Nyquist rate of a function convolved with itself From Schaum's Outline, Digital Signal Processing, Second Edition, 2012, page 155:
If the Nyquist rate for x(t) is:
$$\Omega_s$$
what is the Nyquist rate for:
$$ y(t) = x^{2}(t)$$
Solution:
$$y(t)= x(t)x(t)$$
Multiplication in time domain equals convulsion in frequency domain.  Therefore:
$$Y(j\Omega) = \frac{1}{2\pi}X(j\Omega) * X(j\Omega)$$
Then the book say because of this relationship the highest frequency in y(t) will be twice the frequency of x(t).
My question is: 
How can you make this inference from the convulution of X with itself in the frequency domain?
A few definitions:
$$ F(j\Omega) = \int_{-\infty}^{\infty}f(t)e^{-j\Omega t} dt$$

In the convolution integral
$$ Y(j \Omega)=\int_{-\infty}^{\infty}X(j\Omega') X(j(\Omega-\Omega')) \,d\Omega',$$
both arguments are nonzero and there is a nonzero contribution to the integral when $\Omega\in (2\Omega_0-\epsilon,\Omega_0]$$\Omega'\in (\Omega_0-\epsilon,\Omega_0],$ where $\Omega_0$ is the Nyquist frequency of $X(j\Omega).$
