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Does conjugation of a Dirac delta function makes sense: $\delta^*(x-y)$? Namely is $\delta^*_y = \delta_y$? I am asking specifically in the context of this identity (which is the same as Plancherel's identity, however $\delta$ is not in $L_1$):

$$\int_{\mathbb{R}^d}f(x)\delta^*(x-y)\,dx = \langle f, \delta_y \rangle = f(y) = \mathcal{F}^{-1}[\hat{f}](y) = \langle\hat{f},\hat{\delta}_y \rangle = \int_{\mathbb{R}^d}\hat{f}(u)\exp(2\pi i \langle u,y\rangle)\,du$$

Note that:

$$\mathcal{F}[\delta_y](u) = \int_{\mathbb{R}}\delta(x-y)\exp(-2\pi i \langle u, x \rangle)\,dx = \exp(-2\pi i \langle u, y \rangle)$$

And then:

$$\hat{\delta}^*_y = \exp(2\pi i \langle u, y \rangle)$$

Is this valid?

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    $\begingroup$ The classical (at least in the context i know) involution of measures in $M(\mathbb{R})$ is $\mu^*(E) = \overline{\mu(-E)}$. Hence, as $\delta_y$ is a Radon measure on $\mathbb{R}$, you have $\delta_y^* = \delta_{-y}$. $\endgroup$ – KeeperOfSecrets Sep 28 at 12:36
  • $\begingroup$ @KeeperOfSecrets I think that would be correct, since it matches some of my other results, where $\delta_y^* = \delta_y$ contradicts to those. Thank you. $\endgroup$ – lightxbulb Sep 28 at 12:38
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    $\begingroup$ You are most welcome :) also note that this definition corresponds to the definition of involution on $L_1(\mathbb{R})$ (which is $f^*(x) = \overline {f(-x)}$) if you consider $f$ as an element of $M(\mathbb{R})$ via identification $f \mapsto f(x) dx$. $\endgroup$ – KeeperOfSecrets Sep 28 at 12:49
  • $\begingroup$ @KeeperOfSecrets This seems to imply that $\langle f, \delta_y^* \rangle = \langle \hat{f}, \hat{\delta}_y \rangle$ then. So the identity is not that similar to Plancherel's identity. $\endgroup$ – lightxbulb Sep 28 at 12:59

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