Convolution with functions and two Dirac deltas I have to calculate the following convolution $\mathcal{J}$:
$$
\mathcal{J} = f(\omega)\delta(\omega-\xi)\ast g(\omega)\delta(\omega-\zeta),
$$
where $f$ and $g$ are functions well behaved to do this operation and $\delta$ is the Dirac delta. According to the definition of convolution, we have
$$
\mathcal{J} = \int_{-\infty}^{\infty}f(\omega-\eta)\delta(\omega-\xi-\eta)g(\eta)\delta(\eta-\zeta)d\eta.
$$
However, the result of $\mathcal{J}$ depends on which $\delta$ I consider to do the integration. I look for answers on this site, and found out this question where it is used (what I think ) the following property:
$$
f(x)\delta(x-a)\ast g(x)\delta(x-b) = f(a)g(b)\delta(x-a-b).
$$
Is this a definition for such type of convolution? If not, how can I calculate $\mathcal{J}?$
 A: Forgive me for using notation I'm more used to. Instead of $\omega$, I'll use $x$ as the free variable and instead of $\delta(x-x_0)$, I'll just write $\delta_{x_0}$. The convolution you're asking for is then:
$$
  f \, \delta_\xi \; * \; g \, \delta_\zeta
$$
Solution using Fourier transform
Initially, I would approach this problem using the properties of Fourier transform. Specifically these two properties will be useful:
$$
  \def\R{\mathbb{R}}
  \def\F{\mathcal{F}}
  \def\e{\mathrm{e}}
  \def\i{\mathrm{i}}
  \def\d{\mathrm{d}}
  \F(fg) = \F(f) * \F(g) \\
  \F(f * g) = \F(f) \, \F(g)
$$
As it turns out, there's a much simpler solution (presented down below), but I'll keep it here just for completeness. Applying the rules to the distributions in question, we get:
$$
\begin{aligned}
  &\F( \; f \, \delta_\xi \; * \; g \, \delta_\zeta \; ) \\
  &= \F( f \, \delta_\xi) \; \F(g \, \delta_\zeta)
  = \Big( \F(f) * \F(\delta_\xi) \Big) \; \Big( \F(g) * \F(\delta_\zeta) \Big)
  = ( \tilde f * \e^{-2\pi\i\xi x} )( \tilde g * \e^{-2\pi\i\zeta x} ) \\
  &= \Bigg( \int \tilde f(y) \; \e^{-2\pi\i\xi \, (x-y)} \d{y} \Bigg)
  \Bigg( \int \tilde g(y) \; \e^{-2\pi\i\zeta \, (x-y)} \d{y} \Bigg) \\
  &= \e^{-2\pi\i\xi x} \;\Bigg( \int \tilde f(y) \; \e^{2\pi\i\xi y} \d{y} \Bigg) \;
  \e^{-2\pi\i\zeta x} \;\Bigg( \int \tilde g(y) \; \e^{2\pi\i\zeta y} \d{y} \Bigg) \\
  &= \e^{-2\pi\i\xi x} \; \F^{-1}(\tilde f)(\xi) \; \e^{-2\pi\i\zeta x} \; \F^{-1}(\tilde g)(\zeta) \\[5pt]
  &= \e^{-2\pi\i \, (\xi + \zeta) \, x} \; f(\xi) \; g(\zeta)
\end{aligned}
$$
Finally we will do the inverse Fourier transform:
$$
\begin{aligned}
  f \, \delta_\xi \; * \; g \, \delta_\zeta
  &\;=\;
  \F^{-1}(\; \e^{-2\pi\i \, (\xi + \zeta) \, x} \; f(\xi) \; g(\zeta) \;)
  \\[5pt]
  &\;=\;
  f(\xi) \; g(\zeta) \; \F^{-1}(\; \e^{-2\pi\i \, (\xi + \zeta) \, x} \;)
  \\[5pt]
  &\;=\;
  f(\xi) \; g(\zeta) \; \delta_{\xi + \zeta}
\end{aligned}
$$
Keep in mind that all the refinements I did were only formal, I didn't check whether all the expressions were well-defined. You should check that yourself.
Simpler solution using properties of delta
Alternatively, we can observe that $f \, \delta_\xi = f(\xi) \, \delta_\xi$ holds for any continuous function $f$ and that $\left\langle T * \delta_\xi, \; \varphi(x) \right\rangle = \left\langle T, \; \varphi(x - \xi) \right\rangle$ holds for any distribution $T$. Specifically, it is true that $\delta_\xi * \delta_\zeta = \delta_{\xi + \zeta}$. Then the solution can be found by application of these rules.
$$
  f \, \delta_\xi \; * \; g \, \delta_\zeta
  \;=\; f(\xi) \, \delta_\xi \; * \; g(\zeta) \, \delta_\zeta
  \;=\; f(\xi) \; g(\zeta) \; \delta_{\xi + \zeta}
$$
