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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?

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  • $\begingroup$ The Fourier transform is a linear operator so the derivative is? $\endgroup$ – Bananach Jul 3 '17 at 6:04
  • $\begingroup$ It is itself, if I'm correct? $\endgroup$ – AIM_BLB Jul 3 '17 at 6:07
  • $\begingroup$ Yoh are correct $\endgroup$ – Bananach Jul 3 '17 at 6:35
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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, $$\frac{\delta}{\delta f(x)} \mathfrak F[f](\xi) = e^{ix\xi}$$

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  • $\begingroup$ This answer uses a wrong definition of the Fourier transform. It is an operator $L^2(\mathbb{R})\to L^2(\mathbb{R})$, not a functional $L^2(\mathbb{R})\to\mathbb{R}$ $\endgroup$ – Bananach Jul 4 '17 at 13:12
  • $\begingroup$ @Bananach. Look at the question! $\endgroup$ – md2perpe Jul 4 '17 at 14:04
  • $\begingroup$ I've edited my answer to make it a Fourier transform and not just the value at $\xi=1$. I haven't taken $\mathfrak F$ to act on distributions though. $\endgroup$ – md2perpe Jul 4 '17 at 14:33
  • $\begingroup$ It is still incorrect as you cannot consider pointwise derivatives. But this is just a matter of rewriting and presenting your calculations in a more formal manner $\endgroup$ – Bananach Jul 4 '17 at 17:24
  • $\begingroup$ I know that it is formally incorrect, but I use this notation in a similar way as one sometimes writes $u(x)$ for a distribution; it's not meant as the value at $x$ but rather to denote what variable belongs to $u$. An example of such notation: $\langle u * v, \phi \rangle = \langle u(x) \otimes v(y), \phi(x+y) \rangle$. $\endgroup$ – md2perpe Jul 4 '17 at 17:44
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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 !

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  • $\begingroup$ Quick question, what are $\mathfrak{I},msthfrak{T}$ symbolise? $\endgroup$ – AIM_BLB Jul 3 '17 at 6:06
  • $\begingroup$ @CSA sorry i'm not very much adept with latex formatting i'm just learning those , the first one seems to be for imaginary part , i don't know about others $\endgroup$ – Siddhartha Jul 3 '17 at 6:17
  • $\begingroup$ He was asking for the functional derivative of the Fourier transform, not the derivative of the Fourier transform. en.wikipedia.org/wiki/Functional_derivative $\endgroup$ – md2perpe Jul 3 '17 at 11:19
  • $\begingroup$ sorry for wrong answer , i'm new to integral transforms and couldn't interpret his query correctly , sorry for that @md2perpe $\endgroup$ – Siddhartha Jul 3 '17 at 12:55
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    $\begingroup$ @Lelouch.D.Light. No problem. It's easy to misunderstand a question. $\endgroup$ – md2perpe Jul 3 '17 at 15:42

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