Generalized Fourier Transform: Tempered Distribution

\[\DeclareMathOperator{\sgn}{sgn}\]

A tempered distribution is a linear functional:

\[f: \mathcal{S} \to \mathbb{C}\]

• The space of all tempered distributions denoted \(\mathcal{S}'\), is the dual space of \(\mathcal{S}\).

• The value in \(\mathbb{C}\) that the distribution \(f \in \mathcal{S}'\) assigns to \(\phi \in \mathcal{S}\) is usually denoted by:

  \[\langle f, \phi \rangle\]

Properties without proof:

• Linearity

  \[a \langle f, \phi \rangle + b \langle f, \psi \rangle = \langle f, a \phi + b \psi \rangle\]

• Continuity

  \[\lim_{n \to \infty} \phi_n = \phi \implies \lim_{n \to \infty} \langle f, \phi_n \rangle = \langle f, \phi \rangle\]

Delta Distribution

The \(\delta\) distribution, aka the unit impulse, is a generalized function or distribution.

Operationally, the effect of \(\delta\) is to evalute the function at the origin (i.e. pull out the value of origin).

Classical Interpretation

\[\begin{split}\delta (x) = \begin{cases} \infty & x = 0 \\ 0 & x \ne 0 \end{cases}\end{split}\]

\[\int_{-\infty}^{+\infty} \delta (x) dx = 1\]

It can be viewed as a sequence of bump functions:

That is really operationally how it appear: by limiting process you were concentrating things and you were just pulling out the value at the origin.

\[\delta(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}\]

\[\int_{-\infty}^{+\infty} \delta (x) \phi (x) dx = \phi(0)\]

Distribution Interpretation

It is a mathematical modus operandi by turning the solution of a problem into a definiation

We define that \(\delta\) distribution maps every continuous function to its value at zero of its domain:

\[\langle \delta, \phi \rangle = \phi(0)\]

\(\delta\) is also linear and continuous:

\[a \langle \delta, \phi \rangle + b \langle \delta, \psi \rangle = \langle \delta, a \phi + b \psi \rangle\]

\[\lim_{n \to \infty} \phi_n = \phi \implies \lim_{n \to \infty} \langle \delta, \phi_n \rangle = \langle \delta, \phi \rangle\]

Define a new distribution \(\delta_a\)

\[\delta_a: \phi \in \mathcal{S} \mapsto \langle \delta_a, \phi \rangle = \phi(a)\]

Function Induced Distribution

you give me a (test) function, I have to tell you how a distribution (generalized function) operates on it (by integraion for Fourier transform).

Lemma: if \(f : \mathbb{R}^n \to \mathbb{C}\):

• polynomially bounded

• Riemann integrable on \([−M, M]\) for each \(M > 0\).

Then:

\[f: \phi \in \mathcal{S} \mapsto \langle f, \phi \rangle = \int_{-\infty}^{+\infty} f(x) \phi(x) dx\]

is a tempered distribution (generalized function) induced by \(f\).

As the rapidly decreasing functions are “good enough”, most weird functions can be considered generalized functions (aka distributions) by defining the pairing as:

\[\langle f, \phi \rangle = \int_{-\infty}^{+\infty} f(x) \phi(x) dx\]

For example, the integral of \(sin(2 \pi x)\) does not make any sense, but \(\int_{-\infty}^{+\infty} sin(2 \pi x) \phi(x) dx\) will converge as \(\phi\) is a rapidly decreasing function. Thus \(sin(2 \pi x)\) can be considered as a generalized function as I can tell you how it can operate on test functions (i.e. by integraion).

\[\langle sin(2 \pi x), \phi \rangle = \int_{-\infty}^{+\infty} sin(2 \pi x) \cdot \phi (x) dx\]

Similarly:

\[ \begin{align}\begin{aligned}\langle 1, \phi \rangle = \int_{-\infty}^{+\infty} 1 \cdot \phi (x) dx\\\langle \Pi, \phi \rangle = \int_{-\infty}^{+\infty} \Pi(x) \cdot \phi (x) dx\end{aligned}\end{align} \]

Fourier Transform of Distribution

Properties:

\[T \in \mathcal{S}' \implies \mathcal{F} T \in \mathcal{S}'\]

\[T \in \mathcal{S}' \implies \mathcal{F}^{-1} T \in \mathcal{S}'\]

Theorem

\[\phi \in \mathcal{S}, T \in \mathcal{S}' \implies \langle \mathcal{F} T, \phi \rangle = \langle T, \mathcal{F} \phi \rangle\]

