This image is in the book "Basic partial differential equations" by Treves.
Hi. I have a doubt with respect to the Fourier transform of an differential equations. In the image, The Fourier transform of $\partial_t E$ is $\partial_t \bar{E}$. The verification of this is the following (my attempt):
If $\mathcal{F}_{\xi}$ indicates the Fourier transform with respect to the spatial variable, then: \begin{align} ([\mathcal{F}_{\xi}(\partial_t (E(\xi,t))](x,t),\varphi(x,t))&=(\partial_t E(x,t),[\mathcal{F}_{\xi}\varphi(\xi,t)](x,t))\\ &=(E(x,t),\partial_t [\mathcal{F}_{\xi}\varphi(\xi,t)](x,t))\\ &=(E(x,t), [\mathcal{F}_{\xi}( \partial_t\varphi(\xi,t))](x,t)\\ &=([\mathcal{F}_{\xi}E(\xi,t)](x,t)\partial_t \varphi(x,t))\\ &=(\partial_t [\mathcal{F}_{\xi}E(\xi,t)](x,t),\varphi(x,t)) \end{align} So, $$([\mathcal{F}_{\xi}(\partial_t (E(\xi,t))](x,t),\varphi(x,t))=(\partial_t [\mathcal{F}_{\xi}E(\xi,t)](x,t),\varphi(x,t))$$ for all $\varphi(x,t)$. This implies that $$[\mathcal{F}_{\xi}(\partial_t (E(\xi,t))](x,t)=\mathcal{F}_{\xi}E(\xi,t)](x,t)$$ (i.e. $\widehat{\partial_t E}=\partial_t \widehat{E})$
Question 1. Is the above correct? (Maybe I'm missing a factor $(-i)$ but I mean the idea.
Question 2. Why Fourier of $\delta$ is $\delta(t)$? I don't understanding this. I tried the following:
\begin{align} (\mathcal{F}_{\xi}\delta,\phi)&=(\delta,[\mathcal{F}_{\xi}(\phi(\xi,t)](x,t))\\ &=(\mathcal{F}_{\xi}(\phi(\xi,t))\Bigg|_{(x,t)=(0,0)}?\\ &=\left(\int \mathrm{e}^{-ix\cdot \xi}\phi(\xi,t)d\xi\right)\Bigg|_{(x,t)=(0,0)\text{ or } (x,t)=(0,t)}? \end{align}