In my adv. Analysis course, we have studied the following Sobolev Embedding Theorem:
Let $m\in\mathbb{N}$ and $s>m+d/2$. Then $$H^s(\mathbb{R}^d)\hookrightarrow C_0^m(\mathbb{R}^d)$$That is: $H^s(\mathbb{R}^d)$ embeds into $C_0^m(\mathbb{R}^d)$
The proof we've studied basically starts by noticing that the Schwartz space $\mathcal{S}(\mathbb{R}^d)$ is dense in $H^s$, and then it goes on proving that the inclusion map $$i:H^s\overset{\mathrm{dense}}{\supseteq}\mathcal{S} \longrightarrow C_0^m$$ is continuous. So by existence (and uniqueness) of an extended (injective) linear bounded operator, we have in fact an embedding from the Sobolev Space $H^s(\mathbb{R}^d)$ into $C_0^m(\mathbb{R}^d)$.
But I'm asking myself the nature of such extended embedding, more precisely:
- Does it mean that, under the hypotheses of the theorem, the Sobolev space $H^s$ is a "subset" of $C_0^m$? In the sense that every function in $H^s$ has a representative (of the a.e. equivalence class) in $C_0^m$? In other words, does the extension of the inclusion behaves as an inclusion?
Thanks
As Giusseppe has stated, this does, indeed, mean that you can view the $H^s$ as a subset of $C^m_0$, but, as you guessed, exactly in the best possible sense: Any function class in $H^s$ has a representative time $C^m_0$.
To see this, let $u\in H^s$ (or rather, let $u$ be a representative of such a class) and pick a sequence $(u_n)_{n\in\mathbb{N}}\subseteq \mathcal{S}$ such that $u_n\to u$ in $H^s$. Since $i$ is a bounded operator on $H^s$, we see that $u_n$ must also be convergent in $C^m_0$. This, in particular, implies that $u_n$ is pointwise convergent with a limit $f\in C^m_0$. However, $u_n$ being convergent in $H^s$ implies, in particular, that $u_n$ converges to $u$ in $L^2$. This, in turn, implies that there is a subsequence $u_{n_k}$ which converges to $u$ almost everywhere.
However, this implies that $f=u$ almost everywhere, and hence, any element of $H^s$ admits a classically differentiable representative.