# Question about definition of Sobolev spaces

I'm trying to understand the following definition:

which can also be found here on page 136.

Question 1: Closure with respect to what norm? It's not given in the definition.

Question 2: Do I have this right: I can view $C^\infty$ as a dense subspace of $H^k$ via the map (embedding) $f \mapsto (D^\alpha f)_\alpha$ where the tuple $f$ is mapped to consists of all derivatives $D^\alpha f$ such that $|\alpha| \leq k$. Then this is cool because if we have this we can extend any linear operator $T: C^\infty \to C^n$ continuously to all of $H^k$ so that anything we can do to smooth functions we can also do to Sobolev functions. That is, even if the functions don't have a strong $\alpha$-th derivative we can treat them as if they did.

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The picture is not very clear:) – Paul Aug 12 '12 at 11:37
I'll hazard a wild guess that we probably want to use the $L^2$ norm to complete to something we'll be calling an $L^2$ space. – Kevin Carlson Aug 12 '12 at 11:39
By definition $V$ is a product of $K(k)$ $L^2$-spaces. It should be a Hilbert space, so there's only one candidate. Give it the norm that it should have, find the definition also given in Def. 3.50. – t.b. Aug 12 '12 at 11:45
@t.b. Thank you! So the $\ell^2$-norm, I see. And what about question 2? – Rudy the Reindeer Aug 12 '12 at 11:47

1. We can endow $V$ with the norm $$\lVert (v_j)_{1\leq j\leq K(k)}\rVert_V:=\sqrt{\sum_{j=1}^{K(k)}\lVert v_j\rVert_{L^2(\mathbb T^d)}^2}$$ (it's an usual norm in product of $L^2$ spaces).
2. Yes, by definition $C^{\infty}(\mathbb T^d)$ is dense in its closure. When you say "anything we can do with smooth function, we can do it for Sobolev functions", careful. The property true for smooth functions has to be preserved by taking the limit in $\lVert\cdot\rVert_V$.
Thank you. Just to be sure: does taking the limit mean that we have a sequence of $f_n$ in $C^\infty$ such that all the $\alpha$-th derivatives with $|\alpha| \leq k$ are in $L^2$, that is, $(D^\alpha f)_\alpha \in H^k$ and then we want $f$ to have the property that all $f_n$ have where $\| f_n - f \|_\infty \to 0$? – Rudy the Reindeer Aug 12 '12 at 15:05
I would take $\lVert f_n-f\rVert_V$ to $0$ rather than the supremum norm, but yes that's what I meant. – Davide Giraudo Aug 12 '12 at 15:07
Sorry I am very confused. There is a theorem that if $A$ is a dense subspace of $B$ and $T: A \to C$ is a continuous and linear operator then $T$ uniquely and continuously extends to all of $B$. Now if we take $A= C^\infty$, $B= H^k$ and $C=C^l$ then this would mean that I can do anything I can do to smooth functions $C^\infty \to C^l$, I can also do to Sobolev functions $H^k \to C^l$. What am I missing? – Rudy the Reindeer Aug 18 '12 at 12:12