Hilbert Space

Hilbert Space and Separability

Hilbert Space

A Hilbert space is a real or complex inner product space that is also complete with respect to the metric induced by the inner product.

Remark

All separable Hilbert Spaces are isometrically isomorphic to $l^2$.

Separable Normed Space

A normed space is separable if it has a countable dense subset. i.e. it contains a countable subset whose closure is the entire space.

e.g.

• The finite dimensional complex spaces \newcommand{\H}{\mathcal{H}}\C^{n} are separable Hilbert spaces, with the inner product being $⟨(a_1,\dots ,a_n),(b_1,\dots ,b_n)⟩={\sum }_{i=1}^n{a}_i{\bar{b}}_i.$
• An infinite-dimensional analogue of the above example is the space ${\ell }^2(\mathbb{Z}):=\left\{(\dots ,a_{-2},a_{-1},a_0,a_1,a_2,\dots ):a_i\in \text{C},{\sum }_{i=-\infty }^{\infty }|a_i{|}^2<\infty \right\}$with the inner product $⟨(a_n{)}_{n\in \mathbb{Z}},(b_n{)}_{n\in \mathbb{Z}}⟩={\sum }_{i=-\infty }^{\infty }{a}_i{\bar{b}}_i.$
• The space of all complex-valued measurable function $f$ on ${\mathbb{R}}^d$ such that ${\int }_{{\mathbb{R}}^d}|f(x){|}^2<\infty$, modulo the equivalence relation of being equal almost everywhere, denoted as $L^2({\mathbb{R}}^d)$, is a prime example of a separable Hilbert space. The inner product is defined as $⟨f,g⟩={\int }_{{\mathbb{R}}^d}f(x)\overline{g(x)}\text{dd}x.$We shall first check that $L^2({\mathbb{R}}^d)$ is a vector space, and $⟨\cdot ,\cdot ⟩$ is a well-defined inner product. For any $f,g\in L^2({\mathbb{R}}^d)$, we have ${\int }_{{\mathbb{R}}^d}|f+cg{|}^2\le 4{\int }_{{\mathbb{R}}^d}|f{|}^2+4|c{|}^2{\int }_{{\mathbb{R}}^d}|g{|}^2<\infty ,$so $f+cg\in L^2({\mathbb{R}}^d)$ for all $c\in \text{C}$, hence $L^2({\mathbb{R}}^d)$ is a vector space. Moreover, $\begin{array}{rl}⟨f+cg,h⟩& ={\int }_{{\mathbb{R}}^d}(f+cg)\overline{h}={\int }_{{\mathbb{R}}^d}f\overline{h}+c{\int }_{{\mathbb{R}}^d}g\overline{h}=⟨f,h⟩+c⟨g,h⟩,\\ ⟨f,\lambda g⟩& ={\int }_{{\mathbb{R}}^d}f\overline{\lambda g}=\overline{\lambda }{\int }_{{\mathbb{R}}^d}f\overline{g}=\overline{\lambda }⟨f,g⟩,\\ ⟨f,g⟩& ={\int }_{{\mathbb{R}}^d}f\overline{g}={\int }_{{\mathbb{R}}^d}\overline{\overline{f}g}=\overline{⟨g,f⟩}.\end{array}$So the inner product is well-defined. Now we verify that $L^2({\mathbb{R}}^d)$ is complete.
• Indeed, $L^p({\mathbb{R}}^d)$ is a Hilbert space if and only if $p=2$. For $p\ne 2$, $L^p({\mathbb{R}}^d)$ is only a Banach space.

e.g. Consider the space $L^2(S^1)$, the set $\{e^{in\theta }{\}}_{n\in \mathbb{Z}}$ is an orthonormal basis, and for each function $f\in L^2(S^1)$, the corresponding $n$th coefficient is the $n$th Fourier coefficient: $a_n(f)=\frac{1}{2\pi }{\int }_{\pi }^{\pi }f(x)e^{-inx}\text{dd}x.$This can be seen by the following two facts:

• The Parseval’s identity holds for $L^2(S^1)$: ${\sum }_{n=-\infty }^{\infty }|a_n(f){|}^2=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }|f(x){|}^2\text{dd}x.$
• The Fourier series converges to $f$ in the $L^2$ norm: ${\lim }_{N\to 0}\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left|f(x)-{\sum }_{n\le N}{a}_n{e}^{inx}\right|}^2\text{dd}x=0$ Indeed, the mapping $f\mapsto \{a_n(f){\}}_{n\in \mathbb{Z}}$ makes $L^2(S^1)$ (unitarily) “equivalent” to ${\ell }^2(\mathbb{Z})$, we shall seriously define this equivalence, and prove this in the next section. (See the proposition.)

Pre-Hilbert Spaces

Pre-Hilbert Space

A Pre-Hilbert space is a real or complex inner product space that is not necessarily complete.

Every Pre-Hilbert Space Can be Completed

Suppose we are given a pre-Hilbert space (\H_{0},\langle\cdot,\cdot\rangle_{0}). Then there exists a unique (up to unitarily equivalence) Hilbert space $(\stackrel{˝}{,}⟨\cdot ,\cdot ⟩)$, such that \H_{0}\subset\H is dense, and $⟨\cdot ,\cdot {⟩}_0$ is the restriction of $⟨\cdot ,\cdot ⟩$ to \H_{0}.

Proof