studying for my PDE course I had to face the following object and its properties (I clarify that it was an evaluation that I already submitted), I wanted to know what you think, if it is correct or if you have a better idea
Definition:
Let $f:\mathbb{R}\to\mathbb{C}$ a function in $S(\mathbb{R})$. We define the Gabor transform as \begin{equation*} G[f](a,b) = \frac{1}{\pi^{1/4}}\int_{-\infty}^{+\infty} f(t)e^{-iat}e^{-(t-b)^{2}/2}dt\quad a,b\in\mathbb{R}. \end{equation*} Furthermore, for $h:\mathbb{R}\times\mathbb{R}\to\mathbb{C}$ we define the inverse Gabor transformation by \begin{equation*} G^{-1}[h](t) = \frac{1}{2\pi^{5/4}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} > h(a,b)e^{iat}e^{-(t-b)^{2}/2}dadb. \end{equation*}
Next I will test some properties of this transform. Next I will test some properties of this transform, I would like to know what you think, if they are correct.
Linear property: Let $f,g:\mathbb{R}\to\mathbb{C}$ and $\lambda,\mu\in\mathbb{R}$, then \begin{eqnarray*} G[\lambda f + \mu g](a,b) &=& \frac{1}{\pi^{1/4}}\int_{-\infty}^{+\infty} (\lambda f + \mu g)(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \frac{1}{\pi^{1/4}}\int_{-\infty}^{+\infty} \lambda f(t)e^{-iat}e^{-(t-b)^{2}/2} + \mu g(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \frac{\lambda}{\pi^{1/4}}\int_{-\infty}^{+\infty} f (t)e^{-iat}e^{-(t-b)^{2}/2}dt + \frac{\mu}{\pi^{1/4}}\int_{-\infty}^{+\infty} g(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \lambda G[f](a,b) + \mu G[g](a,b). \end{eqnarray*}
Shifting property: Let $\beta\in\mathbb{R}$, and $f,g:\mathbb{R}\to\mathbb{C}$ such that $g(t) = f(t-\beta)$, then \begin{eqnarray*} G[g](a,b) &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} g(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(t-\beta)e^{-iat}e^{-(t-b)^{2}/2}dt\\ (\text{With }t-\beta = s) &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(s)e^{-ia(s+\beta)}e^{-(s+\beta-b)^{2}/2}ds\\ &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(s)e^{-ias}e^{-ia\beta}e^{-(s-(b-\beta))^{2}/2}ds\\ &=& \frac{e^{-ia\beta}}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(s)e^{-ias}e^{-(s-(b-\beta))^{2}/2}ds\\ &=& e^{-ia\beta}G[f](a,b-\beta), \end{eqnarray*} in short we have to \begin{equation} e^{-ia\beta}G[f](a,b-\beta) = G[g](a,b)\text{ when }g(t) = f(t-\beta). \end{equation}
Modulation property: Let $\alpha \in \mathbb{R}$, let us consider $f,g:\mathbb{R}\to\mathbb{C}$ such that $g(t) = f(t) e^{i \alpha t} $, then we calculate the transform of Gabor and we have to: \begin{eqnarray*} G[g](a,b) &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} g(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(t)e^{i\alpha t}e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(t)e^{-i(a-\alpha)t}e^{-(t-b)^{2}/2}dt\\ &=& G[f](a-\alpha,b). \end{eqnarray*} Therefore, we have seen that \begin{equation} G[f](a-\alpha,b) = G[g](a,b) \text{ when } g(t) = f(t)e^{i\alpha t}. \end{equation}
Power integration property: Note that we can write \begin{eqnarray*} G[f](a,b) &=& \frac{1}{\pi^{1/4}} \int_{-\infty}^{+\infty} f(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \int_{-\infty}^{+\infty} \frac{1}{\pi^{1/4}}f(t)e^{-iat}e^{-(t-b)^{2}/2}dt\\ &=& \text{FT}\left(\frac{1}{\pi^{1/4}}f(t)e^{-(t-b)^{2}/2}\right)(a), \end{eqnarray*} then we say that \begin{equation} \label{Ecuacion01} G[f](a,b) = \widehat{F_{b}[f]}(a),\text{ with }F_{b}[f](t) = \frac{1}{\pi^{1/4}}f(t)e^{-(t-b)^{2}/2}. \end{equation} Finally using the Plancherel theorem on $F_{b}$ it follows that \begin{eqnarray*} \frac{1}{\pi^{1/2}}\int_{-\infty}^{+\infty} |f(t)|^{2} e^{-(t-b)^{2}}dt &=& \int_{-\infty}^{+\infty} \left|\frac{1}{\pi^{1/4}}f(t)e^{-(t-b)^{2}/2}\right|^{2}dt\\ &=& \int_{-\infty}^{+\infty} |F_{b}[f](t)|^{2}dt\\ &=& \int_{-\infty}^{+\infty} |\widehat{F_{b}[f]}(a)|^{2}da \\ &=& \int_{-\infty}^{+\infty} |G[f](a,b)|^{2}da, \end{eqnarray*} therefore we conclude that \begin{equation*} \frac{1}{\pi^{1/2}}\int_{-\infty}^{+\infty} |f(t)|^{2}e^{-(t-b)^{2}}dt = \int_{-\infty}^{+\infty} |G[f](a,b)|^{2}da. \end{equation*}
Energy sum property: Let $f,g:\mathbb{R}\to\mathbb{C}$ be such that $f,g\in S(\mathbb{R})$, then note that \begin{eqnarray*} \int_{-\infty}^{+\infty} f(t)\overline{g(t)}dt &=& \int_{-\infty}^{+\infty} \left(f(t)e^{-(t-b)^{2}/2}\right)\overline{\left(g(t)e^{-(t-b)^{2}/2}\right)}dt\\ &=& \sqrt{\pi}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \text{FT}\left(f(t)e^{-(t-b)^{2}/2}\right)\overline{\text{FT}\left(g(t)e^{-(t-b)^{2}/2}\right)}dadb\\ &=& \sqrt{\pi}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} G[f](a,b)\overline{G[g](a,b)}dadb \end{eqnarray*}
Recovery property: Let $f\in S(\mathbb{R})$, note that: \begin{eqnarray*} \int_{-\infty}^{+\infty} G[f](a,b)e^{iat}da &=& \int_{-\infty}^{+\infty} \widehat{F_{b}[f]}(a)e^{iat}da \\ &=& 2\pi\left(\frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{F_{b}[f]}(a)e^{iat}da\right) \\ &=& 2\pi F_{b}[f](t)\\ &=& 2\pi\left(\frac{1}{\pi^{1/4}}f(t)e^{-(t-b)^{2}/2}\right)\\ &=& 2\pi^{3/4}e^{-\frac{(t-b)^{2}}{2}}f(t). \end{eqnarray*}
The source of these properties is the following Wikipedia link. The Energy Decay Property I did not manage to test, or at least not in a way that convinces me, if anyone has any ideas it is welcome