# 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.

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.

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