Necessary condition for embedding of a homogeneous Sobolev space into the space of distributions Let $\dot{H}^1(\mathbb{R}^n)$ denote the completion of $C^\infty_0(\mathbb{R}^n)$ with respect to the norm
$$C^\infty_0(\mathbb{R}^n) \ni \varphi \mapsto \| \varphi \|^2_1 \equiv \int_{\mathbb{R}^n} |\nabla \varphi(x)|^2dx. $$
So, strictly speaking, $\dot{H}^1(\mathbb{R}^n)$ is a set of equivalence classes of sequences $\{\varphi_m \} \subseteq C^\infty_0(\mathbb{R}^n)$ which are Cauchy with respect to the $\| \cdot \|_1$-norm.
Let's say that $\dot{H}^1(\mathbb{R}^n)$ is embedded into the space of distributions $\mathcal{D}'(\mathbb{R}^n)$ if there is a linear, injective, continuous map $i : \dot{H}^1(\mathbb{R}^n) \to  \mathcal{D}'(\mathbb{R}^n)$. Here, let's take continuity to mean that if $u \in \dot{H}^1(\mathbb{R}^n)$ and $u_n$ is a sequence in $\dot{H}^1(\mathbb{R}^n)$, then $u_n \to u$ in the $\|\cdot\|_1$-norm implies that $iu_n(\psi) \to iu(\psi)$ for all $\psi \in C^\infty_0(\mathbb{R}^n)$. 

I would like to show that if $\dot{H}^1(\mathbb{R}^n)$ is embedded into $\mathcal{D}'(\mathbb{R}^n)$, then, for each $\psi \in C_0^\infty(\mathbb{R}^n)$, the functional $l_{\psi} : C_0^\infty(\mathbb{R}^n) \to \mathbb{C}$ defined by
  $$l_{\psi}(\varphi) = \int_{\mathbb{R}^n} \varphi \psi dx$$
  has the property that $|l_\psi(\varphi)| \le C_{\psi} \|\varphi\|_1$, all $\varphi \in C_0^\infty(\mathbb{R}^n)$, where $C_\psi$ is some positive constant which may depend on $\psi$ (but is independent of $\varphi$).

This is one direction of lemma 1 as stated in section 15.2 in Maz'ya's Sobolev Spaces. 
The typical strategy to achieve a boundedness result like this is to assume not, and then we get a sequence $\varphi_n \in C_0^\infty(\mathbb{R}^n)$, with $\|\varphi_n \|_1 = 1$ such that $|l_\psi(\varphi_n/n)| \ge 1$, all $n$. Setting $u_n = \varphi_n/n$, my thought is then to look that the sequence $iu_n(\psi)$ of numbers (which must converge to zero) and try to derive a contradiction, but here is where I have gotten stuck. 
If we also require an additional, somewhat reasonable property for the embedding $i$, which is that $i$ acts as the identity on $C_0^\infty (\mathbb{R}^n) \subseteq \dot{H}^1(\mathbb{R}^n)$ in the sense that $i\varphi(\psi) = \int \varphi \psi dx$, each $\psi \in C_0^\infty(\mathbb{R}^n)$, then the desired conclusion is obtained straight-away. However, it is not clear that the author is intending for an embedding to have this extra property. 
Is there a way to show that such a $C_{\psi}$ exists without needing this extra requirement on an embedding? 
 A: If you only want an injective continuous map (which is not required to act in the expected way on $C_c^\infty $), you can find a lot of such maps:
Simply note that $C_c^\infty \to \nabla C_c^\infty \subset L^2(\Bbb{R}^n ; \Bbb{C}), f \mapsto \nabla f $ is an isometry, which thus extends to an isometry (in particular injective) $\iota : \dot {H}^1 \to V \subset L^2 (\Bbb{R}^n ; \Bbb{C}^n) \cong L^2 (\Bbb{R}^n) \times \cdots \times L^2 (\Bbb{R}^n)$, where $V $ denotes the image of this isometry.
As a subspace of a separable Hilbert space, $V$ is a separable Hilbert space, and thus so is $\dot{H}^1$. Clearly, $\dot{H}^1$ is infinite dimensional. Hence, we can choose countable orthonormal bases $(f_k)_k$ in $\dot{H}^1$ and $(g_k)_k$ in $L^2 (\Bbb{R}^n)$, so that $\Phi : \dot{H}^1 \to L^2 (\Bbb{R}^n), \sum \alpha_k f_k \mapsto \sum \alpha_k g_k$ is an isometric isomorphism.
Then, you can choose any continuous injective map $\theta : L^2 \to D'$ (for example $f \mapsto g \cdot f$, where $g : \Bbb{R}^n \to \Bbb{C}$ is (polynomially) bounded) and obtain a continuous injection $\theta \circ \Phi : \dot {H}^1 \to D'$.
In view of the variety of possible maps $\theta $ and also of the choice of the orthonormal bases, there is no meaningful necessary condition which can be derived from existence of this injective, linear, continuous embedding of $\dot{H}^1$ into $D'$.