Proof:

\[\begin{split}\langle \mathcal{F} T, \phi \rangle & = \int_{-\infty}^{+\infty} \mathcal{F} T (x) \phi (x) dx \\ & = \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} e^{-2 \pi i x y} T(y) dy \right) \phi (x) dx \\ & = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-2 \pi i x y} \phi(x) T(y) dy dx \\ & = \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} e^{-2 \pi i x y} \phi(x) T(y) dx \right) dy \\ & = \int_{-\infty}^{+\infty} T(y) \cdot \mathcal{F} \phi (y) dy \\ & = \langle T, \mathcal{F} \phi \rangle\end{split}\]

Similarly:

\[\phi \in \mathcal{S}, T \in \mathcal{S}' \implies \langle \mathcal{F}^{-1} T, \phi \rangle = \langle T, \mathcal{F}^{-1} \phi \rangle\]

Applications

Fourier Transform of shifted delta function:

\[\begin{split}\langle \mathcal{F} \delta_a, \phi \rangle & = \langle \delta_a, \mathcal{F} \phi \rangle \\ & = \mathcal{F} \phi (a) \\ & = \int_{-\infty}^{+\infty} e^{-2 \pi i a x} \phi(x) dx \\ & = \langle e^{-2 \pi i a x}, \phi \rangle\end{split}\]

\[\therefore \mathcal{F} \delta_a = e^{-2 \pi i a x}\]

Specifically when \(a = 0\):

\[\mathcal{F} \delta = \mathcal{F} \delta_0 = 1\]

\(\delta\) is infinitely concentrated, its Fourier transform \(F \delta\) is uniformly spread out

Fourier Transform of complex exponential function:

\[\begin{split}\langle \mathcal{F} e^{2 \pi i a x}, \phi \rangle & = \langle e^{2 \pi i a x}, \mathcal{F} \phi \rangle \\ & = \int_{-\infty}^{+\infty} e^{2 \pi i a x} \mathcal{F} \phi(x) dx \\ & = \mathcal{F}^{-1} \mathcal{F} \phi(a) \\ & = \phi(a) \\ & = \langle \delta_a, \phi \rangle\end{split}\]

\[\therefore \mathcal{F} e^{2 \pi i a x} = \delta_a\]

Fourier Transform of cosine function:

\[\begin{split}\mathcal{F} cos(2 \pi a x) & = \mathcal{F} (\frac{1}{2} (e^{2 \pi i a x} + e^{-2 \pi i a x})) \\ & = \frac{1}{2} (\mathcal{F} e^{2 \pi i a x} + \mathcal{F} e^{-2 \pi i a x}) \\ & = \frac{1}{2} (\delta_a + \delta_{-a})\end{split}\]

Fourier Transform of sine function:

\[\begin{split}\mathcal{F} sin(2 \pi a x) & = \mathcal{F} (\frac{1}{2i} (e^{2 \pi i a x} - e^{-2 \pi i a x})) \\ & = \frac{1}{2i} (\mathcal{F} e^{2 \pi i a x} - \mathcal{F} e^{-2 \pi i a x}) \\ & = \frac{1}{2i} (\delta_a - \delta_{-a})\end{split}\]

Derivitive of Distribution

Let \(T'\) be the derivitive of distribution \(T\), then:

\[\begin{split}\langle T', \phi \rangle & = \int_{-\infty}^{+\infty} T' (x) \phi (x) dx \\ & = T(x) \phi (x) |_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} T(x) \phi' (x) dx \\ & = - \langle T, \phi' \rangle\end{split}\]

Again, by turning the solution of a problem into a definiation, we define:

\[\langle T', \phi \rangle = - \langle T, \phi' \rangle\]

By defining how it operates on test function, the derivitive of a distribution can be derived, even for many weird function.

Applications

Derivative of Heaviside Step Function

\[\begin{split}H (x) = \begin{cases} 1 & x \gt 0 \\ 0 & x \le 0 \end{cases}\end{split}\]

\[\begin{split}\langle H', \phi \rangle & = - \langle H, \phi' \rangle \\ & = - \int_{-\infty}^{+\infty} H(x) \phi' (x) dx \\ & = - \int_{0}^{+\infty} \phi' (x) dx \\ & = \phi (0) \\ & = \langle \delta, \phi \rangle\end{split}\]

\[\therefore H' = \delta\]

