Can you move absolute value inside summation?

Question

I have this: $$x(t) = \sum_{n=-\infty}^{\infty} x(nA)y(t-n)$$ and I want to know $|x(t)|^2$. Can I do this:

$$|x(t)|^2 =|\sum_{n=-\infty}^{\infty} x(nA)y(t-n)|^2 = \sum_{n=-\infty}^{\infty} |x(nA)y(t-n)|^2 = \sum_{n=-\infty}^{\infty} |x(nA)|^2|y(t-n)|^2$$ If I then know that $|y(t-n)|^2 \le 1$ $\forall$ $n$, does that mean that: $$|x(t)|^2 \le \sum_{n=-\infty}^{\infty} |x(nA)|^2$$

Answer 1

We can't say $$|x(t)|^2 =|\sum_{n=-\infty}^{\infty} x(nA)y(t-n)|^2 = \sum_{n=-\infty}^{\infty} |x(nA)y(t-n)|^2$$

But we can say

$$|x(t)|^2 =|\sum_{n=-\infty}^{\infty} x(nA)y(t-n)|^2 \leq \sum_{n=-\infty}^{\infty} |x(nA)y(t-n)|^2$$ and if $|y(t-n)|^2 \le 1$ $\forall$ $n$ we have that $$|x(t)|^2 \leq \sum_{n=-\infty}^{\infty} |x(nA)|^2$$