Here is the statement of the problem (Evans PDE 2nd Ed., p.308, exercise 16)
Show that for $n \geq 3$ there exists a constant $C$ so that
$$ \int_{\mathbb {R}^n} \frac{u^2}{\vert x \vert^2} \,dx \leq C \int_{\mathbb{R}^n} \vert Du \vert ^2 dx $$
For all $ u \in H^1(\mathbb{R}^n)$.
Attempt: Well I have shown this result for $u \in C_c^\infty (\mathbb{R}^n)$. How do I extend this result to all of $H^1(\mathbb{R}^n)$ is my question?. I know that since our domain is $\mathbb{R}^n$, we have that $H_0^1=H^1$. Is there some way to invoke Friedrich's mollifiers to finish this problem?. But I don't see how.
Thanks for your help.
Edit Solution for the case when $u\in C_c^\infty(\mathbb{R}^n)$
Take $u\in C_{c}^{\infty}(\mathbb{R}^n)$. Set $F(x)=\dfrac{x}{\vert x \vert ^{2}}$. Then $ \vert F(x) \vert^{2}=\dfrac{1}{\vert x \vert^{2}}$ and $ \operatorname{div}{(F)}= \dfrac{n-2}{\vert x \vert ^{2}}$. Integration by parts gives,
$$
\int_{\mathbb{R}^n} u^{2} \operatorname{div}(F) dx = -\int_{\mathbb{R}^n} D(u^{2}) \cdot F(x) \tag{1} dx
$$
(Note that the boundary terms vanish because our domain is $\mathbb{R}^n$.)
By chain rule we get that
$$
\int_{\mathbb{R}^n} D(u^{2}) \cdot F(x)= 2 \int_{\mathbb{R}^n} uD(u) \cdot F dx = 2 \int_{\mathbb{R}^n} Du \cdot uF \tag{2} dx
$$
By $(1)$ and $(2)$,
$$
\left \vert \int_{\mathbb{R}^n} u^{2} \operatorname{div}(F) dx \right \vert = 2 \left \vert \int_{\mathbb{R}^n} Du \cdot uF \right \vert
$$
By Cauchy-Schwarz inequality we get,
$$
\left \vert \int_{\mathbb{R}^n} u^{2} \operatorname{div}(F) dx \right \vert \leq 2 ||{Du}||_{L^2} \cdot || u F ||_{L^2} \tag{**}
$$
Squaring both sides and replacing $\operatorname{div} F(x)= \dfrac{n-2}{\vert x \vert ^{2}}$, $ \vert F(x) \vert ^{2}= \dfrac{1}{\vert x \vert ^{2}}$ gives us,
$$
\frac{(n-2)^{2}}{4} \left(\int_{\mathbb{R}^n} \frac{u^{2}}{\vert x \vert^{2}} dx \right)^{2} \leq \left ( \int_{\mathbb{R}^n} |D(u)|^{2} dx \right) \left( \int_{\mathbb{R}^n} \frac{u^{2}}{\vert x \vert^{2}} dx \right)
$$
And so,
$$
\frac{(n-2)^{2}}{4} \int_{\mathbb{R}^n} \frac{u^{2}}{\vert x \vert^{2}} dx \leq \int_{\mathbb{R}^n} |D(u)|^{2} dx $$
Best Answer
As you have noted $H_0^1=H^1$, hence, by definition, for each $u\in H^1$, there is $u_n\in C_0^\infty(\mathbb{R}^n)$ such that $u_n\to u$ in $H^1$.
On the other hand, we have that $$\int_{\mathbb{R}^n}\frac{u_n^2}{|x|^2}\leq C\int_{\mathbb{R}^n}|Du_n|^2\tag{1}$$
To proceed, we choose a subsequence of $u_n$, which we will note relabel, such that $u_n\to u$ almost everywhere (see Theorem 4.9. of Brezis). On the right hand side we have convergnce because $u_n\to u$ in $H^1$. On the left hand side, we apply Fatou's Lemma, i.e. $$\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\leq \liminf\int_{\mathbb{R}^n}\frac{u_n^2}{|x|^2}\tag{2}$$
Now we combine $(1)$ and $(2)$ to conclude.