ) ′ derivative = equals μ measure 2 two times ( − T linear map ( f function ) + plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 )
,
\[
f' = {μ}^{2}\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = {μ}^{2}\mathopen{}\left({-}T\mathopen{}\left( f\right)\mathclose{}+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}\right)\mathclose{}
\text{,}
\]
so
f function ′ derivative ∈ element of C 1 space of continuously differentiable functions ( [ interval 0 zero , 1 one ] interval )
\(
f' \in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\)
and
f function ″ second derivative = equals μ measure 2 two times ( − f function + plus 0 zero ) = equals − μ measure 2 two times f function
.
\[
f'' = {μ}^{2}\mathopen{}\left({-}f+0\right)\mathclose{}= {-}{μ}^{2}f
\text{.}
\]
We claim that
f function ( 0 zero ) = equals 0 zero = equals f function ( 1 one )
\(
f\mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{}
\) ,
which amounts to showing
( A self-adjoint operator ( f function ) ) ( 0 zero ) = equals 0 zero = equals ( A self-adjoint operator ( f function ) ) ( 1 one )
\(
\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= 0= \mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 1\right)\mathclose{}
\) . Indeed, from
( A self-adjoint operator ( f function ) ) ( t real number ) = equals −
( T linear map 2 two ( f function ) ) ( t real number )
+ plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 times t real number
\(
\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= {-}
\mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}
+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}t
\) ,
we immediately get
( A self-adjoint operator ( f function ) ) ( 0 zero ) = equals 0 zero
\(
\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= 0
\) ,
and at the other end
A self-adjoint operator ( f function ( 1 one ) ) = equals −
( T linear map 2 two ( f function ) ) ( 1 one )
+ plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 = equals −
∫ integral 0 zero 1 one
( T linear map ( f function ) ) ( x vector )
d x vector
+ plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 = equals −
〈 T linear map ( f function ) , ξ function 0 zero 〉
+ plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 = equals −
〈 f function , T linear map * ( ξ function 0 zero ) 〉
+ plus 〈 f function , ξ function 0 zero - minus ξ function 1 one 〉 = equals 0 zero
.
\[
A\mathopen{}\left( f\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}= {-}
\mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 1\right)\mathclose{}
+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-}
\int _{0}^{1}{}
\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}
\,\mathrm{d}x
+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-}
\mathopen{}\left\langle{}T\mathopen{}\left( f\right)\mathclose{}, {ξ}_{0}\right\rangle\mathclose{}
+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-}
\mathopen{}\left\langle{}f, T^{*}\mathopen{}\left( {ξ}_{0}\right)\mathclose{}\right\rangle\mathclose{}
+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= 0
\text{.}
\]
The general solution to
f function ″ second derivative + plus μ measure 2 two times f function = equals 0 zero
\(
f'' +{μ}^{2}f= 0
\)
is
f function ( x vector ) = equals a real number times cos cosine ( μ measure times x vector ) + plus b real number times sin sine ( μ measure times x vector )
\(
f\mathopen{}\left( x\right)\mathclose{}= a\cos\mathopen{}\left( μx\right)\mathclose{}+b\sin\mathopen{}\left( μx\right)\mathclose{}
\) .
The endpoint condition
f function ( 0 zero ) = equals 0 zero
\(
f\mathopen{}\left( 0\right)\mathclose{}= 0
\)
makes
a real number = equals 0 zero
\(
a= 0
\)
(and thus
b real number ≠ not equal to 0 zero
\(
b\neq 0
\) ).
From
f function ( 1 one ) = equals 0 zero
\(
f\mathopen{}\left( 1\right)\mathclose{}= 0
\) ,
we deduce that μ measure \( μ \) is a nonzero integer multiple of π \( π \) . We conclude that the positive eigenvalues of A self-adjoint operator \( A \) are
λ complex number n integer = equals
1 one
n integer 2 two times π 2 two
\(
{λ}_{n}= \frac{1}{{n}^{2}{π}^{2}}
\)
with normalized eigenfunctions
e unit vector n integer = equals e unit vector n integer ( x vector ) = equals 2 two times sin sine ( n integer times π times x vector )
\(
{e}_{n}= {e}_{n}\mathopen{}\left( x\right)\mathclose{}= \sqrt{2}\sin\mathopen{}\left( nπx\right)\mathclose{}
\) ,
for n integer \( n \) in
{ set 1 one 2 two … } set
\(
\mathopen{}\left\{\, 1, 2, \dotsc\,\right\}\mathclose{}
\) .
(
λ complex number = equals 0 zero
\(
λ= 0
\) )
Ker kernel ( A self-adjoint operator ) = equals 0 zero
\(
\operatorname{Ker}\mathopen{}\left( A\right)\mathclose{}= 0
\)
because
A self-adjoint operator ( f function ) = equals 0 zero
\(
A\mathopen{}\left( f\right)\mathclose{}= 0
\)
implies
T linear map 2 two ( f function ) = equals c real number times ξ function 1 one
\(
{T}^{2}\mathopen{}\left( f\right)\mathclose{}= c{ξ}_{1}
\) ,
which gives
T linear map ( f function ) = equals c real number times ξ function 0 zero
\(
T\mathopen{}\left( f\right)\mathclose{}= c{ξ}_{0}
\) .
But then
T linear map ( f function ( 0 zero ) ) = equals 0 zero
\(
T\mathopen{}\left( f\mathopen{}\left( 0\right)\mathclose{}\right)\mathclose{}= 0
\) ,
so
c real number = equals 0 zero
\(
c= 0
\) .
(
λ complex number < less than 0 zero
\(
λ\lt 0
\) )
Write
λ complex number = equals −
1 one μ measure 2 two
\(
λ= {-}
\frac{1}{{μ}^{2}}
\) .
A self-adjoint operator ( f function ) = equals −
1 one μ measure 2 two
times f function
\(
A\mathopen{}\left( f\right)\mathclose{}= {-}
\frac{1}{{μ}^{2}}
f
\)
gives
f function ( x vector ) = equals α complex number times e Euler's constant μ measure times x vector + plus β complex number times e Euler's constant − μ measure times x vector
\(
f\mathopen{}\left( x\right)\mathclose{}= α{\mathrm{e}}^{μx}+β{\mathrm{e}}^{{-}μx}
\)
(as before) and
f function ( 0 zero ) = equals 0 zero = equals f function ( 1 one )
\(
f\mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{}
\)
forces
α complex number = equals 0 zero = equals β complex number
\(
α= 0= β
\) .
We have diagonalized the operator
A self-adjoint operator \( A \) . That is, the
e unit vector n integer
\(
{e}_{n}
\) 's
form an orthonormal basis for
L 2 Lebesgue space ( [ interval 0 zero , 1 one ] interval )
\(
\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}
\)
(a fact we have already demonstrated less gracefully in an earlier example), and
A self-adjoint operator = equals 1 one π 2 two times ∑ summation n integer = 1 one ∞ infinity
1 one n integer 2 two times e unit vector n integer ⊗ e unit vector n integer
.
\[
A= \frac{1}{{π}^{2}}\sum_{n=1}^{\infty}{}
\frac{1}{{n}^{2}}{e}_{n}\otimes {e}_{n}
\text{.}
\]
Notice that this makes
‖ A self-adjoint operator ‖ = equals ρ Euclidean norm ( A self-adjoint operator ) = equals 1 one π 2 two
\(
\mathopen{}\left\lVert{}A\right\rVert\mathclose{}= ρ\mathopen{}\left( A\right)\mathclose{}= \frac{1}{{π}^{2}}
\) ,
which is not obvious from the definition of
A self-adjoint operator \( A \) .