I'm wondering if the following implications are true. If yes, how can I prove them?
$$ 1. \,\,\,\,\, \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert^2 \, dt \implies \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert \, dt $$
$$ 2. \,\,\,\,\, \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert \, dt \implies \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert^2 \, dt $$
I tried to prove them in this way. Consider two positive real constants $a$ and $b$.
$$ 1. \,\,\,\,\, \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert^2 \, dt = a < + \infty \rightarrow \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert \, dt = \pm \sqrt{a} $$ only the positive solution is correct because the integral of the absolute value is always $\geq 0$, thus:
$$ 1. \,\,\,\,\, \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert^2 \, dt = a < + \infty \rightarrow \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert \, dt = \sqrt{a} < + \infty $$
$$ 2. \,\,\,\,\, \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert \, dt = b < + \infty \rightarrow \displaystyle{\int \limits_{- \infty }^{+ \infty }} \lvert x(t) \rvert^2 \, dt = b^2 < + \infty $$
Thank you for your time.
You cannot write $\int |x(t)|^{2}$ as $(\int |x(t)|)^{2}$, so your arguments are wrong. Let $f (x)=\frac 1 {\sqrt {x}}$ for $0<x<1$ and $0$ elsewhere. Let $g (x)=\frac 1 x$ for $1<x<\infty $ and $0$ elsewhere. These two examples show that both implications are false.