Lecture Notes in Functional Analysis
by William L. Paschke
edition 0.9
image/svg+xml
A. Completeness
Definition I.28
A function
d metric : maps S set × Cartesian product S set → to R real numbers \( d : S\times S \to \mathbb{R} \) metric on a set S set \( S \)
d metric ( x element y element ) ≥ greater than or equal to 0 zero \( d\mathopen{}\left( x, y\right)\mathclose{}\geq 0 \)
d metric ( x element y element ) = equals 0 zero \( d\mathopen{}\left( x, y\right)\mathclose{}= 0 \) x element = equals y element \( x= y \)
d metric ( x element y element ) = equals d metric ( y element x element ) \( d\mathopen{}\left( x, y\right)\mathclose{}= d\mathopen{}\left( y, x\right)\mathclose{} \)
d metric ( x element z element ) ≤ less than or equal to d metric ( x element y element ) + plus d metric ( y element z element ) \( d\mathopen{}\left( x, z\right)\mathclose{}\leq d\mathopen{}\left( x, y\right)\mathclose{}+d\mathopen{}\left( y, z\right)\mathclose{} \)
for all
x element \( x \) ,
y element \( y \) ,
and
z element \( z \) in
S set \( S \) .
Definition I.29
A set with a metric is a metric space .
Proposition I.30
A normed linear space X normed linear space \( X \) metric space with metric given by
d metric ( x vector y vector ) = equals ‖ x vector - minus y vector ‖ \(
d\mathopen{}\left( x, y\right)\mathclose{}= \mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}
\)
We prove one condition is satisfied and leave the remaining conditions as Exercise I.31 .
d metric ( w vector x vector ) + plus d metric ( x vector y vector ) = equals ‖ w vector - minus x vector ‖ + plus ‖ x vector - minus y vector ‖ ≥ greater than or equal to ‖ ( w vector - minus x vector ) + plus ( x vector - minus y vector ) ‖ = equals ‖ w vector - minus y vector ‖ = equals d metric ( y vector w vector ) . \[
d\mathopen{}\left( w, x\right)\mathclose{}+d\mathopen{}\left( x, y\right)\mathclose{}= \mathopen{}\left\lVert{}w-x\right\rVert\mathclose{}+\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}\geq \mathopen{}\left\lVert{}\mathopen{}\left(w-x\right)\mathclose{}+\mathopen{}\left(x-y\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}w-y\right\rVert\mathclose{}= d\mathopen{}\left( y, w\right)\mathclose{}
\text{.}
\]
Unless otherwise specified, convergence of sequences (and therefore of series)
in a normed linear space is with respect to the metric of Proposition I.30 , and thus the topology (open sets, closed
sets, etc.) comes from the norm .
Definition I.32
A sequence ( sequence α metric space element n integer ) sequence n integer = 1 one ∞ infinity \( \mathopen{}\left({α}_{n}\right)\mathclose{}_{n=1}^{\infty} \) metric space M matrix \( M \) metric d metric \( d \) Cauchy
if
lim limit n integer → ∞ infinity
sup supremum m integer ≥ greater than or equal to n integer d metric ( α metric space element n integer α metric space element m integer ) = equals 0 zero \(
\lim_{n\to\infty}{}
\sup_{m\geq n}{}d\mathopen{}\left( {α}_{n}, {α}_{m}\right)\mathclose{}
= 0
\)
Proposition I.34
A normed linear space X normed linear space \( X \) norm ‖ · ‖ \( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \) complete if and only if
∑ summation n integer = 1 one ∞ infinity x vector n integer \( \sum_{n=1}^{\infty}{}{x}_{n}
\) X normed linear space \( X \)
∑ summation n integer = 1 one ∞ infinity ‖ x vector n integer ‖ < less than ∞ infinity \(
\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty
\)
Assume first that X normed linear space \( X \) complete and suppose x vector 1 one \( {x}_{1} \) x vector 2 two \( {x}_{2} \) X normed linear space \( X \) ∑ summation n integer = 1 one ∞ infinity ‖ x vector n integer ‖ < less than ∞ infinity \( \sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty \) s vector n integer = equals ∑ summation j integer = 1 one n integer x vector j integer \( {s}_{n}= \sum_{j=1}^{n}{}{x}_{j} \) m integer > greater than n integer \( m\gt n \)
