Introduction to Mathematical Physics/Dual of a topological space

From Wikibooks, open books for an open world
< Introduction to Mathematical Physics
Jump to: navigation, search

Definition[edit]

Definition:

Let E be a topological vectorial space. The set E^\prime of the continuous linear form on E is a vectorial subspace of E^* and is called the topological dual of E.


Distributions[edit]

chapdistr

Distributions\index{distribution} allow to describe in an elegant and synthetic way lots of physical phenomena. They allow to describe charge "distributions" in electrostatic (like point charge, dipole charge). They also allow to genereralize the derivation notion to functions that are not continuous.

Definition:

L. Schwartz distributions are linear functionals continuous on {\mathcal D}, thus the elements of the dual {\mathcal D}' of {\mathcal D}.

Definition:

A function is called locally summable if it is integrable in Lebesgue sense over any bounded interval.

Definition:

To any locally summable function f, a distribution T_f defined by:


\forall \phi \in {\mathcal D}, <T_f|\phi>=\int f(x)\phi(x)dx

can be associated.

Definition:

Dirac distribution,\index{Dirac distribution} noted \delta is defined by:


\forall \phi \in {\mathcal D}, <\delta|\phi>=\phi(0)

Remark:

Physicist often uses the (incorrect!) integral notation:


\int \delta(x)\phi(x)dx=\phi(0)

to describe the action of Dirac distribution \delta on a function \phi.


Convolution of two functions f and g is the function h if exists defined by:


h(x)=\int_{-\infty}^{+\infty}f(u)g(x-u)du

and is noted:


h=f * g.

Convolution product of two distributions S and T is (if exists) a distribution noted S*T defined by:


\forall \phi(x) \in {\mathcal D}, <S*T,\phi>=<S(x)T(y),\phi(x+y)>

Here are some results:

  • convolution by \delta is unity of convolution.
  • convolution by \delta^{\prime} is the derivation.
  • convolution by \delta^{(m)} is the derivation of order m.
  • convolution by \delta(x-a) is the translation of a.

The notion of Fourier transform of functions can be extended to distributions. Let us first recall the definition of the Fourier transform of a function:

Definition:

Let f(x) be a complex valuated function\index{Fourier transform} of the real variable x. Fourier transform of f(x) is the complex valuated function \hat
f(\sigma)<math> of real variable \sigma</math> defined by:


\hat f(\sigma)=\int_{-\infty}^{+\infty}f(x)e^{-i2\pi \sigma x}dx,

if it exists.

A sufficient condition for the Fourier transform to exist is that f(x) is summable. The Fourier transform can be inverted: if


\hat f(\sigma)=\int_{-\infty}^{+\infty}f(x)e^{-i2\pi \sigma x}dx,

then


f(x)=\int_{-\infty}^{+\infty}\hat f(\sigma)e^{i2\pi \sigma x}dx,

Here are some useful formulas:


f(-x)\rightarrow \hat f(-\sigma)


\bar f(x)\rightarrow \bar{\hat f}(-\sigma)


f(x-a)\rightarrow e^{-2i\pi\sigma a}\hat{f}(\sigma)


f(ax)\rightarrow \frac{1}{a}\hat f(\frac{\sigma}{a})


f'(x)\rightarrow 2i\pi \sigma \hat f(\sigma)


f \ast g \rightarrow \hat f \hat g

Let us now generalize the notion of Fourier transform to distributions. The Fourier transform of a distribution can not be defined by


\forall \phi\in{\mathcal D}, <\hat T|\phi>=<T|\hat \phi>

Indeed, if \phi\in{\mathcal D}, then \phi\notin{\mathcal D} and the second member of previous equality does not exist.

Definition:

Space {\mathcal S} is the space of fast decreasing functions. More precisely, \phi\in{\mathcal S} if

  • its m-derivative \phi^{(m)}(x) exists for any positive integer m.
  • for all positive or zero integers p and m, x^p\phi^{(m)}(x) is bounded.

Definition:

Fourier transform of a tempered distribution T is the distribution \hat T defined by


\forall \phi\in{\mathcal S}, <\hat T|\phi>=<T|\hat \phi>

The Fourier transform of the Dirac distribution is one:


\delta(x) \rightarrow 1(x)

Distributions\index{distribution} allow to describe in an elegant and synthetic way lots of physical phenomena. They allow to describe charge "distributions" in electrostatic (like point charge, dipole charge). They also allow to genereralize the derivation notion to functions that are not continuous.

