application of the inequality $\|fg\|_1 \leq \|f\|_p\|g\|_q$ application of the inequality $\|fg\|_1 \leq \|f\|_p\|g\|_q$ where $1/p + 1/q = 1$

I know this is a straight application of the inequality, but how am I assured that the integral of $(1+|t|)^2|f(t)|^2$ and $(1+|t|)^{-2}$ are finite? I understand the first one I think, but I am not sure for the second.
 A: For the convergence of $\int_{\mathbb R}\frac1{(1+|t|)^2}\,dt$, on any interval the integral is finite because the function is continuous. So you only care about what happens at $\infty$ (and $-\infty$, but the function is even). So, for any $a>0$
$$
\int_a^\infty\frac1{(1+|t|)^2}\,dt\leq\int_a^\infty\frac1{|t|^2}\,dt<\infty.
$$
The integral $\int_{\mathbb R}(1+|t|)^2\,|f(t)|^2\,dt$ might not converge, but the inequality still holds if the right-hand-side is infinite. 
A: The first integral is converging because it is the integral of a continuous function on a compact set. In fact for all the $t\geq R$ the function is zero. Indeed the first integrand is a compactly supported function.
The second integral is finite since its integrand is a continuous function which behaves as $\frac{1}{|t|^2}$
A: f is continuous over whole real line, and it is zero for $|x| \ge R$, so f is bounded by some constant M. so
$\int_{-\infty}^{\infty} ( 1 + |t|)^2 f^2(t) dt =
\int_{-R}^{R} ( 1 + |t|)^2 f^2(t) dt  \le 2MR(1+R)^2$
is finite. 
$\int_{-\infty}^{\infty} ( 1 + |t|)^{-2} dt \le  \int_{-\infty}^{\infty}\frac{1}{1 + t^2} dt$
$= 2 \int_0^{\infty} \frac{1}{1 + t^2} dt $
$ = \pi$
please check correctness of last integral using $t = tan (s)$
and 
$ (1 + |t|)^2 > 1 + |t|^2$ 
