Q:

Functional derivatives of Fourier transform

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?

A:

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,
$$\frac{\delta}{\delta f(x)} \mathfrak F[f](\xi) = e^{ix\xi}$$

