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 $k$.
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.