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?


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 $D'$.

  • $\begingroup$ Thank you for the answer. Am I correct in thinking that, when you write $\nabla C_c^\infty \subseteq L^2$ above, you more explicitly mean $\{ \nabla \varphi : \varphi \in C_0^\infty(\mathbb{R}^n) \} \subseteq L^2(\mathbb{R}^n) \times \cdots \times L^2(\mathbb{R}^n)$? ($n$ total copies of $L^2(\mathbb{R}^n)$). In that case I am having trouble thinking of an injective continuous map $(L^2(\mathbb{R}^n))^n \to \mathcal{D}'(\mathbb{R}^n)$. $\endgroup$ – JZS Jun 13 '17 at 11:27
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    $\begingroup$ @JZShapiro: Oh, damn, you are right. I somehow overlooked that the gradient is a vector. This is a good example of what can happen if you are sloppy with your notation. My bad! If I find a way to fix my answer in the next few hours, I will edit it; otherwise, I will delete it. $\endgroup$ – PhoemueX Jun 13 '17 at 12:54
  • $\begingroup$ @JZShapiro: I edited my post. Please let me know if you have any further questions :) $\endgroup$ – PhoemueX Jun 13 '17 at 21:09

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