Equip $C_0^\infty(\mathbb{R}^n)$ with the norm $$ \|\varphi\|^2_1 := \int_{\mathbb{R}^n}| \nabla \varphi|^2 \text{d} x.$$
Indeed, $\| \cdot \|_1$ is a norm on $C_0^\infty(\mathbb{R}^n)$ because any constant function with compact support must be identically zero. Let $\tilde{H}_0^1(\mathbb{R}^n)$ denote the completion of $C_0^\infty(\mathbb{R}^n)$ with respect to the above norm. Note that $\tilde{H}_0^1(\mathbb{R}^n)$ is not quite the usual space $H_0^1(\mathbb{R}^n)$, which is the closure of $C_0^\infty(\mathbb{R}^n)$ with respect to the norm $(\int_{\mathbb{R}^n} |\varphi|^2 + | \nabla \varphi|^2 \text{d} x)^{\frac{1}{2}}$.
Are there any references that describe properties of $\tilde{H}_0^1(\mathbb{R}^n)$? In particular, is it true that $\tilde{H}_0^1(\mathbb{R}^n) \subseteq L^2(\mathbb{R^n})$?
I have noticed that
$$\| \varphi \|^2_1 = \int_{\mathbb{R}^n} | \xi|^2 |\hat{\varphi }(\xi)|^2 \text{d} x.$$
Which has led me to conjecture that functions in $\tilde{H}_0^1(\mathbb{R}^n)$ may be allowed to have some blow-up near zero, hence it may be that $\tilde{H}_0^1(\mathbb{R}^n)$ is not contained in $L^2(\mathbb{R}^n)$.
Solutions or reference suggestions are greatly appreciated.
The answer is no in general: Take for instance $u(x)= \eta(x) |x|^{-n/p}$, where $\eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $\nabla u\in L^p$, but $u\notin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $\psi_R(x)=\psi(x/R)$ for $\psi\in C_c^\infty$ that is identically one in the unit ball).
What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.