I have a misconception when applying the Fourier Transform to a compacted-supported function and the characteristics of the function obtained.
Intro
I am going to list what I believe is true so you can identify were I am making my conceptual mistake:
- The Fourier Transform of a square-integrable function which is compact-supported in the real line must be an entire analytic function due the Paley–Wiener theorem.
- An entire function is a complex-valued function that is holomorphic on the whole complex plane, so it is complex differentiable everywhere, so it is satisfying being Analytic, infinitely differentiable or Smooth, and its real-imaginary decomposition constituents fulfill the Cauchy–Riemann equations.
- If a function $g(z)$ is complex differentiable with the complex variable being described as $z=\sigma+iw$ so the function could be expressed as: $g(\sigma+iw)=u(\sigma,w)+iv(\sigma,w)$ with $u,\,v$ real-valued functions, then each constituents fulfill the Cauchy–Riemann equations as mentioned, which imply that each constituent function is individually an Harmonic function so both functions $u,\,v$ fulfill $\nabla^2 u = \frac{\partial^2 u}{\partial \sigma^2}+\frac{\partial^2 u}{\partial w^2}=0$ and $\nabla^2 v = \frac{\partial^2 v}{\partial \sigma^2}+\frac{\partial^2 v}{\partial w^2}=0$.
- The Fourier Transform $\hat{f}(iw)$ of an "even function" $f(t)=f(-t)$ is a real-valued function $\hat{f}(iw)\in \mathbb{R}$.
- The Fourier Transform $\hat{f}(iw)$ of an real valued function $f(t)$ fulfill that $\hat{f}(iw)^* = \hat{f}(-iw)$ is an Hermitian function.
I am using the electrician notation for the Fourier transform using the angular frequency including the imaginary unit as part of the variable.
Main text
With this, I am going to use the following example to present the problem:
$$f(t) = \left(\frac{1-t^2+|1-t^2|}{2}\right)^4 \tag{Eq. 1}\label{Eq. 1}$$
which is ploted here fulfilling is a real-valued function supported on the real line, but different from zero only in $t \in [-1,\,1]$ so it has compact-support. Also it can be seen is an "even function" due $(-t)^2 \equiv t^2$ for real-valued $t$, and it is also square-integrable since $\int\limits_{-\infty}^{\infty}\left|\left(\frac{1-t^2+|1-t^2|}{2}\right)^4\right|^2 dt = \frac{65536}{109395}\approx 0.6 \ll \infty$. So the function should be fulfilling all the $5$ points of the introduction.
As it can be seen here, the Fourier Transform of $f(t)$ of \eqref{Eq. 1} could been described using the Bessel function of First kind $J_{\nu}(w)$ and the Gamma function $\Gamma(w)$ as:
$$ \begin{array}{r c l}
\hat{f}(iw) = \int\limits_{-\infty}^{\infty} f(t)\ e^{-iwt}dt & = & \text{sgn}(w)\sqrt{\pi}\left(\frac{2}{w}\right)^{4+\frac{1}{2}}\Gamma(4+1)J_{4+\frac{1}{2}}(w)\\
& = & 24\sqrt{\pi}\,\text{sgn}(w)\left(\frac{2}{w}\right)^{\frac{9}{2}}J_{\frac{9}{2}}(w) \tag{Eq. 2}\label{Eq. 2}
\end{array}$$
Since this Fourier Transform $\hat{f}(iw)$ should be holomorphic, but is a real-valued function, I am confused how it is verified is being complex differentiable:
How is analyzed a real-valued function for complex differentiability?
Since the Fourier Transform $\hat{f}(iw) = \hat{f}(\{\sigma\equiv 0\}+iw)$ it will imply that $u(\sigma,w) = 24\sqrt{\pi}\,\text{sgn}(w)\left(\frac{2}{w}\right)^{\frac{9}{2}}J_{\frac{9}{2}}(w)$ and $v(\sigma,w)=0$: naturally the trivial solution zero function will fulfill $\nabla^2 v = 0$, but for reviewing $u(\sigma,w)$ I don´t know how to take the derivative respect to $\sigma$ since it is zero, neither I have a function related to $\sigma$ to make some manipulation. Even so, if I take $\frac{\partial^2}{\partial w^2}u(\sigma,w) \neq 0$ is definitely not zero as it can be seen here, so if the Fourier Transform is complex differentiable as is stated on the initial points, surely must be $\sigma$-related component that is missing.
Surely I am having a conceptual mistake, but I cannot figure it out so far. Hope you could explain what I am doing wrong, displaying which functions are going to be $u(\sigma,w)$ and $v(\sigma,w)$ for the Fourier Transform of \eqref{Eq. 2} showing in detail how is proved is complex differentiable. Beforehand, thanks you very much.
Best Answer
Computing directly, we can find the Fourier transform of your function to be
$$\int_{-1}^1e^{-i\omega t}\:dt = 2\operatorname{sinc}(\omega)$$
$$ \implies \int_{-1}^1(1-4t^2+6t^4-4t^6+t^8)e^{-i\omega t}\:dt = 2\left(1+\left(\frac{d}{d\omega}\right)^2\right)^4\operatorname{sinc}(\omega)$$
Since this Fourier transform is a linear combination of derivatives of an analytic function, it too must be analytic. Notably this formula does not contain any Bessel functions so there must be an error in your calculation or source.
The conditions of the Paley-Wiener theorem treat $\omega\in\Bbb{C}$, not $\omega\in\Bbb{R}$ so this does indeed describe an entire function in $\omega = \mu + i\nu$, not $s=\sigma+i\omega$.