I have this problem:
Let $\Omega\subset\mathbb{R}^2$ be a bounded open set. Let $u\in\mathbb{W}_2^{(5)}(\Omega)$
$$ \| u \|_1 = \sum_{|\alpha| \leq 5} \| \mathcal{D}^\alpha u \|_{L_2(\Omega)} $$
and
$$ \| u \|_2 = \left( \sum_{|\alpha| \leq 5} \int_\Omega | \mathcal{D}^\alpha(x)|^2 dx\right)^\frac{1}{2} $$
As you can see $\|u\|_1$ is a norm from the sobolev space. I'm supposed to show equivalence between these 2.
To prove that I need to show that there $\exists a,b \in \mathbb{R}$ such that $a\|u\|_2 \leq \|u\|_1 \leq b\|u\|_2$
I managed to show the first inequality but I can't figure out the second one. I found a similar question asked in an even more general case Norm equivalence Sobolev space but I can't figure out how to use the advice given there.
This really is just a trivial consequence of equivalence of norms in finite dimensions. In general, say we're considering $W^{k,p}(\Omega)$, where $\Omega\subset\Bbb{R}^n$. Define $N$ to be size of the set of multindices $\{\alpha\,:\, |\alpha|\leq k\}$; this is a positive integer depending on $k$ and the dimension $n$.
For each $u\in W^{k,p}(\Omega)$, let us associate the vector $\tilde{u}\in\Bbb{R}^N$, defined as $\tilde{u}=\left(\|D^{\alpha}u\|_{L^p(\Omega)}\right)_{|\alpha|\leq k}$. In words, given the Sobolev function $u$, we look at all the weak derivatives $D^{\alpha}u$, take their $L^p$ norm, and arrange these numbers into a tuple in $\Bbb{R}^N$ (the order is irrelevant). Now, the space $\Bbb{R}^N$ is finite-dimensional, so all the norms on it are equivalent. We can now define the Sobolev norm of the function $u$ to be any norm of the associated tuple $\tilde{u}\in\Bbb{R}^N$. So, for example, on the space $\Bbb{R}^N$, you can consider the $\ell^q$ norm: \begin{align} \|u\|_{W^{k,p}(\Omega);\,\ell^q}:=\|\tilde{u}\|_{\ell^q(\Bbb{R}^N)}= \begin{cases} \left(\sum_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^p(\Omega)}^q\right)^{1/q}&\text{if $q\in [1,\infty)$}\\ \max\limits_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^p(\Omega)}&\text{if $q=\infty$.} \end{cases} \end{align} Your first example used $(n,k,p,q)=(2,5,2,1)$ while your second example used $(n,k,p,q)=(2,5,2,2)$.
Now, we can go further and say that there's nothing special about the $\ell^q$ norms on $\Bbb{R}^N$, and that we could simply consider an arbitrary norm $\rho$ on $\Bbb{R}^N$, and simply define \begin{align} \|u\|_{W^{k,p}(\Omega);\,\,\rho}&:=\rho(\tilde{u}). \end{align} Since $\Bbb{R}^N$ is finite-dimensional, any two norms $\rho_1,\rho_2$ on $\Bbb{R}^N$ are equivalent, and consequently, the associated norms $\|\cdot\|_{W^{k,p}(\Omega);\, \rho_1}$ and $\|\cdot\|_{W^{k,p}(\Omega);\, \rho_2}$, first of all, really are norms on $W^{k,p}(\Omega)$, and secondly, they are equivalent (in fact, if $c\rho_1\leq \rho_2\leq C\rho_1$, then $c\|\cdot\|_{W^{k,p}(\Omega);\, \rho_1}\leq \|\cdot\|_{W^{k,p}(\Omega);\, \rho_2}\leq C\|\cdot\|_{W^{k,p}(\Omega);\, \rho_1}$).
If the above is still too abstract, then you should start off by proving that the $\ell^1$ and $\ell^2$ norms on $\Bbb{R}^N$ are equivalent (this is a very elementary exercise, for which you don't need any abstract results (and the proof idea is already there in the comments)).