Weak derivative and Locally summable functions I have three question regarding the appearance of the space of locally summable functions in the definition of weak derivatives and sobolev spaces.
The deifinition of weak derivatives from Evans:
Suppose $v, u \in L^1_{loc}(U)$, and $\alpha$ is a multiindex. We say that v is the $\alpha$:th weak partial derivative of $u$, written $D^{\alpha}u=v$, provided 
$\int_U uD^\alpha\phi dx =(-1)^{|\alpha|}\int_U v\phi dx $, 
for all test functions $\phi \in C^\infty_c(U)$
1) Why is it necessary to use the space $L^1_{loc}(U)$ and not for example $L^1(U)$?

In the definition of a Sobolev space $W^{k,p}(U)$ in Evans:
The Sobolev space $W^{k,p}(U)$ consists of all locally summable functions $u$: $U \rightarrow \mathbb{R} $  such that for each multiindex $\alpha$ with $|\alpha| \leq k, D^\alpha u $ exists in the weak sense and belongs to $L^p(U)$.
As I understand, locally summable means that $u \in L^1_{loc}(U)$. 
2) Does the usage of $L^1_{loc}(U)$ in this definition comes from the fact that it is used in the definition of the weak derivative above?

Last, the Sobolev norm $\|u\|_{W^{k,p}(U)}$ is defined as:
$\|u\|_{W^{k,p}(U)} = (\sum_{
|\alpha| \leq k}\int_U|D^\alpha u|^p dx)^{\frac{1}{p}}$,
3) But for a function $u \in W^{k,p}(U)$ how do we now that the term $\int_U|u|^p dx$ (that appears in the Sobolev norm when $\alpha = (0,0, ...,0)$) is $<\infty$ since all we know is that $u \in L^1_{loc}(U)$ and we dont have that $u \in L^p (U)$?
Thanks :)
 A: The answer to 1) is that in order to make sense of the expression $\int u \phi$ for $\phi$ a $C^\infty$ function with compact support, it is enough that $u \in L_\mathrm{loc}^1$ (you do not need that $u \in L^1$). 
What is the added value of allowing $u$ to be only in $L_\mathrm{loc}^1$ instead of asking that $u \in L^1$? Well, you want to be able to say that weak derivatives are a generalization of usual derivatives. But $C^1$ functions are not in general in $L^1(\mathbb R^n)$, though they are always in $L_\mathrm{loc}^1(\mathbb R^n)$. So if you want to talk about the weak derivative $u(x)=x^2$ on $\mathbb R$, and prove that it is equal to $v(x)=2x$, you need to work with $L_\mathrm{loc}^1(\mathbb R)$.  
A: I had a similar question while reading through Evans. I don't know the answer to all your questions, but I think I can respond to some of them. 
"As I understand, locally summable means that $u \in L^1_{loc}(U)$."
Yes, this is correct. One can verify by looking at the Wikipedia page for locally summable functions, for example. 
As for your question 2, I think this is true. It's a bit strange to me because most books I have seen do not have the locally summable part, but maybe it's a bit more general. 
As for your question 3, the full definition of a Sobolev space in the book is that the Sobolev space $W^{k, p}(U)$ consists of all locally summable functions $u: U \to \mathbb{R}$ s.t. for each multiindex $\alpha$ with $|\alpha| \leq k$, $D^\alpha u$ exists in the weak sense and belongs to $L^p(U)$. 
So taking $\alpha = 0$ in the definition is exactly what you want. 
