Relation between weighted Sobolev space and ^$L^{1}(\mathbb{R}^{N})$ Consider the Hilbert space
$$X(\mathbb{R}^{N})=\left\{u\in H^{1}(\mathbb{R}^{N}); \;\int_{\mathbb{R}^{N}}\left|x\right|^{2}\left|u\right|^{2}dx<\infty\right\}$$
equipped with the norm 
$$\|u\|^{2}_{X}=\int_{\mathbb{R}^{N}}\left(\left|\nabla u\right|^{2}+\left|x\right|^{2}\left|u\right|^{2}\right)dx.$$
For $N=1$, it is easy to show that $X(\mathbb{R}^{})\hookrightarrow L^{1}(\mathbb{R})$. If $N\geq2$,   is it true that  $X(\mathbb{R}^{N})\hookrightarrow L^{1}(\mathbb{R}^{N})$ ? Thank You for any comments.
 A: The statement is false. First, let us look at the $\mathbb{R}^2$ case. Consider the function
\begin{align}
u(x) = 
\begin{cases}
e^{-2} & \text{ if } |x|<e, \\
\frac{1}{|x|^2\log|x|} & \text{ if } |x|\geq e.
\end{cases}
\end{align}
Observe $u \notin L^1(\mathbb{R}^2)$ and $u \in L^2(\mathbb{R}^2)$ since
\begin{align}
\int_{\mathbb{R}^2} |u(x)|\ dx = \pi+ \int_{|x|\geq e} \frac{dx}{|x|^2\log |x|} = \pi+ 2\pi\int^\infty_e \frac{dr}{r \log r} = \infty 
\end{align}
and
\begin{align}
\int_{\mathbb{R}^2} |u(x)|^2\ dx = \frac{\pi}{e^2}+2\pi \int^\infty_e \frac{dr}{r^3(\log r)^2}<\infty
\end{align}
Now, let us show  $u \in X(\mathbb{R}^2)$. Observe
\begin{align}
|\nabla u| = 
\begin{cases}
0 & \text{ if } |x|<e,\\
\frac{2\log |x|+1}{|x|^3(\log |x|)^2} & \text{ if } |x|\geq e.
\end{cases}
\end{align}
which means 
\begin{align}
\int_{\mathbb{R}^2} |\nabla u|^2 \leq 9\int_{|x|\geq e} \frac{dx}{|x|^6(\log |x|)^2}<\infty.
\end{align} 
Lastly, we have
\begin{align}
\int_{\mathbb{R}^2} |x|^2|u(x)|^2\ dx = C + \int_{|x|\geq e} \frac{dx}{(|x|\log |x|)^2} = C + 2\pi \int ^\infty_e \frac{dr}{r (\log r)^2}<\infty.
\end{align}
Indeed, $u \in X(\mathbb{R}^2)$ but $u \notin L^1(\mathbb{R}^2)$. 
For the case $N>2$ just consider the function
\begin{align}
u(x) = 
\begin{cases}
1 & \text{ if } |x|<1,\\
|x|^{-N} & \text{ if } |x|\geq 1
\end{cases}.
\end{align}
Edit: For the case $N=2$ and $0<\epsilon<1$, we see by Holder's inequality
\begin{align}
\int_{\mathbb{R}^2} |u|^{2-\epsilon}\ dx =& \int_{|x|\leq 1} |u|^{2-\epsilon}\ dx + \int_{|x|>1} |x|^{2-\epsilon}\frac{|u|^{2-\epsilon}}{|x|^{2-\epsilon}}\ dx \\
\leq&\ \left(\pi\right)^{\epsilon/2}\left( \int_{|x|\leq 1} |u|^2\right)^{(2-\epsilon)/2}+ \left(\int_{|x|>1} |x|^2|u|^2 \right)^{(2-\epsilon)/2}\left(\int_{|x|>1} \frac{1}{|x|^{(4-2\epsilon)/\epsilon}} \right)^{\epsilon/2}<\infty
\end{align}
since $\frac{4-2\epsilon}{\epsilon}>2$. For $N>2$ we could easily adapt the same proof but we will have a smaller range for $\epsilon$. 
