1. **Problem Statement:** We want to establish the connection between the convolution of the Fourier transforms $F(f) * F(g)$ and the Fourier transform of the product $F(fg)$, given that $f$ and $g$ are integrable functions. 2. **Recall the Fourier Transform and Convolution Definitions:** - The Fourier transform of a function $h(t)$ is defined as: $$F(h)(\\omega) = \int_{-\infty}^{\infty} h(t) e^{-i \\omega t} dt$$ - The convolution of two functions $a$ and $b$ is defined as: $$ (a * b)(t) = \int_{-\infty}^{\infty} a(\tau) b(t - \tau) d\tau $$ 3. **Key Property (Convolution Theorem):** The Fourier transform of the convolution of two functions is the product of their Fourier transforms: $$F(f * g) = F(f) \, F(g)$$ 4. **Establishing the Connection:** We want to relate $F(f) * F(g)$ and $F(fg)$. Using the inverse Fourier transform $F^{-1}$, we know: $$F^{-1}(F(f) * F(g)) = f g$$ This means the convolution of the Fourier transforms corresponds to the Fourier transform of the product of the original functions. 5. **Summary:** - The convolution of the Fourier transforms $F(f) * F(g)$ is the Fourier transform of the product $fg$: $$F(f) * F(g) = F(fg)$$ - This is a fundamental duality: convolution in the frequency domain corresponds to multiplication in the time domain, and vice versa. This completes the connection between the convolution of Fourier transforms and the Fourier transform of the product of functions.