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I have the following statement whith an argumentation which I do not understand.

Fix integers $k,l$ such that $0\leq l\leq k$. Then the identity map on $C^\infty(\mathbb{T}^d)$ extends to the injective continuous operator $$i_{k,l}:H^k(\mathbb{T}^d)\rightarrow H^l(\mathbb{T}^d)$$ of norm one. Moreover, for every $j\in\{1,...,d\}$ the partial derivative $$\partial_j:C^\infty(\mathbb{T}^d)\rightarrow C^\infty(\mathbb{T}^d)$$ extends to the continuous operator $$\partial'_j:H^k(\mathbb{T}^d)\rightarrow H^{k-1}(\mathbb{T}^d)$$ with norm less or equal then one.

The proof consider the embedding of the $C^\infty$ space in $(L^2)^{K(k)}$ (The sobolev space is defined as the closure of the embedding). My problem is more for the norms. I can't see why should be so. And the proof states it is clear... Could someone explain?

We defined the Sobolev space on $L^2(\mathbb{T}^d)$ as the closure of $C^\infty(\mathbb{T}^d)$ given an integer factor $k$. We use the embedding $f\mapsto (f_\alpha)_{\parallel \alpha\parallel_1\leq k}\in L^2(\mathbb{T}^d)^{K(k)}$ where $K(k)$ is the number of multi indices less or equal than $k$.

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    $\begingroup$ What does $(L^2)^{K(k)}$ mean? I have played with Sobolev spaces for a while and I have never seen such notation. Can you define the space and give the embedding, or at least provide a link a document where they can be found? $\endgroup$ – Joonas Ilmavirta Jul 16 '15 at 16:25
  • $\begingroup$ @JoonasIlmavitra We defined the Sobolev space on $L^2(\mathbb{T}^d)$ as the closure of $C^\infty(\mathbb{T}^d)$ given an integer factor $k$. We use the embedding $f\mapsto (f_\alpha)_{\parallel \alpha\parallel_1\leq k}\in L^2(\mathbb{T}^d)^{K(k)}$ where $K(k)$ is the number of multi indices less or equal than $k$. $\endgroup$ – sky90 Jul 17 '15 at 4:19
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By density it is enough to consider smooth functions. The arguments work for general Sobolev functions as well, but there is less to worry about with smooth functions.

Consider first $i_{k,l}$ for $0\leq l\leq k$. Take a function $u\in C^\infty$. Its squared norm is $$ \|u\|_{H^k}^2=\sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^2}. $$ Now $$ \|u\|_{H^l}^2=\sum_{|\alpha|\leq l}\|\partial^\alpha u\|_{L^2}\leq\sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^2}=\|u\|_{H^k}^2 $$ (we added the positive term $\sum_{l<|\alpha|\leq k}\|\partial^\alpha u\|_{L^2}$), so $\|i_{k,l}u\|_{H^l}=\|u\|_{H^l}\leq\|u\|_{H^k}$. This inequality means that $i_{k,l}$ is continuous with norm at most one. To see that the norm is exactly one, we need to come up with a function $u\in H^k$ so that $\|u\|_{H^l}=\|u\|_{H^k}$ (or a sequence of functions so that the norms are arbitrarily close). But this is simple: take the constant function; then $\|u\|_{H^l}=\|u\|_{H^k}=\|u\|_{L^2}$.

Then consider $\partial_j$. Take $u\in C^\infty$ and $k\geq1$. We want to estimate the norm $\|\partial_ju\|_{H^{k-1}}$ by $\|u\|_{H^k}$. Let $M_k$ be the set of all multi-indices of order $k$ or smaller, and let $M_{k,j}$ be its subset of multi-indices that have nonzero $j$th component. Now $$ \|\partial_ju\|_{H^{k-1}}^2 = \sum_{|\alpha|\leq k-1}\|\partial^\alpha\partial_ju\|_{L^2} = \sum_{\beta\in M_{k,j}}\|\partial^\beta u\|_{L^2} \leq \sum_{\beta\in M_k}\|\partial^\beta u\|_{L^2} = \|u\|_{H^k}^2. $$ This estimate shows that the operator norm of $\partial_j$ is at most one.

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  • $\begingroup$ Thanks. You are using the norm of the sobolev space which is the bast argument you can give but in my proof there is no mention to the norm, which is going to be introduced later. So I was asking myself if we could come to such a conclusion also without an explicit rigorous proof $\endgroup$ – sky90 Jul 18 '15 at 7:20
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    $\begingroup$ @sky90, I don't think one could make these proofs without calculations with explicit formulas for the norms. Otherwise getting the constants right sounds next to impossible. The claims sound reasonable and something like that should be true, but I don't see a significantly different way of proving that. Side note: There is another popular norm on $H^k$ (which uses the Fourier series), namely $\|u\|_{H^k}^2=\sum_{\xi\in\mathbb Z^d}(1+|\xi|^2)^{k}|\hat u(\xi)|^2$. The claims are still true and the proof is similar, and it still requires an explicit calculation. $\endgroup$ – Joonas Ilmavirta Jul 18 '15 at 7:43

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