If a function, $\phi (x)$, has compact support and is in $L^2(\mathbb R)$, then is $x\phi (x) \in L^1(\mathbb R)$?

Question

I am trying to do the below question and I have been able to prove that if $\phi (x) \in L^2(\mathbb R)$ and $x\phi (x) \in L^1(\mathbb R)$ then $\phi$ is square integrable $\iff \int \phi(x)dx=0$.

The step that I'm missing is that if a function, $\phi (x)$, has compact support and is in $L^2(\mathbb R)$, then $x\phi (x) \in L^1(\mathbb R)$. The question:

Answer 1

Just apply Holder's inequality on the support of $\phi$. $$\int_{\mathbb{R}}|x\phi|\,dx\le L\int_{-L}^L|\phi|\,dx\le L(2L)^{1/2}\Vert\phi\Vert_{L^2}<\infty,$$ where $\phi=0$ outside $[-L,L]$.

Answer 2

Since $\phi$ has compact support over the real line, there is some finite family of open intervals $(l_k,u_k)$ whose union contains the support of $\phi$. Since $\phi$ is in $L^2$ the integral

$$ \int_{-\infty}^\infty |\phi|^2dx = \sum_k \int_{l_k}^{u_k}|\phi|^2dx $$ is finite. Thus $$ \int_{-\infty}^\infty |x\phi|^2dx = \sum_k \int_{l_k}^{u_k}|x\phi|^2dx \leq \sum_k|u_k|^2\int_{l_k}^{u_k}|\phi|^2dx $$ is also finite. Holder's inequality implies $$ \left(\int_a^b |f(x)|dx\right)^2 \leq \int_a^b |f(x)|^2dx, $$ so we end up with $$ \int_{-\infty}^\infty |x\phi|dx \leq \sqrt{\int_{-\infty}^\infty |x\phi|^2dx}, $$ and the left-hand side is therefore finite.