Functional derivatives of Fourier transform

Question

I've tried to calculate this as well as been looking around online and I can't seem to find the calculation of this. If $\mathfrak{F}$ is the Fourier transform: $$ f\mapsto \int_{-\infty}^{\infty} f(x)e^{ix}dx, $$ then what would the functional derivative $$ \frac{\delta \mathfrak{F}}{\delta f} $$ be?

Answer 1

The functional derivative is defined as the linear mapping $$\left\langle \frac{\delta}{\delta f(x)} F[f](\xi), \phi(x) \right\rangle := \left. \frac{d}{d\lambda} F[f+\lambda\phi](\xi) \right|_{\lambda=0}$$

For $\mathfrak F[f](\xi) := \int_{-\infty}^{\infty} f(x)e^{ix\xi} \, dx$ we get $$\begin{align} \left. \frac{d}{d\lambda} \int_{-\infty}^{\infty} (f+\lambda\phi)(x)e^{ix\xi} \, dx \right|_{\lambda=0} & = \left. \frac{d}{d\lambda} \int_{-\infty}^{\infty} \left(f(x)+\lambda\phi(x)\right)e^{ix\xi} \, dx \right|_{\lambda=0} \\ & = \left. \frac{d}{d\lambda} \left( \int_{-\infty}^{\infty} f(x) e^{ix\xi} \, dx + \lambda \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \right) \right|_{\lambda=0} \\ & = \left. \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \right|_{\lambda=0} \\ & = \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \\ & = \left\langle e^{ix\xi}, \phi(x) \right\rangle \end{align}$$ Thus, $$\frac{\delta}{\delta f(x)} \mathfrak F[f](\xi) = e^{ix\xi}$$

Answer 2

Are you looking for derivative of Fourier transform ?

considering $\omega$ insted of $f$ we can write accroding to the definition

$$\begin{align}\frac{d \mathcal{F}(\omega)}{d\omega}&=\int_{-\infty}^{+\infty}f(t)\frac{d}{d\omega}\cdot e^{i\omega t}\,dt\\&=-i\int_{-\infty}^{+\infty}t.f(t)e^{i\omega t}\,dt\\&=-i \cdot\mathcal{IFT}\{t \,.f(t)\}\end{align}$$

And the reverse process can be

$$\mathcal{FT}\left\{\frac{d f^n(t)}{d^nt}\right\}=(i\omega)^n.F(\omega)$$

Please let me know if somehow i can improve my answer , Thanks !