The Sobolev embedding theorem as stated in my notes says that if we have $k > l + d/2$ then we can continuously extend the inclusion $C^\infty(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$ to $H^k(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$.

We define $H^k$ to be the closure of $C^\infty$ with respect to the Sobolev norm, see my previous question for the definition.

What I'm confused about is, why we need the condition $k > l + d/2$. What exactly does it give us? If $H^k$ is the closure of $C^\infty$ we already get that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. What am I missing? Thanks for your help.


You said that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. This is true, provided that the topology on $C^\infty$ is inherited from that of $H^k$. In other words, one has to use $H^k$-norms in the domain to define continuity of $T:C^\infty\to C^l$. The condition $k>\frac{n}2+l$ guarantees this continuity when $T$ is the canonical inclusion.

  • $\begingroup$ I think what I said is wrong: The Sobolev embedding theorem gives me an inclusion $i: H^k \hookrightarrow C^l$. Which would mean that if I have $T: C^l \to C^l$ I can also apply $T$ to $H^k$. On the other hand, by definition $C^\infty$ is dense in $H^k$ which would mean that I can continuously extend any continuous $T: C^\infty \to C^l$ to $H^k \to C^l$. $\endgroup$ – Rudy the Reindeer Aug 18 '12 at 6:35
  • $\begingroup$ @Matt: Please read the part after "provided". $\endgroup$ – timur Aug 18 '12 at 7:16
  • $\begingroup$ Dear timur. Thanks for your patience, I think I'm getting there. So the inclusion $i: C^\infty \hookrightarrow C^l$ is always continuous? By always I mean for all $l$ and $C^\infty$ with the $\|\cdot\|_\infty$ topology and $C^l$ with the $\|\cdot\| = \max_{0\leq i \leq l} \|\partial_i \cdot \|_\infty$ topology. $\endgroup$ – Rudy the Reindeer Aug 18 '12 at 11:18
  • $\begingroup$ I think it's "clear" that the topology on $C^l$ is coarser or equally coarse as the topology on the domain $C^\infty$. Does that sound right? $\endgroup$ – Rudy the Reindeer Aug 18 '12 at 11:19
  • $\begingroup$ Yes. For continuity, you cannot take the uniform-norm topology on $C^\infty$ unless $l=0$, because you would need to control a derivative in terms of pointwise values. By this reasoning the norm on $C^\infty$ must be at least as strong as that of $C^l$. $\endgroup$ – timur Aug 18 '12 at 16:13

Suppose that the linear embedding $(C^\infty(\mathbb T^d),\|\,\cdot\,\|_{H^k}) \hookrightarrow (C^l(\mathbb T^d),\|\,\cdot\,\|_{C^l})$ is continuous from the incomplete space on the left to the complete space on the right.

Then $i$ lifts to a unique, continuous linear map $\bar i: (H^k,\|\,\cdot\,\|_{H^k}) \rightarrow (C^l(\mathbb T^d),\|\,\cdot\,\|_{C^l})$, where $(H^k,\|\,\cdot\,\|_{H^k})$ is the completion of $(C^\infty(\mathbb T^d),\|\,\cdot\,\|_{H^k})$.

To get an embedding theorem, we need to check that $\bar i$ is injective. This property is not automatic, and fails in general.

  • 2
    $\begingroup$ Are you sure? See Timur's answer. $\endgroup$ – Nate Eldredge Aug 15 '12 at 14:47
  • $\begingroup$ @NateEldredge I assumed that the OP had chosen parameters so that the inclusion $i$ was already continuous when $C^\infty$ is equipped with the $H^k$ norm. Maybe I misunderstood the question. $\endgroup$ – user940 Aug 15 '12 at 16:34

The elements which are in $H^k∖C^{\infty}$ could be mapped by the extension to a function which may not be $C^l$. Sobolev embedding say that it's actually the case with a condition on the dimension and the order of derivatives we want to work with.

  • $\begingroup$ That doesn't quite make sense: the extension, if it exists, will by definition be a map from $H^k$ to $C^l$. $\endgroup$ – Nate Eldredge Aug 15 '12 at 14:49

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