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By the Rellich-Kondrachov theorem, one knows that the embedding $H^1(0,1) \subset L^2(0,1)$ is compact.

On another hand, by Sobolev inequalities, one also has $H^1(0,1) \subset C^0[0,1]$ (in fact, even $C^{0,\frac{1}{2}}$ in this one-dimensional case, by using the fundamental theorem of calculus and some Cauchy-Schwartz arguments).

My question is whether there exists some "intermediate subspace" in the following sense.

Namely, does there exists a Hilbert space $H$ which is compactly embedded in $L^p(0,1)$ for some $p\geq 1$, and which is not a subspace of $C^0[0,1]$?

char
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1 Answers1

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Yes, such Hilbert spaces exist and they are a special case of fractional Sobolev spaces. For $\alpha\in(0,1/2)$ we have $H^\alpha(0,1)\subset L^2(0,1)$ by definition, and one can show that the step function which is $1$ on $(1/2,1)$ and $0$ else is in $H^\alpha(0,1)$. Since this function is not continuous, $H^\alpha(0,1)$ does not embed in $C^0[0,1]$.

See also Proof that the characteristic function of a bounded open set is in $H^{\alpha}$ iff $\alpha < \frac{1}{2}$ and To what fractional Sobolev spaces does the step function belong? (Sobolev-Slobodeckij norm of step function) for more details.

It is also known that $H^\alpha(0,1)$ embeds compactly into $L^2(0,1)$ for $\alpha\in (0,1/2)$. This follows from Theorem 7.1 in this pdf.

supinf
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  • Excellent, thank you! Just to make things clear, are you saying that fractional Sobolev of order $\alpha < \frac12$ compactly embeds into $L^2(0,1)$? – char Dec 11 '20 at 21:30
  • @bgsk good point, I skipped over that. But yes, the embedding is compact, and I added a source. – supinf Dec 11 '20 at 21:57