Isomorphic Hilbert Spaces

Lecture 9

The completeness relation, Eq. (1.13), is a remarkable result! It implies the generalized Fourier expansion

$$f(x)\doteq\sum^\infty_{k=1} c_k u_k$$

with $c_k=\langle u_k,f\rangle$ . Indeed, the completeness relation implies

$$0 = \langle f,f\rangle - \lim_{N\to\infty} \sum^N_{k=1}\overline{c}_k c_k\,.$$

Subtracting and adding the limit of the sum

$$\sum^N_{\ell =1} c_\ell\langle f,u_\ell\rangle = \sum^N_{\ell =1}\,\sum^N_{k =1} c_\ell\overline{c}_k\langle u_k,u_\ell\rangle$$

one has

$$= \langle f,f\rangle -\lim_{N\to\infty}\Big\{ \sum^N_{k=1}\overline{c}_k \langle u_k,f\rangle -\sum^N_{\ell =1}c_\ell\langle f,u_\ell\rangle$$ 0

$$~\qquad~~+\sum^N_{\ell =1}\sum^N_{k=1} c_\ell\overline{c}_k\langle u_k,u_\ell \rangle\Big\}$$

$$= \langle f-\lim_{N\to\infty}\sum^N_{k=1}c_k u_k,f-\lim_{N\to\infty} \sum^N_{\ell =1}c_\ell u_\ell\rangle$$ 0

$$= \lim_{N\to\infty}\Vert f-\sum^N_{k=1}c_k u_k\Vert^2\,,$$ 0

which is what is meant by

$$f\doteq\sum^\infty_{k=1}c_ku_k\,;~~c_k=\langle u_k,f\rangle~\quad~\forall\, f\in{\cal H}\,.$$

But there is more. The geometrical significance of this generalized Fourier expansion is astonishing. It is a one-to-one linear correspondence - let us call it ${\cal F}$ - between $f\in{\cal H}=L^2(a,b)=$ the Hilbert space of square integrable functions on $[a,b]$ and $\{ c_1,c_2,\dots ,c_k,\dots\}\in\ell^2=$ the Hilbert space of square summable series (``functions on the integers''). The correspondence

$$f \mathop{\to}\limits^{\cal F} {\cal F}(f)=\{ c_k\}\equiv\{ c_1,c_2,\dots ,c_k,\dots\}$$

$$g \mathop{\to}\limits^{\cal F} {\cal F}(g)=\{ d_k\}\equiv\{ d_1,d_2,\dots ,d_k,\dots\}$$

$$\alpha f+\beta g \mathop{\to}\limits^{\cal F} {\cal F}(\alpha f+\beta g)= \alpha\{ c_k\} +\beta\{ d_k\}$$

(i) is one-to-one and onto, which means it has an inverse:

$${\cal F}^{-1}(\{ c_k\})=f\equiv \sum^\infty_{k=1}c_ku_k \stackrel{{\mathcal F}^{-1}}{\leftarrow}\{ c_k\}~~;$$

(ii) is linear, which means it takes closed triangles in $L^2$ into closed triangles in $\ell ^2$ :

$$\{ c_k+d_k\} = \{ c_k\} +\{ d_k\}\,;$$

Figure 1.7: Linear map between $L^2$ , the space of square integrable functions, and $l^2$ , the space of square summable sequences. This map is an isometry which is induced by the generalized Fourier series.

(iii) preserves lengths. Indeed,

$$f = \sum^\infty_{k=1}c_k u_k$$

$$g = \sum^\infty_{\ell =1} d_\ell u_\ell$$

implies

$$\langle f-g,f-g\rangle = \sum_{k=1}^\infty (\overline{c}_k - \overline{d}_k)(c_k-d_k)$$

$$\Vert f-g\Vert^2 = \sum^\infty_{k=1}\vert c_k-d_k\vert^2~\qquad~\forall\, f,g\in{\cal H}\,.$$

Figure 1.8: The isometry between $L^2$ , the space of square integrable functions, and $l^2$ , the space of square summable sequences, preserves lengths and angles.

It follows that

$$\langle f,g\rangle = \sum^\infty_{k=1}\overline{c}_k d_k$$

$$\Vert f\Vert^2 = \sum^\infty_{k=1}\vert c_k\vert^2 \quad .$$

Consequently, ${\cal F}$ preserves lengths, inner products, and angles (if the Hilbert space is real).

Definition: A linear transformation which is one-to-one and onto is called an isomorphism.

Definition: A distance preserving transformation between two metric spaces is called an isometric transformation, or simply an isometry.

In that case, the two spaces are said to be isometric spaces. This means they look the same from the viewpoint of geometry.

To summarize, the striking feature of the completeness, i.e., Parseval's relation is that it establishes an isometric isomorphism, or more briefly an isometry between the two spaces.

Thus

$$L^2(a,b)~\textrm{and}~\ell^2~\textrm{are~isometric~Hilbert~spaces}\,.$$

They are geometrically the same (right triangles in one space correspond to right triangles in the other space).

Because one can establish a linear isometry between any Hilbert space and one and the same $\ell ^2$ , the space of square summable series, one obtains the

Theorem: (Isomorphism theorem) Any two complex Hilbert spaces are isomorphic. In fact, so are any two real Hilbert spaces.

