The operator $Sf=f(1)$ defined on $L^1[0,1]$ Consider the operator $T: L^1[0,1]\to \mathbb{R}$ defined for each $f\in L^1[0,1]$ by
$$Tf=\int_0^1 f(y)dy$$
Then it is clear that $T$ is a bounded (linear) operator from $L^1[0,1]$ to $\mathbb{R}$.
Now consider the operator $S$ defined by
$$Sf=f(1)$$
In some references such as in this book http://library.lol/main/CEFE6C167A0375440A8C17A2DFFA3D82
to define the operator $S$ on $L^1[0,1]$, they seem to consider it as an unbounded operator with domain the Sobolev space $W^{1,1}[0,1]$.
Why the choice of $W^{1,1}[0,1]$? Is it because in general $f(1)$ doesn't make sense for functions $f\in L^1[0,1]$. In this case, why does it make sense if we consider $f\in W^{1,1}[0,1]$?
 A: The elements of $L^1[0,1]$ are not functions, but equivalence classes of functions (up to a.s. equality). So, saying $f \in L^1[0,1]$ for some fuction $f$ is in fact abuse of notation. It is no problem, if you only perform operations, which do not depend on the representative of these equivalence classes. However, evaluation at a specific point in NOT independent of the class, hence expressions like $f(1)$ are not well-defined. (This is because $\{1\}$ is a Lebesgue nullset)
It can be shown that $W^{1,1}[0,1]$ is the space of absolutely continuous functions $[0,1] \to \mathbb{R}$. Or more exactly: the elements of $W^{1,1}[0,1]$ are still equivalence classes, but every class contains exactly one absolutely continuous function. So, one has a cannonical representative of each clas and one can define certain operations w.r.t. this representative. E.g. the expression $f(1)$ is then defined as evalueating the unique absolutely continuous representative at the point $1$.
Remark: Especially in more dimensions this connected to Sobolev embedding Theorems, e.g. $W^{d,1}(\mathbb{R}^d)$ can be embedded into $C(\mathbb{R}^d)$ and this embedding is continuous.