Statistical description[edit]

Random variables[edit]

Distribution theory generalizes the function notion to describe\index{random variable} physical objects very common in physics (point charge, discontinuity surfaces,\dots). A random variable describes also very common objects of physics. As we will see, distributions can help to describe random variables. At section secstoch, we will introduce stochastic processes which are the mathemarical characteristics of being nowhere differentiable.

Let B a tribe of parts of a set \Omega of "results" \omega. An event is an element of B, that is a set of \omega's. A probability P is a positive measure of tribe B. The faces of a dice numbered from 0 to 6 can be considered as the results of a set \Omega=\{\omega_1,\dots,\omega_6\}. A random variable X is an application from \Omega into R (or C). For instance one can associate to each result \omega of the de experiment a number equal to the number written on the face. This number is a random variable.

Probability density[edit]

Distribution theory provides the right framework to describe statistical "distributions". Let X be a random variable that takes values in R.

Definition:

The density probability function f(x) is such that:


P(X\in \mathrel{]}x,x+dx\mathrel{]})=f(x)dx

It satisfies: \int f(x)dx=1

Example:

The density probability function of a Bernoulli process is:


f(x)=p\delta (x)+q\delta (x-1)

Moments of the partition function[edit]

Often, a function f(x) is described by its moments:

Definition:

The p^{th} moment of function f(x) is the integral


m_p=\int x^pf(x)dx

Definition:

The mean of the random variable or mathematical expectation is moment m_1

Definition:

Variance v is the second order moment:


v=\sigma^2=\int (x-m_1)^2f(x)dx

The square root of variance is called ecart-type and is noted \sigma.

Generating function[edit]

Definition:

The generatrice function\index{generating function} of probability density f is the Fourier transform off.

Example:

For the Bernouilli distribution:


\hat{f}(\nu)=p+qe^{-2i\pi \nu}

The property of Fourier transform:


\hat{f}^{(p)}(\nu)=\int(-2i\pi x)^p f(x)e^{-2i\pi\nu x}

implies that:


m_p=\frac{1}{(-2i\pi)^p}\hat{f}^{(p)}(0)

Sum of random variables[edit]

Theorem:

The density probability f_x associated to the sum x=x_1+x_2 of two {\bf independent} random variables is the convolution product of the probability densities f_{x_1} and f_{x_2}.

We do not provide here a proof of this theorem, but the reader can on the following example understand how convolution appears. The probability that the sum of two random variables that can take values in i,\dots,N with N>n is n is, taking into account all the possible cases:


P_{(x=n)}=P_{(x_1=0)}P_{(x_2=n)}+
P_{(x_1=1)}P_{(x_2=n-1)}+\dots+P_{(x_1=0)}P_{(x_2=n)}

This can be used to show the probability density associated to a binomial law. Using the Fourier counterpart of previous theorem:

\begin{matrix}
\hat{f}(\nu)&=&(p+qe^{-2i\pi \nu})^n \\
&=&\sum_{k=0}^{n}C_n^k p^{n-k}q^{k}e^{-2i\pi \nu k}
\end{matrix}

So


f(x)=\sum_{k=0}^{n}C_n^k p^{n-k}q^{k}\delta(x-k)

Let us state the central limit theorem.

Theorem:

The convolution product of a great number of functions tends\footnote{ The notion of limit used here is not explicited, because this result will not be further used in this book. } to a Gaussian. \index{central limit theorem}


[f(x)]^{*n} \longrightarrow \frac{1}{\sigma\sqrt{2\pi
n}}e^{-\frac{x^2}{2n\sigma^2}}

Proof:

Let us give a quick and dirty proof of this theorem. Assume that \hat{f}(\nu) has the following Taylor expansion around zero:


\hat{f}(\nu)=\hat{f}(0)+\nu\hat{f}^{\prime}(0)+\frac{\nu^2}{2}\hat{f}^{\prime\prime}(0)

and that the moments m_p with p\geq 3 are zero. then using the definition of moments:


\hat{f}(\nu)=m_0+\nu(-2i\pi )m_1+\frac{\nu^2}{2}(-2i\pi)^2m_2

This implies using m_0=1 that:


\ln((\hat{f}(\nu))^n)=n\ln(1-2i\pi \nu m_1-\frac{\nu^2}{2}(-2i\pi)^2m_2)

A Taylor expansion yields to


\ln((\hat{f}(\nu))^n)\sim n(-2i\pi \nu m_1-\frac{\nu^2}{2}(-2i\pi)^2m_2)

Finally, inverting the Fourier transform:


(\hat{f}(\nu))^{*n}\sim \delta(x-nm_1)*\frac{1}{\sigma\sqrt{2\pi
n}}e^{-\frac{x^2}{2n \sigma^2}}