Subspace Approximation vs. the Riesz-Fischer Theorem

Lecture 8

The previous theorem on page took $f\in L^2(a,b)$ as given and then constructed a sequence $\{ c_1,c_2, \cdots\}$ , which by Bessel's inequality

$$\sum_{k=1}^N \vert c_k \vert^2 \le \Vert f \Vert^2 \quad \textrm{for all}~N$$

is square-summable, i.e.

$$\{c_i \} \in l^2\quad .$$

By contrast, the Riesz-Fischer theorem allows us to turn this theorem around: Given a square summable sequence,

$$\{c_i \} \in l^2\quad ,$$

the R-F theorem considers the concomitant Cauchy sequence $\sum_{k=1}^N c_k u_k,~N=1,2,\cdots$ in $\mathcal{H}=L^2(a,b)$ , and then guarantees the existence of a square-intergrable function

$$f\in L^2(a,b)\quad .$$

This function is related to $\{c_i \} \in l^2$ by virtue of the property that

$$\langle u_k,f \rangle=c_k \quad \quad k=1,2,\cdots$$

with the result that

$$\lim_{N\rightarrow \infty}\vert f- \sum_{k=1}^N c_k u_k\vert ^2=0\quad .$$ (110)

If the square-summable sequence $\{c_i \}$ yields a function which happens to be continuous, then

$$f(x)=\sum_{k=1}^\infty c_k u_k(x) \quad .$$ (111)

However, if the square-summable sequence yields a function which has one or more discontinuities, then one does not have pointwise equality, Eq.(1.11). Instead, one has the weaker condition, Eq.(1.10). This condition does not specify the value of $f$ at the point(s) of discontinuity. Instead, it specifies an equivalence class of functions, all having the same graph everywhere except at the point(s) of discontinuity.

In summary, the converse of the ``approximation via subspace theorem'' is as follows:

Theorem 15.2 (Infinite Fourier series as an element of $\cal H$ )
Given an orthonormal system, $\{ u_k\}$ , of vectors in the Hilbert space $\cal H$ , let $c_1, c_2,\cdots ,c_k, \cdots$ be a sequence of complex numbers such that

$$\sum\limits^\infty_{k=1}\vert c_k\vert^2$$

converges. Then there exists in $\cal H$ an element $f$ such that

1. $f$ has Fourier coefficients which coincide with the given numbers $c_k$ , $k=1,2,\dots$ .
2. and $f$ is such that
   

   $$\lim_{N\to\infty}\Vert f-\sum^N_{k=1}c_ku_k\Vert^2 = 0\,,$$ (112)

   
   which is equivalent to
   $$\Vert f\Vert^2 =\lim_{N\to \infty}\sum^N_{k=1}\vert c_k\vert^2 ~~\left( =\sum^\infty_{k=1}\vert c_k\vert^2 \right)\,.$$

Comments:

1. The validity this theorem is nearly obvious. From the given sequence of numbers $c_1,c_2,\dots ,c_k,\dots$ one constructs the sums
   $$w^\ast_N = \sum^N_{k=1}c_ku_k~~\qquad~~ N=1,2,\dots$$
   They belong, of course, to ${\cal H}$ , where they form a Cauchy sequence. This readily follows from the given convergence of
   $$\sum^\infty_{k=1}\vert c_k\vert^2\,.$$
   (Why? Hint: Consider $\Vert w_M^* -w_N^*\Vert ^2$ ) The Hilbert space ${\cal H}$ is complete. Consequently, that Cauchy sequence converges to an element in ${\cal H}$ , say $f$ . It is easy to show that $\langle u_k,f\rangle = \lim\limits_{N\to\infty}\langle u_k,w^\ast_N \rangle = c_k$ . It also is easy to show that Eq.(1.12) holds.

2. The statement that
   $$\lim_{N\to\infty} \Vert f-\sum^N_{k=1}c_ku_k\Vert =0$$
   is often written as

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

   
   
   and one says that ``$f$ equals $\sum\limits^\infty_{k=1} c_k u_k$ in the least square sense or in the mean''.