Derivative of Signum Function

\[\begin{split}\sgn (x) = \begin{cases} -1 & x \lt 0 \\ 0 & x = 0 \\ 1 & x \gt 0 \end{cases}\end{split}\]

\[\begin{split}\langle \sgn', \phi \rangle & = -\langle \sgn, \phi' \rangle \\ & = -\int_{0}^{+\infty} \phi' dx + \int_{-\infty}^0 \phi' dx \\ & = 2 \phi (0) \\ & = \langle 2 \delta, \phi \rangle\end{split}\]

\[\therefore \sgn' = 2 \delta\]

Fourier Transform of Signum Function

\[\begin{split}(\mathcal{F} \sgn) (s) & = \frac{1}{2 \pi i s} (\mathcal{F} \sgn') (s) \\ & = \frac{1}{2 \pi i s} \left(\mathcal{F} (2 \delta) \right) (s) \\ & = \frac{2}{2 \pi i s} \\ & = \frac{1}{\pi i s}\end{split}\]

Fourier Transform of Heaviside Step Function

\[\begin{split}H (x) & = \begin{cases} 1 & x \gt 0 \\ 0 & x \le 0 \end{cases} \\ & = \frac{1}{2}(1 + \sgn(x))\end{split}\]

\[\begin{split}\therefore \mathcal{F} H (s) & = \mathcal{F} (\frac{1}{2} + \frac{\sgn{x}}{2}) (s) \\ & = \frac{1}{2} \mathcal{F} 1 (s) + \frac{1}{2} \mathcal{F} \sgn (s) \\ & = \frac{1}{2} \delta + \frac{1}{2 \pi i s}\end{split}\]

Warning

The following derivation is WRONG:

\[\begin{split}\mathcal{F} H (s) & = \frac{1}{2 \pi i s} (\mathcal{F} H') (s) \\ & = \frac{1}{2 \pi i s} \left(\mathcal{F} \delta \right) (s) \\ & = \frac{1}{2 \pi i s}\end{split}\]

As the highest score answer [1] suggested:

… There’s a family of functions that differ by additive constants and all have the same derivative. Their Fourier transforms differ by \(\delta\) s at the origin (proportional to the additive constants), so it can’t be the case that you get the Fourier transform of all of them by dividing the transform of the derivative by \(i \omega\)

Multiplication of Distribution

\[\langle Tf, \phi \rangle = \int_{-\infty}^{+\infty} T(x) f(x) \phi(x) dx = \langle T, f\phi \rangle\]

Note

• Arbitrary distributions \(S\) and \(T\) can not multiply as \(ST\) is undefined.

• A distribution multiplied by a function \(f\) is valid if \(f \phi \in \mathcal{S}\)

Convolution of Distribution

Note

• It is often invalid to convolute two arbitrary distributions, as well as to convolute with an arbitrary function.

• With the distribution interpretation, convolution \(f * T\) is defined in terms of pairing not integral. It mimics the classical integral operation of convolution.

\[f \in \mathcal{S}, \psi \in \mathcal{S}' \implies \forall \phi \in \mathcal{S}: \langle \psi * f, \phi \rangle = \langle \psi, \phi * f^- \rangle\]

Proof:

\[\begin{split}\langle \psi * f, \phi \rangle & = \int_{-\infty}^{+\infty} (\psi * f)(x) \phi (x) dx \\ & = \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} f(x - y) \psi (y) dy \right) \phi (x) dx \\ & = \int_{-\infty}^{+\infty} \psi (y) \left( \int_{-\infty}^{+\infty} f(x - y) \phi (x) dx \right) dy \\ & = \int_{-\infty}^{+\infty} \psi (y) \left( \int_{-\infty}^{+\infty} f^- (y - x) \phi (x) dx \right) dy \\ & = \int_{-\infty}^{+\infty} \psi (y) \left( \phi * f^- \right) (y) dy \\ & = \langle \psi, \phi * f^- \rangle\end{split}\]

\[\tag*{$\blacksquare$}\]

Extending to the general case:

\[\langle S * T, \phi \rangle = \langle S, \phi * T^- \rangle\]

Note

This formula holds only if \(\phi * T^- \in \mathcal{S}\).

Convolution Theorem of Fourier Transform:

\[ \begin{align}\begin{aligned}\mathcal{F} (S * T) = \mathcal{F} S \cdot \mathcal{F} T\\\mathcal{F} (S \cdot T) = \mathcal{F} S * \mathcal{F} T\end{aligned}\end{align} \]

Note

Again, these formulas hold only if \(\forall \phi \in \mathcal{S}, \phi * T^- \in \mathcal{S}\).

Proof can be found at Convolution of operations section.

Reference

[1]

https://math.stackexchange.com/questions/73922/fourier-transform-of-unit-step

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