Here what I've done to prove that Sobolev Spaces $W^{(m,p)}(\Omega)$ are uniformly convex for $1<p<\infty$
Given integers $n\geq 1,k\geq0$, we define $N(n,m)$ as the number of multi-indices $\alpha=(\alpha_1,\alpha_2,...,\alpha_n)$ such that $|\alpha|\leq m$. I then proceed to label each such multi-index $\alpha=\alpha^{(m)}$, where $m$ runs from $1$ to $N(n,m)$. Then, I create a mapping $$P:W^{m,p}(\Omega)\rightarrow {(L^p(\Omega))}^N$$ $$u\mapsto(u,D^{\alpha^(1)},D^{\alpha^(2)},...,D^{\alpha^(N)})$$
Then, in many references it is written that $P$ is an isometric, but how can we know it is an isometric mapping when we even don't know the metric on ${(L^p(\Omega))}^N$ ? So, what is the metric ${(L^p(\Omega))}^N$ so we can prove that the mapping is an isometric mapping?
Thanks
It depends on your norm on $W^{m,p}(\Omega)$. You could choose \begin{equation*} \| u \|_{m,p}^2 = \sum_\alpha \|D^\alpha u \|_p^2, \end{equation*} or \begin{equation*} \| u \|_{m,p}^p = \sum_\alpha \|D^\alpha u \|_p^p. \end{equation*} Then, you similarly choose \begin{equation*} \| u \|_{p}^2 = \sum_{i=1}^N \|u_i \|_p^2, \end{equation*} or \begin{equation*} \| u \|_{p}^p = \sum_{i=1}^N \|u_i \|_p^p \end{equation*} on $(L^p(\Omega))^N$.