THC Science

Hilbert apaces

A Hilbert space is an inner product space which is a complete metric space with respect to the metric generated by the inner product. A metric space is said to be complete if every sequence in the space converges to an element of the space. It can also be noticed that all the functions that form the Hilbert space are square integrable.

A function ƒ(t) is said to be square integrable on an interval [a, b] if

\int _{a}^{b}|f(t)|^{2}\,dt<{\infty }.

L^p space

Let us examine an important linear function space, the $L^{p}$ space. $L^{p}[a,b]$ denotes the set of real or complex valued functions. ƒ(t) is defined on the interval $\scriptstyle t\,\in \,[a,b]$ such that

\int _{a}^{b}|f(t)|^{p}\,dt<{\infty }.

In the $L^{p}[a,b]$ space , the norm is defined as

\left[\int _{a}^{b}|f(t)|^{p}\,dt\right]^{1/p}.

By the famous work of F.Reisz , we know that the Lp space is complete. This class of normed vector spaces that are complete are known as Banach spaces. Let's define an inner product for the $L^{p}[a,b]$ space as

\langle f,g\rangle =\int _{a}^{b}f(t){\overline {g}}(t)\,dt.

The above definition of inner product is in fact valid as it satisfies all the axioms of inner product. In particular for $L^{2}[a,b]$ , the norm is defined as

\|f\|_{2}=\left[\int _{a}^{b}|f(t)|^{2}\,dt\right]^{1/2}.

Let us consider the inner product

\langle f,f\rangle =\int _{a}^{b}|f(t)|^{2}\,dt.

We find that

\langle f,f\rangle =\|f\|_{2}\cdot \|f\|_{2}=\|f\|_{2}^{2}.

We find that the definition of inner product induces the norm in case of $L^{2}$ spaces only . Any other inner product definition does not induce norm in any of the $L^{p}$ spaces. So of all the $L^{p}$ spaces, $L^{2}$ alone is a Hilbert space

Properties of Hilbert space

For proving that $L^{p}$ is a normed space, we make use of Holder's inequality which is stated as $\|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}$ where 1/p + 1/q = 1. The numbers p and q are called as Holder's conjugate. In case of Hilbert spaces, p = q = 2 , Holder's inequality becomes

\|fg\|_{1}\leq \|f\|_{2}\|g\|_{2}.

That is,

\int _{a}^{b}|f(t).g(t)|dt\leq \left[\int _{a}^{b}|f(t)|^{2}\,dt\right]^{1/2}\cdot \left[\int _{a}^{b}|g(t)|^{2}\,dt\right]^{1/2}

The above inequality is called the Cauchy-Schwartz inequality. The $L^{q}$ space is the dual space of the $L^{p}$ space. Particularly in the case of Hilbert spaces where p=q=2 , the Linear transformation from the space to its dual is isometric(norms are equal in both the spaces) and isomorphic(one to one).

Another important theorem in Hilbert Spaces is the Reisz Representation theorem which states that any bounded Linear functional can be expressed in terms of its inner product. That is,

$\langle f,g\rangle =\int _{a}^{b}f(t)\cdot g(t)\,dt.$ is a linear functional. One cannot miss the similarity between the above expression and the one to obtain the Fourier coefficients.

Finally, the Parseval's identity uses the fact that the linear transformations between $L^{2}$ space and its dual are isometric to come up with the following result

$\sum _{k=0}^{\infty }|\langle f,f_{k}\rangle |^{2}=\|f\|^{2}$ where $\left[f_{k}\right]_{k=0}^{\infty }$ is a sequence of orthonormal basis vectors.

Fourier analysis and Hilbert spaces

With the above theory in background, we will look into the theory of Fourier analysis. Fourier's work states that any arbitrary function f(t) on the interval $[-\pi ,\pi ]$ can be decomposed as The Fourier series expansion of f is

f(t)={\frac {1}{2}}a_{0}+\sum _{n=1}^{\infty }[a_{k}\cos(kt)+b_{k}\sin(kt)]

where,

$a_{k}={\frac {2}{\pi }}\int _{-\pi }^{\pi }f(t)\cos(kt)\,dt$
$b_{n}={\frac {2}{\pi }}\int _{-\pi }^{\pi }f(t)\sin(kt)\,dt$ .

The validity of the above decomposition should be verified. From the theory that we have seen before, we need to prove the following to validate the Fourier decomposition

1. The functions $\left[\cos(kt)\right]_{k=0}^{\infty }$ and $\left[\sin(kt)\right]_{k=0}^{\infty }$ form an orthonormal basis for the space $L^{p}[-\pi ,\pi ].$

2.The system of the above trigonometric functions is complete in the space $L^{2}[-\pi ,\pi ]$

3.The limitations of the decomposition must be looked into. Also the conditions when the infinite sum does not converge must be examined.

