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!


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.

  • $\begingroup$ Sorry, I put the parenthesis in the wrong place. I edited my OP and put the definition of the norm right $\endgroup$ – George Oct 10 '19 at 19:46
  • $\begingroup$ Well, this can be handled very similarly. Check answer. $\endgroup$ – WoolierThanThou Oct 10 '19 at 20:09

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