What does $\|D^\alpha u\|$ and $|D^\alpha u|$ mean in Evans' PDE book? For example, in the following proof from Evans' book, what do $\|Du\|$ and $|Du|$ mean? 
According to the appendix, the $|D^\alpha u|$ could mean the square root of the sum of all the squares of all derivatives of order $\alpha$. If this is the case here, it make sense that they become the same at the end of the proof.  However if this is the case, it has nothing to do with the Sobolev norm after all?



 A: In the notation of Evans we have

and

so we would have
$$\|Du\|_{L^p(\mathbb{R}^n)}=\||Du|\|_{L^p(\mathbb{R}^n)}=\left(\int_{\mathbb{R}^n}|Du|^p\right)^\frac{1}{p}$$
exactly as in the proof. Writing it out gives us
$$\|Du\|_{L^p(\mathbb{R}^n)}=\left(\int_{\mathbb{R}^n}|Du|^p\right)^\frac{1}{p}=\left(\int_{\mathbb{R}^n}\left(\sum_{|\alpha|=1}|D^\alpha u|^2\right)^\frac{p}{2}\right)^\frac{1}{p}=\left(\int_{\mathbb{R}^n}\left(\sum_{i=1}^n|u_{x_i}|^2\right)^\frac{p}{2}\right)^\frac{1}{p}.$$
This theorem does not contain any Sobolev norms, but it will be used to prove similar inequalites that do contain Sobolev norms.

Note that $D^\alpha u$ is a single function (of a specific derivative of $u$) and $D^k u$ a vector (of all derivatives of order $k$), as Evans considers $\alpha$ as a vector and $k$ as a number.
The second inequality in the proof comes from the fact that $|u_{x_i}|=\left(|u_{x_i}|^2\right)^\frac{1}{2}\leq \left(\sum_{i=1}^n|u_{x_i}|^2\right)^\frac{1}{2}=|Du|$ applied to the first inequality, and so the proof continues.
