Banach and Hilbert spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 18 May 2011

Banach and Hilbert spaces

Let $ℂ$ be the field of complex numbers. A complex vector space is an abelian group $V$ with a function $ℂ\times V\to V$ such that

(a)   If $c\in ℂ$ and $v_1,v_2\in V$ then $c(v_1+v_2)=c{v}_1+c{v}_2$,

(b)   If $c_1,c_2\in ℂ$ and $v\in V$ then $(c_1+c_2)v=c_{1}v+c_{2}v$,

(c)   If $c_1,c_2\in ℂ$ and $v\in V$ then $c_1\left(c_{2}v\right)=\left(c_1{c}_2\right)v$,

(d)   If $v\in V$ then $1v=v$.

Let $X$ and $Y$ be complex vector spaces. A linear transformation from $X$ to $Y$ is a function $T:X\to Y$ such that

if $c_1,c_2\in ℂ$ and $x_1,x_2\in X$     then     $T(c_1{x}_1+c_2{x}_2)=c_{1}T\left(x_1\right)+c_{2}T\left(x_2\right)$.

The morphisms in the category of vector spaces are linear transformations.

A topological vector space is a complex vector space $V$ with a topology such that addition and scalar multiplication are continuous maps. The morphisms in the category of topological vector spaces are continuous linear transformations.

Let $V$ be a complex vector space. A set $C\subseteq V$ is convex if $C$ satisfies

if $x,y\in C,t\in [0,1]$ then $tx+(1-t)y\in C$.

A topological vector space $V$ is locally convex if $V$ has a basis of neighbourhoods of 0 consisting of convex sets.

Let $V$ be a complex vector space and let $C\subseteq V$ be a convex subset of $V$. A function $f:C\to ℝ$ is convex if $f$ satisfies

if $x,y\in C,t\in [0,1]$ then $f(tx+(1-t\left)y\right)\le tf\left(x\right)+(1-t)f\left(y\right)$.

HW: Show that the exponential function $\exp :(a,b)\to ℝ$ is convex.

A normed linear space is a complex vector space $V$ with a function $\Vert \Vert :V\to {ℝ}_{\ge 0}$ such that

(a)   if $x,y\in V$ then $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$,

(b)   if $c\in ℂ$ and $v\in V$ then $\Vert cv\Vert =\left|c\right|\Vert v\Vert$,

(c)   if $v\in V$ and $\Vert v\Vert =0$ then $v=0$.

HW: Show that if $V$ is a normed linear space then the map $\Vert \Vert :V\to {ℝ}_{\ge 0}$ is uniformly continuous, $V$ is a metric space with respect to the metric $d:V\times V\to {ℝ}_{\ge 0}$ defined by

$d(x,y)=\Vert x-y\Vert$

and, with the metric space topology, $V$ is a topological vector space.

A Banach space is a normed linear space $X$ such that $X$ is a complete metric space with respect to the metric $d:X\times X\to {ℝ}_{\ge 0}$ defined by

$d(x,y)=\Vert x-y\Vert$.

Let $X$ and $Y$ be normed linear spaces. An isometry from $X$ to $Y$ is a linear transformation $T:X\to Y$ such that

if $x\in X$     then     $\Vert Tx\Vert =\Vert x\Vert$.

Let $X$ and $Y$ be normed linear spaces. The norm of a linear transformation $T:X\to Y$ is

$\Vert T\Vert =\sup \left\{\Vert Tx\Vert \right|x\in X\text{such that}\Vert x\Vert \le 1\}$.

A linear transformation is bounded if $\Vert T\Vert <\infty$.

HW: If $X$ and $Y$ are normed linear spaces such that points are closed then a linear transformation $T:X\to Y$ is continuous iff it is bounded (reference???)

HW: Show that if $X$ and $Y$ are normed linear spaces then $B(X,Y)=\{\text{bounded linear transformations}\phi :X\to Y\}$ with $\Vert \Vert$ is a normed linear space and that if $Y$ is a Banach space then $B(X,Y)$ is a Banach space.

A Hilbert space is a complex vector space $V$ with a function $⟨,⟩:V\times V\to ℂ$ such that

(a)   if $v_1,v_2\in V$ then $⟨v_1,v_2⟩=\overline{⟨v_2,v_1⟩}$,

(b)   if $c_1,c_2\in ℂ$ and $v_1,v_2,v_3\in V$ then $⟨c_1{v}_1+c_2{v}_2,v_3⟩=$ c_1⟨v_1,v_3⟩+c_2⟨v_2,v_3⟩$,$

(c)   if $v\in V$ and $⟨v,v⟩=0$ then $v=0$,

(d)   $V$ is a Banach space with norm $\Vert \Vert :V\to {ℝ}_{\ge 0}$  given by      ${\Vert v\Vert }^2=⟨v,v⟩$.

Let $V$ be a Hilbert space and let $T:V\to V$ be a linear transformation. The adjoint of $T$ is the linear transformation

$T^{*}:V\to V$    defined by     $⟨T{v}_1,v_2⟩=⟨v_1,T^{*}{v}_2⟩$,

for $v_1,v_2\in V$. The linear transformation $T:V\to V$ is unitary if $T$ satisfies

if $v_1,v_2\in V$     then     $⟨T{v}_1,T{v}_2⟩=⟨v_1,v_2⟩$.

Duals

Let $X$ with $\Vert \Vert :V\to {ℝ}_{\ge 0}$ be a normed linear space. Define

$X^{*}=\{\phi :X\to ℂ|\phi \text{is a linear transformation and}\Vert \phi \Vert <\infty \},$

where

$\Vert \phi \Vert =\sup \left\{\Vert \phi \left(x\right)\Vert \right|x\in X\text{such that}\Vert x\Vert \le 1\}$.

Then, see [Ru, 5.21 and Ch. 5 Ex. 8],

(a)   $X^{*}$ is a Banach space.

(b)   $X^{*}$ separates points on $X$, i.e. if $x_1,x_2\in X$ and $x_1\ne x_2$ then there exists $\phi \in X^{*}$ such that $\phi \left(x_1\right)\ne \phi \left(x_2\right)$.

(c)   The map

$\begin{array}{rccl}\iota :& X& ⟶& X^{**}\\ & \begin{array}{c}x\\ \\ \end{array}& \begin{array}{c}⟼\\ \\ \end{array}& \begin{array}{cccc}{\iota }_x:& X^{*}& \to & ℂ\\ & \phi & \mapsto & \phi \left(x\right)\end{array}\end{array}$

is an injective linear map such that $\Vert {\iota }_x\Vert =\Vert x\Vert$.

The construction of $f$ in part (b) is a special case (or corollary) of the Hahn-Banach theorem, see [Ru, 5.21 and Theorem 5.20].

If $M$ is a subspace of a normed linear space $X$ and $\phi :M\to ℂ$ is a bounded linear functional then there exists a bounded linear functional $\Phi :X\to ℂ$ such that

(a)   if $m\in M$ then $\Phi \left(m\right)=\phi \left(m\right)$,

(b)   $\Vert \Phi \Vert =\Vert \phi \Vert$.

The proof of this theorem is essentially by induction, where the induction step extends $\phi$ from $M$ to $M+ℂx_0$ for a vector $x_0$ which is not in $M$.

Notes and References

These notes were synthesized from [Ru], [Kirillov], ... ????? They have evolved over the years through graduate courses in "Representation Theory" at University of Wisconsin, Madison and a course in "Measure Theory" at the Masters level at University of Melbourne. This presentation follows [Ru, Chapters 4 and 5].

References

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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