# Compact embedding of $H_0^1$ into $C([0,1])$

We can compactly embed the Lebesgue space $L^p([0,1])$ into $C([0,1])$.

For this, we define the integral operator $$T \colon g \mapsto \int_0^x g(t) dt.$$ By the use of the Arzela-Ascoli theorem, one can show that this is indeed a compact operator.

On wikipedia it is stated that

The case $p = 2$ can be seen as a simple instance of the fact that the injection from the Sobolev space $H_{0}^{1}(\Omega )$ into $L^2(\Omega)$, for $\Omega$ a bounded open set in $\mathbb{R}^d$, is compact.

What is meant by injection of $H_0^1(\Omega)$ into $L^2(\Omega)$ (maybe the trivial injection, since each Sobolev function lies in a corresponding $L^p$-space)?

How does this statement follow for $p=2$ from the compact embedding which was shown at the beginning?

A Banach space $X$ is compactly embedded in $Y$ if $X \subset Y$ and the inclusion $i:X \to Y$ is compact. Since $C([0,1]) \subset L^p([0,1])$, and not the other way around, $L^p([0,1])$ cannot be compactly embedded into $C([0,1])$.
The wikipedia article is exhibiting a compact operator $T:L^p([0,1]) \to C([0,1])$. This does not mean $L^p([0,1])$ is compactly embedded in $C([0,1])$, since this is not the inclusion (and it can't be since $L^p([0,1])$ is much larger than $C([0,1])$).
The Rellich–Kondrachov compactness theorem says that $H^1([0,1])$ is compactly embedded into $L^2([0,1])$. This means the inclusion $i:H^1([0,1]) \to L^2([0,1])$ is compact. In other words, bounded sequences in $H^1([0,1])$ admit subsequences converging in $L^2([0,1])$.
As to the comment in the wikipedia article, the operator $T$ is a bounded linear operator $T:L^2([0,1]) \to H^1([0,1])$. If we compose $T$ with the inclusion $i:H^1([0,1]) \to L^2([0,1])$, then $i \circ T$ is a compact mapping from $L^2([0,1])$ to $L^2([0,1])$, since $i$ is compact. Thus, no need to use Arzela-Ascoli directly here (incidentally, the Arzela-Ascoli theorem is an integral part of the proof of the Rellich-Kondrachov Theorem, so there is no getting away from it!). This is what the wikipedia article is getting at, I believe.
I should note there is nothing special about $p=2$ here. The Rellich-Kondrachov theorem holds for $1 \leq p \leq \infty$, and so the comment above does as well.