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Introduction to Linear Algebra with Mathematica

Glossary

Preface

This section is devoted to the Fourier transform of functions that vanish for x < 0 (often called causal functions). These results are classically tied to the Paley–Wiener theorem, analytic continuation of the Fourier transform into a half‑plane, and the connection with the Laplace transform.

Casual Functions

Let
\[ f \in 𝔏^1 (\mathbb{R}), \qquad f(t) = 0 \quad\mbox{for } t < 0 . \]
Define its Fourier transform (on the real line) by
\[ ℱ[f] = \hat{f}(\xi ) = \int_{\mathbb{R}} f(t)\,e^{\mathbf{j}t\xi}\,{\text d}t = \int_0^{+\infty}f(t)\,e^{\mathbf{j}t\xi}\,{\text d}t , \quad \xi\in\mathbb{R}. \]
Here j or ⅉ is the imaginary unit on the complex plane ℂ, so ⅉ² = −.

This integral actually makes sense for complex arguments in a half‑plane, and that extends to a holomorphic function there. This is the basic Paley–Wiener phenomenon for functions supported in [0, ∞).

We extend the Fourier integral to complex arguments by taking a complex number

\[ z = \xi + \mathbf{j}\,\eta , \qquad \xi, \eta \in \mathbb{R} . \]
Consider the integral
\begin{equation} \label{EqPW.1} F(z) = \int_0^{+\infty}f(t)\,e^{\mathbf{j}t\,z}\,{\text d}t . \end{equation}
Write the exponential explicitly:
\[ e^{\mathbf{j}t\,z} = e^{\mathbf{j}t\left( \xi + \mathbf{j}\eta \right)} = e^{\mathbf{j}t\xi}\,e^{-t\,\eta}. \]
So
\[ F(z) = \int_0^{\infty} f(t)\,e^{\mathbf{j}t\xi}\,e^{-t\,\eta}\,{\text d}t . \]
We want to know for which η this integral converges absolutely.

Compute absolute value

\[ \left\vert f(t)\,e^{\mathbf{j}t\,z} \right\vert = \left\vert f(t) \right\vert \cdot \left\vert e^{\mathbf{j} t\xi} \right\vert \cdot \left\vert e^{-\eta t} \right\vert = \left\vert f(t) \right\vert e^{-\eta t} . \]
since \( \displaystyle \quad \left\vert e^{\mathbf{j} t\xi} \right\vert = 1 . \ \) Thus,
\[ \int_0^{\infty} \left\vert f(t)\,e^{\mathbf{j}t\,z} \right\vert {\text d}t = \int_0^{\infty} \left\vert f(t) \right\vert e^{-\eta t} {\text d}t . \]
If η ≥ 0, then \( \displaystyle \quad e^{-\eta t} \le 1 \quad \) for all t ≥ 0, so
\[ \int_0^{\infty} \left\vert f(t)\,e^{\mathbf{j}t\,z} \right\vert {\text d}t \le \int_0^{\infty} \left\vert f(t) \right\vert {\text d}t = \| f \|_1 < \infty . \]
Hence the integral defining F(z) converges absolutely for all z with &Imz = η ≥ 0.

If η < 0, we cannot guarantee convergence from f ∈ 𝔏¹ alone, because eηt grows. So we have:

For every z with ℑz ≥ 0, the integral

\[ F(z) = \int_0^{\infty} f(t)\,e^{\mathbf{j}t\,z} \,{\text d}t < \infty \]
converges absolutely. In particular, for real z (i.e., η = 0), we recover the usual Fourier transform:
\[ F(\xi ) = \hat{f}(\xi ) = \int_0^{\infty} f(t)\,e^{\mathbf{j}t\,\xi} \,{\text d}t , \qquad \xi \in \mathbb{R}. \]

Now fix any compact set K ⊂ { z ∈ ℂ : ℑz > 0 }. There exists η₀ > 0 such that ℑz ≥ η₀ > 0 for all z ∈ K. Then for z ∈ K and t ≥ 0,

\[ \left\vert f(t) \,e^{\mathbf{j}tz} \right\vert = \left\vert f(t) \right\vert \cdot e^{-\Im z \cdot t} \le \left\vert f(t) \right\vert e^{-\eta_0} . \]
Since η₀ > 0, the function t ↦ e−η₀t is integrable on [0, ∞) when multiplied by f ∈ 𝔏¹. More precisely,
\[ \int_0^{\infty} \left\vert f(t) \,e^{- \eta_0 t} \right\vert \,{\text d}t \le \int_0^{\infty} \left\vert f(t) \right\vert {\text d} t = \| f \|_1 < \infty . \]
Thus, for all z ∈ K, the integrand \( \displaystyle \quad fIt)\,e^{\mathbf{j}tz} \quad \) is dominated by an integrable function independent of z. Moreover, for each fixed t, the map
\[ z \,\mapsto \,f(t)\,e^{\mathbf{j}zt} \]
is entire (holomorphic on all of ℂ.

By the dominated convergence theorem in its holomorphic version (Morera’s theorem or differentiation under the integral sign), we conclude:

The function

\[ F(z) = \int_0^{\infty} f(t)\,e^{\mathbf{j}zt} \,{\text d}t \]
is holomorphic on the open half‑plane { z ∈ ℂ : ℑz > 0 }.

So the Fourier transform ℱ[f](ξ) is the boundary value (on the real axis) of a holomorphic function defined on the upper half‑plane.

If we write z = ⅉλ, then ℑz > 0 corresponds to ℜλ > 0. Indeed,

\[ z = \mathbf{j}\lambda \quad \Longrightarrow \quad \Im z = \Re \lambda . \]
So ℑz > 0 is equivalent to ℜλ > 0.
\[ z = \mathbf{j}\lambda \quad \Longrightarrow \quad \Im z = \Re \lambda . \]
Define the Laplace transform
\[ ℒ[f]_{t\to\lambda} \, = f^L (\lambda ) = \int_0^{\infty} f(t)\, e^{-\lambda t}\,{\text d}t , \quad \Re\lambda > 0. \]
Then
\[ F(z) = \int_0^{\infty} f(t)\,e^{\mathbf{j}zt} {\text d}t = \int_0^{\infty} f(t)\,e^{\mathbf{j}\left( \mathbf{j}\lambda \right) t} {\text d}t = ℒ[f]_{t\to\lambda} , \qquad z= \mathbf{j}\lambda . \]
On the real axis (z = ξ, i.e., ξ = ⅉλ), we have
\[ \hat{f}(\xi ) = F(\xi ) = ℒ[f]_{t\to\lambda} , \qquad \lambda = \mathbf{j}\xi . \]

So for causal f (vanishing for t < 0), the Fourier transform is just the Laplace transform evaluated on the imaginary axis, and the analyticity in a half‑plane is exactly the Laplace‑transform analyticity. We summarize results in the following statement.

Theorem 1 (Paley–Wiener): Let f ∈ 𝔏¹(ℝ) with f(t) = 0 for t < 0. Define \[ \hat{f}(\xi ) = \int_0^{\infty} f(t)\,e^{\mathbf{j}t\zi} \,{\text d}t , \qquad \xi \in \mathbb{R} . \] Then:
  1. The function \[ F(z) = \int_0^{\infty} f(t)\, e^{\mathbf{j}z t}\,{\text d}t \] is well defined and holomorphic on the half‑plane ℑz > 0.
  2. For real ξ, F(ξ) = ℱ[f](ξ).
  3. Equivalently, ℱ[f](ξ) = ℒ[f](ⅉξ), where ℒ[f] is the Laplace transform of f.

 

 

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