Show that the triangle inequality remains for the following norm in $W^{m,p}(\Omega)$ Let be $\Omega$ an open set in $\mathbb{R}^n$. I'm trying to show that
$$||u||_{m,p} = \left( \sum\limits_{|\alpha| \leq m} \int_{\Omega} |D^{\alpha} u|^p dx \right)^{\frac{1}{p}}$$
is a norm in $W^{m,p}(\Omega)$, but I'm stuck in the triangle inequality. I would like some help to prove it.
My attempt in order to show the triangle inequality:
$\begin{align*}
\left( \sum\limits_{|\alpha| \leq m} \int_{\Omega} |D^{\alpha} (u+v)|^p dx \right)^{\frac{1}{p}} &\leq \left( \sum\limits_{|\alpha| \leq m} \int_{\Omega} 2^p (|D^{\alpha} u|^p + |D^{\alpha} v|^p) dx \right)^{\frac{1}{p}}\\
&= 2 \left( \sum\limits_{|\alpha| \leq m} \int_{\Omega} (|D^{\alpha} u|^p + |D^{\alpha} v|^p) dx \right)^{\frac{1}{p}}
\end{align*}$
I think that I need use some inequality which should be well known, as this inequality that I used above, but I don't have idea what inequality is that.
Thanks in advance!
 A: Answer to the original question:
You're just being overcautious. For each multi-index $\alpha$, by Minkowski's inequality, we have
\begin{align}
\left(\int_{\Omega}|D^{\alpha} (u+v)|^p\textrm{d}x\right)^{1/p}&=\left(\int_{\Omega}|D^{\alpha} u+D^{\alpha}v|^p\textrm{d}x\right)^{1/p}\\
&\leq \left(\int_{\Omega}|D^{\alpha} u|^p\textrm{d}x\right)^{1/p}+\left(\int_{\Omega}|D^{\alpha} v|^p\textrm{d}x\right)^{1/p}
\end{align}
Use this for each term in the sum, and you're done.
In general, the sum of semi-norms is, again, a semi-norm.
Answer to the current question:
Now you have a composition of norms, but it isn't actually much of a problem:
It's probably easier to write
$$
||u||_{m,p}=\left(\sum_{|\alpha|\leq m} ||D^{\alpha}u||_p^p\right)^{1/p}
$$
Thus, applying Minkowski
$$
||u+v||_{m,p}=\left(\sum_{|\alpha|\leq m} ||D^{\alpha}u+D^{\alpha}v||_p^p\right)^{1/p}\leq \left(\sum_{|\alpha|\leq m}(||D^{\alpha}u||_p+||D^{\alpha}v||_p)^p\right)^{1/p}
$$
Now, for each $n,$ $(\sum_{j=1}^n |x_j|^p)^{1/p}$ defines a norm on $\mathbb{R}^n$ (it's $L^p$ of the counting measure on $\mathbb{R}^n$ if you will).
Thus, applying the triangle inequality of this norm, we arrive at
$$
||u+v||_{m,p}\leq \left(\sum_{|\alpha|\leq m}||D^{\alpha}u||_p^p\right)^{1/p}+ \left(\sum_{|\alpha|\leq m}||D^{\alpha}v||_p^p\right)^{1/p}=||u||_{m,p}+||v||_{m,p},
$$
which was what we wanted.
