$f(x)$ and $xf(x)\in L^2(\mathbb{R})$ then $f(x)\in L^1(\mathbb{R})$ If $f(x)$ and $xf(x)\in L^2(\mathbb{R})$ then $f(x)\in L^1(\mathbb{R})$.
I know that if $E$ is of finite measure, then we can infer from $f(x)\in L^2(E)$ to get $f(x)\in L^1(E)$. However, now $E=\mathbb{R}$, I don't know how to get the result now. How to apply $xf(x) \in L^2$?
 A: Split into two: 
$$\begin{split}
\int_\mathbb{R} |f(x)|dx &= \int_{|x|\le 1} |f(x)|dx + \int_{|x|>1} |f(x)| dx \\
&= \int_{|x|\le 1} |f(x)|dx + \int_{|x|>1} \frac{1}{|x|}{|x|}\cdot |f(x)| dx
\end{split}$$
Then apply Cauchy-Schwarz on both terms. 
A: Let $A = [0,1]$ and $B = [1, \infty]$. Note that
$$ \int_\mathbb{R} |f| \ dm = \int_A |f| \ dm + \int_B |f| \ dm$$
Observe that $m(A) = 1 < \infty$, use this to show that $f \in L^1 (A)$.
Next denote $g(x) = x f(x)$ and note that
$$ |f(x)| = \left| \frac{g(x)}{x}\right|$$
Let us write the second integral in a more suggestive format
$$ \int_B |f| \ dm = \int_B |g| \left| \frac{1}{x} \right| \ dm$$
Now apply the Hölder inequality (which in this case is Cauchy-Schwarz) and conclude that $f \in L^1(B)$. Now combine both results and conclude that $f \in L^1 (\mathbb{R})$.
A: Since you already know this is true for finite measure sets it's sufficient to consider $f$ on $X=[1,\infty)$, say (the negative number case is similar). There you have
$$\int_X |f|dx =\int_X |f| \frac{|x|}{|x|}dx$$
Now apply Cauchy Schwarz and use your assumptions. 
