The question asks to identify the true relations involving the Fourier transform ($\tilde{f}$), convolution ($f * g$), and the product of functions ($fg$).

Analyzing Convolution Properties

We examine each option based on the provided definitions:

• Option A: Check if the convolution operation is commutative ($f * g = g * f$).
  The convolution $(f * g)(x)$ is defined as $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(y)g(x - y)dy$.
  The convolution $(g * f)(x)$ is $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} g(y)f(x - y)dy$.
  By substituting $u = x - y$ in the expression for $(g * f)(x)$, we find that $(g * f)(x) = (f * g)(x)$. Thus, convolution is commutative. Option A is true.
• Option B: Check if $\tilde{f * g} = \tilde{g * f}$.
  Since $f * g = g * f$ (from Option A), their Fourier transforms must be identical. Therefore, $\tilde{f * g} = \tilde{g * f}$. Option B is true.
• Option C: Check if $\tilde{f * g} = \tilde{fg}$.
  This statement implies that the Fourier transform of a convolution is equal to the Fourier transform of the product. This is generally not true. The Convolution Theorem states otherwise. Option C is false.
• Option D: Check if $\tilde{f * g} = \tilde{f}\tilde{g}$.
  This is the standard Convolution Theorem for the given definition of the Fourier Transform.

  $\tilde{f * g}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} (f * g)(x) e^{i\omega x} dx$

  $= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(y)g(x - y)dy \right) e^{i\omega x} dx$

  Changing variables ($u = x - y$) and rearranging, we can show this equals $\tilde{f}(\omega)\tilde{g}(\omega)$. Option D is true.

Conclusion on True Relations

Based on the analysis, the true relations are:

• $f * g = g * f$ (Commutativity)
• $\tilde{f * g} = \tilde{g * f}$ (Commutativity of Transforms)
• $\tilde{f * g} = \tilde{f}\tilde{g}$ (Convolution Theorem)

Therefore, options A, B, and D are correct.