It is known that

\left\langle {\frac {\cos(kt)}{\sqrt {\pi }}},{\frac {\cos(lt)}{\sqrt {\pi }}}\right\rangle ={\frac {1}{\pi }}\int _{-\pi }^{\pi }\cos(kt)\cos(lt)\,dt=0

for

k\neq l

and = 1 for

k=l

and similarly

\left\langle {\frac {\sin(kt)}{\sqrt {\pi }}},{\frac {\sin(lt)}{\sqrt {\pi }}}\right\rangle ={\frac {1}{\pi }}\int _{-\pi }^{\pi }\sin(kt)\sin(lt)\,dt=0

for

k\neq l

and = 1 for

k=l

Thus the trigonometric system of functions forms an orthonormal sequence for the inner product function space $L^{2}[-\pi ,\pi ].$

For showing that the trigonometric system of functions form is complete, we make use of Bessel's inequality which states that if $\left[f_{k}\right]_{k=0}^{\infty }$ is an orthonormal sequence in an inner product space , the series of real numbers $\sum _{k=0}^{\infty }|\langle f,f_{k}\rangle |^{2}=\|f\|^{2}$ converges and

$\sum _{k=0}^{\infty }|\langle f,f_{k}\rangle |^{2}\leq \|f\|^{2}$ . So for the trigonometric system of functions

\left[{\frac {1}{{\sqrt {(}}2\pi )}},{\frac {\cos(t)}{\sqrt {\pi }}},{\frac {\sin(t)}{\sqrt {\pi }}},\dots ,{\frac {\cos(kt)}{\sqrt {\pi }}},{\frac {\sin(kt)}{{\sqrt {\pi }})}}\right],\|1\|={\sqrt {(}}2\pi ),\|\cos kt\|=\|\cos kt\|{\sqrt {(}}\pi ).

Therefore, the Bessel's inequality for the trigonometric system of functions becomes

{\frac {{a}_{0}^{2}}{4}}+\sum _{k=1}^{\infty }{a}_{k}^{2}+{b}_{k}^{2}\leq \int _{-\pi }^{\pi }f(t)^{2}\,dt<\infty ,

where ƒ(t) is a square-integrable function in $L^{2}[-\pi ,\pi ]$ . Therefore , the infinite sum ${\frac {{a}_{0}^{2}}{4}}$ + $\sum _{k=1}^{\infty }{a}_{k}^{2}+{b}_{k}^{2}$ converges and hence the system is complete.This type of convergence is called convergence in mean square. This is a weaker form of convergence.

Fourier decomposition is unique or the transformation is isomorphic. This is a direct implication of Parseval's identity.Let us consider a signal f(t) in time domain .What Parseval's identity means is that the transformation of the signal from time to frequency domain causes no loss of information about the signal and one can recover the signal completely from frequency back to the time domain.It is also noted that time and frequency are dual forms of representation of an arbitrary function in L2 space.

Geometrical representation of Fourier coefficients

The Fourier coefficients $a_{k}$ and $b_{k}$ can be thought of geometrically as the projection of an arbitrary function f(t) onto the subspaces spanned by the orthonormal basis functions $\left[\cos(kt)\right]_{k=0}^{\infty }$ and $\left[\sin(kt)\right]_{k=0}^{\infty }$ .

Dirichlet's conditions and Gibbs phenomenon

Fourier decomposition fails if a function is not square integrable. We have seen that for a function to form a hilbert space, it must be square integrable. All the results that are shown do not hold good and the infinite sum could diverge.

Dirichlet's conditions states that discontinuous functions could be expressed in terms of Fourier series provided the it has finite number of discontinuties in a period. This might seem to be an anamoly as we have been discussing the cases of only continuous functions so far.Let us consider the Fourier series representation of a square wave. We find that it consists of extraneous peaks which do not disappear even if value of k is increased.This Phenomenon is called Gibbs Phenomenon.

This phenomenon reflects the difficulty in approximating a discontinuous function by a finite series of continuous sine and cosine waves. This phenomenon shows that Fourier coefficients of a function that is smooth and continuous decay rapidly for larger values of k resulting in rapid convergence whereas discontinuous functions will have very slowly decaying Fourier coefficients (causing the Fourier series to converge very slowly). This provides a partial explanation of the Gibbs phenomenon, since Fourier series with absolutely convergent Fourier coefficients would be uniformly convergent .Similarly, it is impossible for a discontinuous function to have absolutely convergent Fourier coefficients

THC Science

Bringing Science to the Cannabis Conversation!

Cannabis Indica