I want to check if the Function
$u(x)=|x-1|$
is in $H^{2}(\Omega) , \Omega = (0,2) $
My idea: I want to check if the function has a weak derivative. But that would only proof $L^{1}$ what should I do next? Checking if there is a weak derivative for the weak derivate? Then I would have at least proven $L^{2}$. How do I proof that the set is in $H^{2}$ now?
I think you're a little confused about what all this is saying. Or at least, I'm confused reading your question. Let's unravel everything and make sure we're on the same page. $H^2(\Omega)$ is the function space $W^{2,2}(\Omega)$ consisting of all functions on $\Omega$ with all weak-partials of order $l \leq 2$ being $L^2$ functions. Since $\Omega$ is a region in $\mathbb{R}$, we just drop the word 'partial.' So it has to have two weak derivatives, and they both need to be $L^2$ functions. There's two steps then clearly, one is to calculate the weak derivatives, and the other is to observe they are in $L^2(\Omega)$.
For both steps, I see no reason why we should use $\Omega$ as you've been given it, rather than translate it so that things are at the origin, just to make the algebra easier. So let's move everything so that $f(x) = |x|$ and $\Omega = (-1,1)$. From here, I provide a sketch of the argument, leaving the details to you - there is really no good way to get a grip on Sobolev spaces other than doing some of these calculations.
I claim a weak-derivative of this function is the Heaviside step function $H(x)$. I leave the proof to you - this should be not so hard since the definition of the weak derivative is basically just integration by parts against test functions, and on each ray, the derivative is $\pm 1$, only being undefined at $0$. Is this function $L^2$? Yes. What's $|H(x)|$? Why is it clearly $L^p$?
What is the second weak derivative? Well it's a weak derivative of $H(x)$. If this function exists, again, it satisfies an integration by parts formula, by definition. But this one is a bit more problematic. One way to prove that such a function does not exist is to take a sequence of test functions, with $\varphi_1(0) = 1$, and then define $\varphi_n(x) = \varphi_1(nx)$. Then since all of these functions have their support contained in $\Omega$, and we are compressing them by a factor of $n$, they go to $0$ in the $L^2$ norm. On the other hand, if you write $\int_\Omega f(x) \varphi_n(x)$ where $f$ denotes a candidate for weak derivative, then can you evaluate this for each $n$? Take the limit, and obtain a contradiction.