‖ s vector m integer - minus s vector n integer ‖ = equals ‖ ∑ summation j integer = n integer + plus 1 one m integer x vector j integer ‖ ≤ less than or equal to ∑ summation j integer =
n integer + plus 1 one
m integer
‖ x vector j integer ‖ ≤ less than or equal to ∑ summation j integer =
n integer + plus 1 one
∞ infinity
‖ x vector j integer ‖ , \[
\mathopen{}\left\lVert{}{s}_{m}-{s}_{n}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{j=n+1}^{m}{}{x}_{j}\right\rVert\mathclose{}\leq \sum_{j=
n+1
}^{
m
}{}
\mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}
\leq \sum_{j=
n+1
}^{
\infty
}{}
\mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}
\text{,}
\] n integer \( n \) ∞ infinity \( \infty \) ( sequence s vector n integer ) sequence n integer = 1 one ∞ infinity \( \mathopen{}\left({s}_{n}\right)\mathclose{}_{n=1}^{\infty} \) Cauchy and has a limit, s vector \( s \) X normed linear space \( X \) s vector \( s \) ∑ summation n integer = 1 one ∞ infinity x vector n integer \( \sum_{n=1}^{\infty}{}{x}_{n}
\) s vector \( s \)
Suppose next that
∑ summation n integer = 1 one ∞ infinity ‖ x vector n integer ‖ < less than ∞ infinity \(
\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty
\) ∑ summation n integer = 1 one ∞ infinity x vector n integer \( \sum_{n=1}^{\infty}{}{x}_{n}
\) X normed linear space \( X \)
( sequence x vector n integer ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}
\) X normed linear space \( X \)
( sequence y vector k positive integer ) sequence k positive integer = 1 one ∞ infinity \(
\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}
\) Cauchy sequence in X normed linear space \( X \)
( sequence y vector k positive integer ) sequence k positive integer = 1 one ∞ infinity \(
\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}
\) Cauchy , there exists a strictly increasing sequence
( sequence k positive integer n integer ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left({k}_{n}\right)\mathclose{}_{n=1}^{\infty}
\)
‖ y vector k positive integer n integer - minus y vector k positive integer n integer + plus 1 one ‖ < less than 1 one 2 two n integer \(
\mathopen{}\left\lVert{}{y}_{{k}_{n}}-{y}_{{k}_{n+1}}\right\rVert\mathclose{}\lt \frac{1}{{2}^{n}}
\) n integer \( n \) n integer \( n \)
x vector n integer = equals y vector k positive integer n integer - minus y vector k positive integer n integer + plus 1 one \(
{x}_{n}= {y}_{{k}_{n}}-{y}_{{k}_{n+1}}
\)
∑ summation n integer = 1 one ∞ infinity ‖ x vector n integer ‖ < less than ∞ infinity \(
\sum_{n=1}^{\infty}{} \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty
\) ∑ summation n integer = 1 one ∞ infinity x vector n integer \( \sum_{n=1}^{\infty}{}{x}_{n} \) X normed linear space \( X \) ∑ summation n integer = 1 one ∞ infinity x vector n integer \( \sum_{n=1}^{\infty}{}{x}_{n} \)
( y vector k positive integer 1 one - minus y vector k positive integer 2 two ) + plus ⋯ + plus ( y vector k positive integer n integer - minus 1 one - minus y vector k positive integer n integer ) = equals y vector k positive integer 1 one - minus y vector k positive integer n integer . \[
\mathopen{}\left({y}_{{k}_{1}}-{y}_{{k}_{2}}\right)\mathclose{}+\dotsb+\mathopen{}\left({y}_{{k}_{n-1}}-{y}_{{k}_{n}}\right)\mathclose{}= {y}_{{k}_{1}}-{y}_{{k}_{n}}
\text{.} \]
( sequence y vector k positive integer n integer ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left({y}_{{k}_{n}}\right)\mathclose{}_{n=1}^{\infty}
\) Cauchy sequence with a convergent subsequence converges, the original sequence
( sequence y vector k positive integer ) sequence k positive integer = 1 one ∞ infinity \(
