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##\mathcal{F}\left\{f \ast g\right\} = \mathcal{F}\left\{f\right\} \cdot \mathcal{F}\left\{g\right\}##.

Considering this equation:

##g\left(x, y\right) = h\left(x, y\right) \ast f\left(x, y\right)##

Are these steps valid if I were to compute for ##f\left(x, y\right)##?

##\mathcal{F}\left\{g\left(x, y\right)\right\} = \mathcal{F}\left\{h\left(x, y\right) \ast f\left(x, y\right)\right\} \\

\mathcal{F}\left\{g\left(x, y\right)\right\} = \mathcal{F}\left\{h\left(x, y\right)\right\} \cdot \mathcal{F}\left\{f\left(x, y\right)\right\} \\

\frac{\mathcal{F}\left\{g\left(x, y\right)\right\}}{\mathcal{F}\left\{h\left(x, y\right)\right\}} = \mathcal{F}\left\{f\left(x, y\right)\right\} \\

\mathcal{F}^{-1}\left\{\frac{\mathcal{F}\left\{g\left(x, y\right)\right\}}{\mathcal{F}\left\{h\left(x, y\right)\right\}}\right\} = f\left(x, y\right)##

Thank you in advance.