Follow-up on Sobolev embedding theorem

I asked this question in a comment when I realised that the answerer is away until September so I am posting it here in a new thread.

I've been thinking about the Sobolev embedding theorem, given as follows: If $k > l + d/2$ then we can continuously extend the inclusion $C^\infty (\mathbb T^d) \hookrightarrow C^l (\mathbb T^d)$ to $H^k (\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$ where $\mathbb T^d$ is the $d$-dimensional torus and $H^k$ is the closure of $C^\infty$ with respect to the norm $\|(D^\alpha f)_\alpha \| = \sqrt{ \sum_\alpha \|D^\alpha f\|^2}$.

Can you tell me if this is correct?

(i) By definition of $H^k$ we can uniquely and continuously extend any continuous linear operator $T$ that has domain $C^\infty (\mathbb T^d)$ to all of $H^k$.

(ii) What the Sobolev embedding theorem gives us is a continuous inclusion $i: H^k \hookrightarrow C^l$ so that given a continuous linear operator $T: C^l \to X$ (to any linear normed space $X$) we can apply $T$ to $H^k$ via $T \circ i$.

I think I used to mix up (i) and (ii) and now I think that these are two different facts, independent of each other. Thanks for your help.

I'm assuming you are thinking of $C^\infty$ as a normed space (with the Sobolev norm) in i). Then i) is correct yes, but has nothing to do with Sobolev embedding.
• Thank you, Thomas! Yes, I am thinking of $C^\infty$ as sequences in $\prod_{N(k)} L^2$ of the form $(f, 0, 0 \dots)$ where $f \in C^\infty$ with the Sobolev norm I gave in the theorem. Did you see Davide's comment in the thread I linked and my question in response to it? I'm asking because if (i) is correct, his comment seems incorrect. Sorry, I forgot to include this in the question. – Rudy the Reindeer Aug 18 '12 at 12:42
• Perhaps I should say vectors not sequences since $N(k) < \infty$. – Rudy the Reindeer Aug 18 '12 at 12:47
• @Matt: No, I guess Davide is correct, if I got the intent of your question correctly. If you have a property which is true in $C^\infty$ and you want to have that property in the Sobolev space, you have to prove it (most naturally using approximation in the norm with respect to which you took the closure, showing that the property is preseved when taking limits in that norm). Take, as a simple example that, for any $k$, the $k-th$ derivative of a $C^\infty$ function exists and is Lipshitz (on a compact domain). This fails to be true rather soon for the Sobolev space if $k$ increases. – user20266 Aug 18 '12 at 13:02
• What I wrote above is non-sense. We have $f \mapsto (D^\alpha f)_\alpha$ and then we take the closure of this set. – Rudy the Reindeer Aug 20 '12 at 5:43