\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}
\)
Recall (an excellent reference for rudiments of measure theory is chapter 11 of Rudin's Principles [10 ] , also known as the blue Rudin ) for a measure space
( tuple Ω measure space , μ measure ) tuple \( \mathopen{}\left(Ω, μ\right)\mathclose{} \)
g non-negative measurable function : maps Ω measure space → to [ interval 0 zero , ∞ infinity ] interval \(
g : Ω \to \mathopen{}\left[0, \infty\right]\mathclose{}
\)
∫ integral Ω measure space g non-negative measurable function d μ measure \(
\int _{Ω}{}g\,\mathrm{d}μ
\) [ interval 0 zero , ∞ infinity ] interval \( \mathopen{}\left[0, \infty\right]\mathclose{} \) Ω measure space \( Ω \) μ measure \( μ \)
( sequence g non-negative measurable function n integer ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left({g}_{n}\right)\mathclose{}_{n=1}^{\infty}
\)
∑ summation n integer = 1 one ∞ infinity
∫ integral Ω measure space g non-negative measurable function n integer d μ measure = equals ∫ integral Ω measure space
∑ summation n integer = 1 one ∞ infinity g non-negative measurable function n integer d μ measure . \[
\sum_{n=1}^{\infty}{}
\int _{Ω}{}{g}_{n}\,\mathrm{d}μ
= \int _{Ω}{}
\sum_{n=1}^{\infty}{}{g}_{n}
\,\mathrm{d}μ\text{.}
\] h complex-valued measurable function \( h \) Ω measure space \( Ω \)
∫ integral Ω measure space | modulus h complex-valued measurable function | modulus d μ measure < less than ∞ infinity \(
\int _{Ω}{}{\mathopen{}\left\lvert{}h\right\rvert\mathclose{}}_{}\,\mathrm{d}μ\lt \infty
\)
∫ integral Ω measure space h complex-valued measurable function d μ measure \(
\int _{Ω}{}{h}_{}\,\mathrm{d}μ
\) C complex numbers \( \mathbb{C} \)
Definition I.39
For a measure space ( tuple Ω measure space , μ measure ) tuple \( \mathopen{}\left(Ω, μ\right)\mathclose{} \) p real number \( p \) 1 one ≤ less than or equal to p real number < less than ∞ infinity \( 1\leq p\lt \infty \) M p real number space of measurable functions ( Ω measure space ) \( \mathrm{M}^{p}\mathopen{}\left( Ω\right)\mathclose{} \) M p real number space of measurable functions \( \mathrm{M}^{p} \) f function \( f \) Ω measure space \( Ω \)
∫ integral Ω measure space | modulus f function | modulus p real number d μ measure < less than ∞ infinity . \[
\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{p}\,\mathrm{d}μ\lt \infty
\text{.}
\]
Exercise I.40
Prove that M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \)
Hint
Prove as an intermediate step that for functions f function \( f \) g function \( g \) M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \)
( ∫ integral Ω measure space | modulus f function + plus g function | modulus 2 two d μ measure ) 1 one 2 two ≤ less than or equal to
( ∫ integral Ω measure space | modulus f function | modulus 2 two d μ measure ) 1 one 2 two + plus
( ∫ integral Ω measure space | modulus g function | modulus 2 two d μ measure ) 1 one 2 two . \[
{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f+g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}+{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}
\text{.}
\]
Exercise I.41
Show that
〈 f function , g function 〉 M 2 space of measurable functions = equals ∫ integral Ω measure space f function times g function ¯ complex conjugate d μ measure \[
\mathopen{}\left\langle{}f, g\right\rangle\mathclose{}_{\mathrm{M}^{\mathrm{2}}}= \int _{Ω}{}f\overline{g}\,\mathrm{d}μ
\] positive sesquilinear form on M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \)
Hint
As an intermediate step, as in the proof of Proposition I.3 , prove that for functions f function \( f \) g function \( g \) M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \)