Lecture 10

Comment: The isometric isomorphism is a unitary transformation whose elements are $\{ u_1(x),u_2(x),\dots\}$ . Indeed, consider the equation

$$\sum^\infty_{k=1}u_k(x)c_k =f(x)~~.$$

The coefficients $c_k$ are the components of an infinite dimensional column vector in $\ell ^2$ . The function $f$ is an infinite dimensional column vector whose components $f(x)$ are labelled by the continuous index $x$ . It follows that $\{ u_k(x) \}$ are the entries of a matrix whose columns are labelled by $k$ and whose rows are labelled by $x$ . The orthogonality conditions

$$\langle u_i, u_j\rangle = \delta_{ij}$$

expresses the orthonormality of the colums of this matrix. The completeness relation

$$\sum^\infty_{k=1}u_k(x) \overline{u}_k (x) = \frac{\delta (x-x')}{\rho (x)}$$

expresses the orthonormality of the rows. It follows that $\{ u_k(x) \}$ represents a unitary transformation which maps $\mathcal{H}=L^2(a,b)$ onto $\ell ^2$ .

Exercise 15.1 (SQUARED LENGTHS AND INNER PRODUCTS)
An isometry between the Hilbert space $\cal{H}$ of square integrable functions $f$ , and the Hilbert space $\ell ^2$ of square summable sequences $\{c_k\}_{k=1}^\infty$ is a linear one-to-one and onto transformation $f\rightarrow \{c_k\}$ with the property that it preserves squared lengths:

$$\langle f,f \rangle =\sum_{k=1}^\infty \vert c_k \vert^2~,~~~ \quad \forall f \in \cal{H}~~.$$

SHOW that

$$\langle f,g \rangle =\sum_{k=1}^\infty \overline{c_k} d_k ~,~~~d_k =\langle u_k,g\rangle \quad \forall f,g \in \cal{H}~~.$$

where

$$f\rightarrow \{c_k\} ~\textrm{and} ~g\rightarrow \{d_k\}~.$$

Exercise 15.2
Let $g$ be a fixed and given square integrable function, i.e.

$$0<\int^\infty_{-\infty}\overline g(x) g(x)~dx\equiv\Vert g \Vert^2<\infty$$

One can think of $g$ as a function whose non-zero values are concentrated in a small set around the origin $x=0$ .

Consider the concomitant ``windowed'' Fourier transform on $L^2(-\infty ,\infty )$ , the space of square integrable functions,

$$T:~~L^2(-\infty,\infty) \rightarrow \cal{R} (T)$$

$$f \rightarrow Tf(\omega,t)\equiv \int^\infty_{-\infty}\overline{g}(x-t)e^{-i\omega x}f(x)~dx$$

Let $h(\omega,t)$ be an element of the range space $\cal{R} (T)$ . It is evident that

$$\langle h_1,h_2 \rangle =\int^\infty_{-\infty}\int^\infty_{-\infty} \overline h_1(\omega,t) h_2(\omega,t)~d\omega dt$$

is an inner product on $\cal{R} (T)$ .

FIND a formula for $\langle Tf_1,Tf_2 \rangle$ in terms of the inner product

$$(f_1,f_2 )\equiv \int^\infty_{-\infty}\overline f_1(x) f_2(x)~dx$$

on $L^2(-\infty ,\infty )$ .

Exercise 15.3 (SHANNON'S SAMPLING FUNCTIONS)
By

(a)

starting with the orthonormality relation

$$\int\limits^\pi_{-\pi} \delta_N \left(t - \frac{2\pi}{2N+1}k\right) \delta_N \left(t - \frac{2\pi}{2N+1}l\right)~dt = \frac{2N+1}{2\pi} \delta_{kl}\,;$$

$$k,l = -N,\dots, N$$

where

$$\delta_N (u) = \frac{1}{2\pi} \frac{\sin (N + \frac{1}{2})u}{\sin... ...u}{2}} \quad \left( = \frac{1}{2\pi} \sum\limits^N_{n = -N} e^{inu}\right)\,,$$

(b)

then rescaling the integration domain by introducing the variable

$$z = \frac{N + \frac{1}{2}}{2\pi w}~t\,,$$

where $w > 0$ is a fixed constant (the "band width"),

(c)

and finally going to the limits $N\to\infty$ .

(i)

Show that the set of functions

$$\left\lbrace \frac{\sin\pi (2wz - k)}{\pi (2wz - k)} \equiv \sin {\rm c} (2wz-k) : \quad k = 0,\pm1, \pm 2,\dots \right\rbrace$$

is an orthogonal set satisfying

$$\int\limits^\infty_{-\infty} \sin {\rm c} (2wz-k)\sin {\rm c} (2wz-l) dz = A\delta_{kl}\,.$$

What is $A$ ?

(ii)

This set of functions

$$\left\{ u_k=\frac{1}{\sqrt A} \,\hbox{sinc}(2wz-k):~k=0,\pm 1,\cdots \right\}$$

is not complete on $L^2(-\infty ,\infty )$ , but it is complete on a certain subset $B\subset L^2 (-\infty,\infty)$ .

What is this subset? i.e. What property must a function $f(t)$ have in order that $f\in B$ ?

This question can be answered with the help of Parseval's (``completeness'') relation as follows: Recall that completeness on $B$ here means that $f\in B$ implies that one can write $f$ as

$$f(z)=\sum_{k=-\infty}^\infty c_ku_k(z),~~~~u_k=\frac{1}{\sqrt A} \,\hbox{sinc}(2wz-k)$$

with $c_k=\langle u_k,f\rangle$ , which we know is equivalent to

$$\langle f,f\rangle =\sum_{k=-\infty}^\infty \vert c_k\vert^2~~.$$ (115)

Thus, to answer the question, we must ask and answer: What property must $f$ have in order to guarantee that Eq.(1.15) be satisfied? Therefore, to give a correct answer, one must (i) identify the property and (ii) then show that Parseval's relation is satisfied by every such function.

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