# Sobolev Embedding Theorems in Dimension One

What exactly is the content of Sobolev Embedding Theorems (compact for Sobolev spaces and Hölder spaces) when we're looking at functions on the real line?

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I am not exactly sure what you are asking about; but Sobolev embedding in one dimension is essentially just fundamental theorem of calculus + Hölder inequality + (optionally) interpolation of Lebesgue spaces. The statement of the theorem can be easily checked on, say, Wikipedia, so I guess you are not just asking for the statement(s). Can you clarify what you mean by "the content"? (What do you seek in an answer?) – Willie Wong Sep 3 '12 at 13:29

The theorem tells you that if $U$ is a bounded open subset of $\mathbb R$ and $k > l + d/2$ then the inclusion $C^\infty (U) \hookrightarrow C^l(U)$ can be continuously extended to $H^k(U) \hookrightarrow C^l(U)$ where $H^k(U)$ is your Sobolev space.

This is the Sobolev embedding theorem for $\mathbb R^d$, so in your question, $d=1$. $C^l(U)$ denotes the set of all continuous functions $f: U \to \mathbb R$ (or $\mathbb C$) such that $f$ has $l$ continuous derivatives.

As a consequence, if $T: C^l \hookrightarrow X$ is a linear operator, you may apply it to $H^k (U)$.

Here are three related threads: 1, 2, 3.

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Thanks for your answer! Unfortunately, I still have some questions: What's d here? Is $H^k(U) = H^\infty,k(U)$ the space with the Sobolev sup-norm? Is $C^l$ a Hölder space, if es, what's the $\alpha$. I would also like to know if there's a version for all of $R$, not just bounded domains, and a version comparing different Sobolev spaces? Thanks for our reply! – corni75 Aug 29 '12 at 12:59
@corni75 I added some of the answers to your questions to the answer. $d$ is the dimension of your space. I don't think that $H^\infty = H^k$. The $k$ in the Sobolev space is the number of derivatives we take. I find it difficult to imagine what $H^\infty$ looks like. What is $k(U)$ in your comment? I don't know whether there is a version for all of $R$ but I suspect one would have to restrict to something like bounded functions whose derivatives are all also bounded. – Rudy the Reindeer Aug 29 '12 at 13:09
Sobolev spaces usually take some p-norm, that's why they're usually indexed by two and not just one letter... – corni75 Aug 29 '12 at 13:23
@corni75: The customary notation is $H^{k} = W^{k,2}$. Matt isn't talking about Hölder spaces at all (no Morrey inequality or anything involved here). This answer only adresses the easiest instance (only continuity, no compactness) of the Sobolev embedding theorems. – t.b. Aug 29 '12 at 13:38
@corni75 Sorry, for some reason your previous comment didn't ping me. If you had read the threads I linked to you would've noticed that $p$ is chosen to be $2$ implicitly here. I was trying to keep this as simple as possible. But of course we can go general: the Sobolev embedding theorem tells you (for $d=1$) that if $U\subset \mathbb R$ is a bounded subset with sufficiently smooth boundary, $1 < p \leq q < \infty$, $m \geq k \in \mathbb N$, such that $$m - \frac{1}{p} \geq k - \frac{1}{q}$$ then $$W^{m,p}(U) \hookrightarrow W^{k,q}(U)$$ is continuous. – Rudy the Reindeer Aug 29 '12 at 14:45