( ∫ integral Ω measure space | modulus f function times g function | modulus 2 two d μ measure ) 1 one 2 two ≤ less than or equal to
( ∫ integral Ω measure space | modulus f function | modulus 2 two d μ measure ) 1 one 2 two times
( ∫ integral Ω measure space | modulus g function | modulus 2 two d μ measure ) 1 one 2 two . \[
{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}fg\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}
\text{.}
\]
Definition I.42
Let N subspace of null vectors \( N \) null vectors (functions) of M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \) L 2 Lebesgue space = equals M 2 space of measurable functions / N subspace of null vectors \( \mathrm{L}^{\mathrm{2}}= \mathrm{M}^{\mathrm{2}}/N \)
The space L 2 Lebesgue space \( \mathrm{L}^{\mathrm{2}} \) M 2 space of measurable functions \( \mathrm{M}^{\mathrm{2}} \)
Theorem I.43
The space L 2 Lebesgue space \( \mathrm{L}^{\mathrm{2}} \) Hilbert space with respect to the inner product
〈 f function , g group element 〉 L 2 Lebesgue space = equals ∫ integral Ω measure space f function times g group element ¯ complex conjugate d μ measure . \[
\mathopen{}\left\langle{}f, g\right\rangle\mathclose{}_{\mathrm{L}^{\mathrm{2}}}= \int _{Ω}{}f\overline{g}\,\mathrm{d}μ
\text{.}
\]
Take a sequence
( sequence f function n integer ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left({f}_{n}\right)\mathclose{}_{n=1}^{\infty}
\) L 2 Lebesgue space \( \mathrm{L}^{\mathrm{2}} \)
∑ summation n integer = 1 one ∞ infinity ‖ f function n integer ‖ L 2 Lebesgue space < less than ∞ infinity \(
\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\lt \infty
\) L 2 Lebesgue space \( \mathrm{L}^{\mathrm{2}} \) L 2 Lebesgue space \( \mathrm{L}^{\mathrm{2}} \) Proposition I.21 ). We may assume that the measure space
is σ-finite (that is, a countable union of sets of finite measure), for we may fix function representatives for each f function n integer \( {f}_{n} \)
⋃ union n integer , k integer = 1 one ∞ infinity
{ set x element of measure space | such that
| modulus f function n integer ( x element of measure space ) | modulus ≥ greater than or equal to 1 one k integer } set . \[
\bigcup_{n,k=1}^{\infty}{}
\mathopen{}\left\{\, x\,\middle\vert\,
, \mathopen{}\left\lvert{}{f}_{n}\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\geq \frac{1}{k},
\,\right\}\mathclose{}
\text{.}
\] g group element ∈ element of L 2 Lebesgue space \( g\in \mathrm{L}^{\mathrm{2}} \) g group element ≥ greater than or equal to 0 zero \( g\geq 0 \)
∑ summation n integer = 1 one ∞ infinity
∫ integral Ω measure space g group element times | modulus f function n integer | modulus d μ measure ≤ less than or equal to ∑ summation n integer = 1 one ∞ infinity
‖ g group element ‖ L 2 Lebesgue space times ‖ f function n integer ‖ L 2 Lebesgue space . \[
\sum_{n=1}^{\infty}{}
\int _{Ω}{}g\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\,\mathrm{d}μ
\leq \sum_{n=1}^{\infty}{}
\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}
\text{.}
\]
∫ integral Ω measure space
∑ summation n integer = 1 one ∞ infinity g group element times | modulus f function n integer | modulus d μ measure < less than ∞ infinity . \[
\int _{Ω}{}
\sum_{n=1}^{\infty}{}g\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}
\,\mathrm{d}μ\lt \infty\text{.}
\]
g group element times ∑ summation n integer = 1 one ∞ infinity | modulus f function n integer | modulus < less than ∞ infinity \(
g\sum_{n=1}^{\infty}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\lt \infty
\)
Setting g group element = equals χ E convex set characteristic function of E convex set \( g= \chi_{E} \) E convex set \( E \)
∑ summation n integer = 1 one ∞ infinity | modulus f function n integer | modulus < less than ∞ infinity \(
