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Introduction to Linear Algebra with Mathematica

Glossary

Preface

Let ℝ be the real line parameterized by x We consider complex-valued functions on ℝ that is integrable in Lebesue sence. The Fourier transform of a function f is denoted by
\begin{equation} \label{EqFunction.1} ℱ_{x\to \xi}\left[ f(x) \right] (\xi ) = \hat{f} (\xi ) = f^F (\xi ) = \int_{-\infty}^{+\nfty} f(x) \,e^{-{\bf j}x\cdot\xi} {\text d} x , \end{equation}
where j is the imaginary unit on the complex plane ℂ, so j² = −1. This integral \eqref{EqFunction.1} is a functional on a space of functions (called dual) parameterized by ξ. The goal is to show that f has a representation as an inverse Fourier transform
\begin{equation} \label{EqFunction.2} f(x) = ℱ_{\xi\to x}^{-1} \left[ \hat{f}(\xi ) \right] (x ) = \mbox{P.V.} \frac{1}{2\pi} \int_{-\infty}^{+\nfty} \hat{f}(\xi ) \,e^{{\bf j}x\cdot\xi} {\text d} \xi , \end{equation}
where «P.V.» is the abbreviation for Cauchy principal value regularization.

There are two problems. One is to interpret the sense in which these integrals converge. The second is to show that the inversion formula actually holds.

 

 

Absolutely Integrable Functions

The space of lebesque integrable functions 𝔏¹(ℝ) is a Banach space. Its dual space is 𝔏∞, the space of essentially bounded functions. The Fourier transform \eqref{EqFunction.1} defines a linear functional that assigns to a bounded function e−ξx ∈ 𝔏∞ a function that belongs to 𝔏¹(ℝ).

Theorem 1: If f ∈ 𝔏¹(ℝ), thenits Fourier transform fF is in 𝔏∞(ℝ) and satisfy
\[ \| \hat{f} \|_{\infty} \leqslant \| f \|_1 . \]
Furthemore, the Fourier transform of f belongs to ℭ0(ℝ), the space of bounded continuous functions that vanish at infinity.

Theorem 2: If f, g ∈ 𝔏¹(ℝ), then their convolution f☆g is another function in 𝔏¹(ℝ), defined by
\[ \left( f \star g \right) (x) = \int_{-\infty}^{+\infty} f(x-y)\,g(y)\,{\text d} x = \left( g \star f \right) (x) . \]
Their Fourier transform is the product of the Fourier transforms:
\[ ℱ_{x\to \xi}\left[ f \star g \right] = \hat{f}(\xi )\cdot \hat{g}(\xi ) . \]

Theorem 3: If f ∈ 𝔏¹(ℝ) and is also continuous and bounded, we have the inversion formula in the form
\[ f(x) = \lim_{\varepsilon\downarrow 0} \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{{\bf j}x\xi} e^{-\varepsilon |\xi |} \hat{f}(\xi ) \,{\text d} \xi . \]
The inverse Fourier transform of this kernel is
\[ ℱ_{\x\to x}^{-1} \left[ e^{-\varepsilon |\xi |} \right] = \frac{1}{\pi}\cdot \frac{\varepsilon}{\varepsilon^2 + x^2} . \]
It is easy to calculate that
\[ \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{{\bf j}x\xi} \hat{\delta}_{\varepsilon} (\xi )\,\hat{f} (\xi )\,{\text d} \xi = \left( \delta_{\varepsilon} \star f \right) (x) , \]
where
\[ \hat{\delta}_{\varepsilon} (\xi ) = e^{-\varepsilon |\xi |} . \]
However δε is an approximation of the delta function. The result follows by taking> ε → 0.

The continuous of the Fourier teansform fF follows from the dominated convergence theorem. Unlike on the torus 𝕋, 𝔏¹(ℝ) does not contain 𝔏²(ℝ), so formula \eqref{EqFunction.2} does not recover f from its Fourier transform: we simply don't know whether fF is integrable---in fact, it is generally not integrable.

 

Square Integrable Functions

Since 𝔏²(ℝ) is a Hilbert space, it provides the most elegant and simple theory of the Fourier transform.

Theorem 4: If f ∈ 𝔏¹(ℝ) ∩ 𝔏²(ℝ), then its Fourier transform fF ∈ 𝔏²(ℝ) and
\[ \| \hat{f} \|_2^2 = \| f \|_2^2 . \]

Theorem 5: Let f ∈ 𝔏²(ℝ). For every positive number 𝑎, let \( f_a = \chi_{[-a, a]} f \) be its product with characteristic function of the interval [−𝑎, 𝑎]. Then f𝑎 is in 𝔏¹(ℝ) ∩ 𝔏²(ℝ) and f𝑎 → f in 𝔏²(ℝ) as 𝑎 → ∞. Furthermore, there exists \( \hat{f} \) from 𝔏²(ℝ) such that \( \hat{f}_a \to \hat{f} \) as 𝑎 → ∞.
The map from a function into its Fourier transform gives a continuous map from 𝔏¹(ℝ) to the part of ℭ0(ℝ). That is, the Fourier transform of an integrable function is continuous and bounded (this is obvious) and approach zero at infinity. Furthermore, this map is one-to-one.

The inverse Fourier transform gives a continuous map from 𝔏¹(ℝ) to ℭ0(ℝ). This is also a one-to-one transformation

Theorem 6: If f ∈ 𝔏¹(ℝ) ∩ 𝔏²(ℝ) and its derivative f' = df/dx exists (in the sense that f is an integral of its derivative) and if f' is also in 𝔏²(ℝ), then the Fourier transform of df/dx is in 𝔏¹(ℝ). As a consequence, f is in ℭ0(ℝ).
\[ \| \hat{f} \|_2^2 = \| f \|_2^2 . \]

 

Schwartz Space

 

\[ f(r) = \int_0^{\infty} F_{\nu} (k)\,J_{\nu} (kr)\,k\,{\text d}k . \]
For Hankel transformations, we have
\[ \int_0^{\infty} r\left( \frac{{\text d}^2 f}{{\text d} r^2} + \frac{1}{r} \, \frac{{\text d} f}{{\text d} r} - \frac{\nu^2}{r^2} \, f \right) J_{\nu} (rk)\,{\text d}r = - k^2 F_{\nu} (k) = - k^2 \int_0^{\infty} f(r)\,J_{\nu} (kr)\,r\,{\text d}r . \]
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