Dirichlet Kernel: Fountainhead of All Subspace Vectors

Consider the space of functions which lie in the subspace

$$W_{2N+1} = \textrm{span}\,\left\{ \frac{1}{\sqrt{2\pi}},\frac{1}{\sqrt{\pi}} \cos nt,\frac{1}{\sqrt{\pi}}\sin nt\colon n=1,\dots ,N\right\}$$

$$= \textrm{span}\,\left\{\frac{1}{\sqrt{2\pi}}e^{int}\colon n=0,\pm1,\dots , \pm N\right\}\,.$$

One can say that each of these functions owes its existence to the Dirichlet kernel

$$\delta_N(t-x)=\frac{1}{2\pi}\sum^N_{n=-N} e^{in(t-x)}.$$ (210)

First, note that $\delta_N(t-x)$ is a vector in $W_{2N+1}$ for every $x$ . Second, note that this vector generates a set of orthonormal basis vectors for $W_{2N+1}$ . They are generated by repeated shifts of the function $\delta _N (t)$ along the $t$ -axis. Indeed the resulting vectors are

$$g_k(t)=\frac{2\pi}{2N+1}\delta_N (t-x_k)$$

where

$$x_k=\frac{\pi}{N+\frac{1}{2}} k\,,~\qquad~k=0,1,\dots ,2N\,$$

is the amount by which the function $\delta _N (t)$ has been shifted to the right. The increment between successive shifts is evidently

$$\triangle t=\frac{\pi}{N+\frac{1}{2}}\quad ,$$

the separation between the successive zeroes of $\delta _N (t)$ in the interval $(0,2\pi )$ . This means that, to obtain the function $g_k (t)$ , the maximum of $\delta _N (t)$ has been shifted to the location of its $k^{\textrm{th}}$ zero. As a consequence, note that

$$g_k(t=x_\ell) = \frac{2\pi}{2N+1}\delta_N (x_\ell -x_k)\quad \ell ,k=0,1,\dots ,2N$$

$$= \left\{ \begin{array}{ll} 0 & \ell \ne k \\ 1 & \ell =k \end{array}\right. \quad .$$

or more briefly,

$$\boxed{g_k(x_l)=\delta_{k\ell}}~.$$ (211)

This is called the sifting property of the function $g_k$ . What is the significance of this important property? To find out compare it with the fact that the functions are orthogonal relative to the given inner product; in particular

$$\frac{2N+1}{2\pi}\int _0^{2\pi} g_k(t)g_{k'}(t)~ dt=\delta_{kk'}~~.$$

That is, except for a normalization factor, the set of elaments $\{ g_k:k=0,1,\cdots,2N\}$ forms an orthonormal basis for the subspace $W_{2N+1}$ . Note that the property

$$g_k(x_\ell )=\delta _{k\ell }$$

does not depend on the inner product structure of the subspace $W_{2N+1}$ at all. Instead, recall that this property is a manifestation of a universal property which all vector spaces have, regardless of what kind of inner product each one may or may not be endowed with. This universal property is, of course, the Duality Principle: Every vector space, in our case $W_{2N+1}$ , has a dual vector space, which is designated by $W_{2N+1}^*$ and which is the space of linear functionals on $W_{2N+1}$ . In particular, this property expresses the duality between the given basis

$$\{ g_k : k=0,1,\dots ,2N \}$$

for $W_{2N+1}$ and the dual basis

$$\{ \omega ^\ell : \ell=0,1,\dots ,2N \}$$

for $W^*_{2N+1}$ , the space of linear functionals on $W_{2N+1}$ . A typical basis functional (``dual basis element'')

$$\begin{array}{cccc} \omega ^\ell : & W_{2N+1} & \rightarrow... ... & f & \leadsto &\omega ^\ell (f)\equiv f(t=x_\ell) \end{array}$$

is the linear map (``evaluation'' map) which assigns to the vector $f$ the value of $f(t)$ (viewed as a function) at $t=x_\ell$ .

By applying this linear functional to each of the basis vectors $g_k$ in $W_{2N+1}$ one finds that

$$\omega ^\ell (g_k) \equiv g_k(x_\ell) =\delta_{kl} \quad .$$

This duality relationship between the two bases, we recall, verifies the duality between $W_{2N+1}$ and $W^*_{2N+1}$ .

The usefulness of this ``evaluation'' duality is that one can use it to solve the following reconstruction problem:

Given:

• a set of samples of the function $f\in W_{2N+1}$
  $$\{ ~(x_0, f(x_0)),~(x_1, f(x_1)), \cdots, (x_k, f(x_k)),\cdots, ~(x_{2N}, f(x_{2N} ))~\} ~~;$$

• a basis $\{ g_k \}$ for $W_{2N+1}$ consisting of functions with the sifting property
  $$g_k(x_\ell )=\delta _{k\ell } ~~.$$

Find: a set of coefficients $\{ \alpha_k \}$ such that

$$\sum^{2N}_{k=0} \alpha_k g_k(t) = f(t)$$

whenever $t=x_0,x_1, \cdots,x_{2N}$ .

This problem has an easy solution. Letting $t=x_0,x_1, \cdots,x_{2N}$ , and using the duality relation, one finds

$$\alpha_k =f(x_k) .$$

Consequently,

$$f(t) = \sum^{2N}_{k=0} f(x_k)g_k(t)\,.$$ (212)

The amazing thing about this equation is that it not only holds for the sampled values but also for any $t$ in the interval $[0,2\pi ]$ .

How is it, that one is able to reconstruct $f(t)$ with 100% precision on the whole interval $[0,2\pi ]$ by only knowing $f(t)$ at the points $\{ x_k \}$ ?

Answer: we are given the fact that the function $f(t)$ is a vector in $W_{2N+1}$ . We also are given that the functions $g_k(t),k=0,1,\cdots, 2N$ , form a basis for $W_{2N+1}$ , and that these functions have the same domain as $f(t)$ . Equation (2.12) is a vector equation. Consequently, its reinterpretation as a function equation is also 100% accurate.

Exercise 21.2 (DIRICHELET BASIS)
Consider the ($2N+1$ )-dimensional space $W_{2N+1}\subset L^2(0,2\pi)$ which is spanned by the O.N. basis $\{ \frac{1}{\sqrt{2\pi}}e^{ikt},k=0,\pm,\cdots,\pm N\}$ :

$$W_{2N+1}=span \{ \frac{1}{\sqrt{2\pi}}e^{ikt} \}_{k=-N}^{k=N}$$

Next consider the set of shifted Dirichelet kernel functions,

$$g_k(t)=\frac{2\pi}{2N+1}\delta_N(t-x_k)\equiv \frac{1}{2N+1} \sum_{n=-N}^N e^{in(t-k\pi/(N+\frac{1}{2}))}~.$$

Show that

$$\{g_k(t):~k=0,\pm 1, \cdots, \pm N \} \equiv B$$

is a basis (``Dirichelet'' basis) for $W_{2N+1}$ . This, we recall, means that one must show that

(a)

the set $B$ is one which is linearly independent, and

(b)

the set $B$ spans $W_{2N+1}$ , i.e. if $f$ is an element of $W_{2N+1}$ , then one must exhibit constants $b_k$ such that

$$f(t)=\sum_{n=-N}^N b_k g_k(t) ~.$$