\sum_{n=1}^{\infty}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\lt \infty
\)
∑ summation n integer = 1 one ∞ infinity f function n integer \(
\sum_{n=1}^{\infty}{}{f}_{n}
\) S sum of functions = equals ∑ summation n integer = 1 one ∞ infinity f function n integer \( S= \sum_{n=1}^{\infty}{}{f}_{n}
\) S sum of functions m integer = equals ∑ summation n integer = 1 one m integer f function n integer \( {S}_{m}= \sum_{n=1}^{m}{}{f}_{n}
\)
( ∫ integral Ω measure space | modulus S sum of functions - minus S sum of functions N integer | modulus 2 two d μ measure ) 1 one 2 two = equals
( ∫ integral Ω measure space lim limit m integer → ∞ infinity | modulus S sum of functions m integer - minus S sum of functions N integer | modulus 2 two d μ measure ) 1 one 2 two ≤ less than or equal to lim inf limit infimum m integer → ∞ infinity
( ∫ integral Ω measure space | modulus S sum of functions m integer - minus S sum of functions N integer | modulus 2 two d μ measure ) 1 one 2 two = equals lim inf limit infimum m integer → ∞ infinity
( ∫ integral Ω measure space | modulus ∑ summation n integer = N integer + plus 1 one m integer | modulus f function n integer | modulus | modulus 2 two d μ measure ) 1 one 2 two ≤ less than or equal to lim inf limit infimum m integer → ∞ infinity
∑ summation n integer = N integer + plus 1 one m integer ‖ f function n integer ‖ L 2 Lebesgue space = equals ∑ summation n integer = N integer + plus 1 one ∞ infinity ‖ f function n integer ‖ L 2 Lebesgue space . \[
{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}S-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}= {\mathopen{}\left(\int _{Ω}{}\lim_{m\to\infty}{}{\mathopen{}\left\lvert{}{S}_{m}-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq \liminf_{m\to\infty}{}
{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}{S}_{m}-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}
= \liminf_{m\to\infty}{}
{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}\sum_{n=N+1}^{m}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}
\leq \liminf_{m\to\infty}{}
\sum_{n=N+1}^{m}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}
= \sum_{n=N+1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}
\text{.}
\]
‖ S sum of functions - minus S sum of functions N integer ‖ L 2 Lebesgue space ≤ less than or equal to ∑ summation n integer = 1 one ∞ infinity ‖ f function n integer ‖ L 2 Lebesgue space \[
\mathopen{}\left\lVert{}S-{S}_{N}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\leq \sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}
\] S sum of functions ∈ element of L 2 Lebesgue space \( S\in \mathrm{L}^{\mathrm{2}} \) S sum of functions = equals S sum of functions - minus S sum of functions N integer + plus S sum of functions N integer \( S= S-{S}_{N}+{S}_{N} \)
lim limit N integer → ∞ infinity ‖ S sum of functions - minus S sum of functions N integer ‖ L 2 Lebesgue space = equals 0 zero \(
\lim_{N\to\infty}{}\mathopen{}\left\lVert{}S-{S}_{N}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}= 0
\)
∑ summation n integer = 1 one ∞ infinity ‖ f function n integer ‖ L 2 Lebesgue space < less than ∞ infinity \(
\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\lt \infty
\)
Definition I.44
If Ω measure space \( Ω \) S set \( S \) μ measure \( μ \) S set \( S \) l 2 little l 2 ( S set ) = equals L 2 Lebesgue space ( S set ) \( \mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( S\right)\mathclose{} \)
l 2 ( S set ) = equals { set
ξ function : maps S set → to C complex numbers | such that
∑ summation s set element ∈ S set | modulus ξ function ( s set element ) | modulus 2 two < less than ∞ infinity } set . \[
\mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}= \mathopen{}\left\{\,
ξ : S \to \mathbb{C}
