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I'm currently working with this function: $$u(x)=\max(1-|x|,0)$$ I am asked to find $u', u''$ in the sense of distributions, and whether $u,u'$ belong to the Sobolev space $W^1(R)$

I found the following results to the first question: $$u'(x)=\chi_{[-1,0]}-\chi_{[0,1]}$$ $$u''(x)=-2\delta_0+\delta_{-1}+\delta_{1}$$ Using my book's definition of a $W^k(R)$ Sobolev space, which summarily defines it as the "space of functions whose derivatives of $n\le k $ order belong to $L^2(R)$", I'm trying to use it to answer the second question, but I'm not sure it's the right way (or the smarter). For $u'$ we thus have: $$\int_{-1}^0|1|^2dx-\int_0^1|1|^2dx=0<\infty$$ which means $u'\in L^2(R)\rightarrow u\in W^1(R)$

Does this make sense? Because if it does, I'm not sure how to implement it with $u''$. If it doesn't, well... I've got bigger problems. Thanks!

EDIT: hold on, I'm dumb. If the order is $n \le 1$ clearly $u'$ can't belong to $W^1(R)$ since its derivative is second order. I guess I'm left wondering if this reasoning makes sense, then. Or, does the k in $W^k(R)$ "start counting" from the considered function? i.e. the first derivative of $u'$ (of which I'm trying to determine the belonging to $W^1(R)$) is in fact $u''$ and that's what I have to test?

Silence
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You computed correctly that $u' \in L^2$. You still need to show $u \in L^2$ and then you have $u \in W^1=\{ v \in L^2 ~|~ v' \in L^2\}$.

Now, as you said, for $u' \in W^1$ you have to show $u' \in L^2$ and $u'' \in L^2$. You have shown that the first statement is true. Now, it is quite standard to show that the Dirac distribution does not belong to $L^2$, see here, and therefore, $u' \notin W^1$. We can also say $u \notin W^2$.

Cahn
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    Great, that's exactly what I needed. Thank you very much! – Silence Jan 23 '20 at 12:30
  • Actually one last question if you will: when I want to prove that $u' \in W^1$, you said I still need to prove that $u'$ itself belongs to $L^2$, which I didn't do for $u$. So I assume that considering $v=u'$ and verifying that $v'=u'' \in L^2$ isn't what I'm doing, but rather considering that even if I'm now working with $u'$, I am still referring to the base function $u$? Not sure if I managed to explain myself properly. I.e. when I want to determine whether a derivative belongs to a Sobolev space of the same order, do I treat it as a new base function or does it remain a derivative? – Silence Jan 23 '20 at 12:37
  • Yes, for $u \in W^1$ you have to show $u \in L^2$ and $u' \in L^2$. So you still have to prove $u \in L^2$. In other words $W^1(R)={u:R \to R | u \in L^2(R) \text{ and } u' \in L^2(R)}$. – Cahn Jan 23 '20 at 18:03