\,\middle\vert\,
, \sum_{s\in S}{}{\mathopen{}\left\lvert{}ξ\mathopen{}\left( s\right)\mathclose{}\right\rvert\mathclose{}}^{2}\lt \infty,
\,\right\}\mathclose{}\text{.}
\] S set \( S \) l 2 = equals l 2 ( S set ) \( \mathrm{l}^{0}= \mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{} \)
We'll see later
that all Hilbert spaces are (isomorphic to) l 2 ( S set ) \( \mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{} \) S set \( S \) S set \( S \) l 2 \( \mathrm{l}^{0} \) l 2 ( N natural numbers (including zero) ) \( \mathrm{l}^{0}\mathopen{}\left( \mathbb{N}\right)\mathclose{} \)
The well-known procedure for completing a metric space adapts readily to normed linear spaces , yielding Banach spaces . In the case of an inner product space , we can extend the inner product to the metric completion to obtain a Hilbert space as follows: Start with an inner product space V inner product space \( V \) S set of all Cauchy sequences \( S \) Cauchy sequences in V inner product space \( V \) positive sesquilinear form on S set of all Cauchy sequences \( S \)
〈 ( sequence x vector n integer ) sequence n integer = 1 one ∞ infinity , ( sequence y vector n integer ) sequence n integer = 1 one ∞ infinity 〉 = equals lim limit n integer → ∞ infinity
〈 x vector n integer , y vector n integer 〉 . \[
\mathopen{}\left\langle{}\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}, \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}\right\rangle\mathclose{}= \lim_{n\to\infty}{}
\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}
\text{.}
\]
| modulus 〈 x vector n integer , y vector n integer 〉 - minus 〈 x vector m integer , y vector m integer 〉 | modulus ≤ less than or equal to | modulus 〈 x vector n integer , y vector n integer 〉 - minus 〈 x vector m integer , y vector n integer 〉 + plus 〈 x vector m integer , y vector n integer 〉 - minus 〈 x vector m integer , y vector m integer 〉 | modulus ≤ less than or equal to ‖ x vector n integer - minus x vector m integer ‖ times ‖ y vector n integer ‖ + plus ‖ x vector m integer ‖ times ‖ y vector n integer - minus y vector m integer ‖ . \[
\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{m}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{n}\right\rangle\mathclose{}+\mathopen{}\left\langle{}{x}_{m}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{m}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}\text{.}
\] ( sequence x vector n integer ) sequence n integer = 1 one ∞ infinity \( \mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty} \) ( sequence y vector n integer ) sequence n integer = 1 one ∞ infinity \( \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty} \) Cauchy , they are bounded, and so
‖ x vector n integer - minus x vector m integer ‖ times ‖ y vector n integer ‖ + plus ‖ x vector m integer ‖ times ‖ y vector n integer - minus y vector m integer ‖ ≤ less than or equal to A real number times ‖ x vector n integer - minus x vector m integer ‖ + plus B real number times ‖ y vector n integer - minus y vector m integer ‖ \[
\mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}\leq A\mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}+B\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}
\] A real number \( A \) B real number \( B \)
( sequence 〈 x vector n integer , y vector n integer 〉 ) sequence n integer = 1 one ∞ infinity \(
\mathopen{}\left(\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}\right)\mathclose{}_{n=1}^{\infty} \) Cauchy . Letting N set of null vectors \( N \) S set of all Cauchy sequences \( S \) Cauchy sequences that converge to 0 zero \( 0 \) V inner product space ¯ completion = equals S set of all Cauchy sequences / N set of null vectors \( \overline{V}= S/N \) V inner product space \( V \)