Intuition behind Sobolev norm This morning I was thinking at the following (simple) fact. Let us consider $[0, 1] \to \mathbb{R}$ functions and define a linear functional 
$$F(u)=u(1)-u(0).$$
$F$ is not continuous on $L^2(0, 1)$ (in fact, it is not even defined everywhere), but it is continuous on $H^1(0, 1)$: 
$$\lvert F(u) \rvert \le \int_0^1\lvert u'(x)\rvert\, dx\le \lVert u \rVert_{H^1}.$$
How would you give an intuitive explanation of this phenomenon? In what sense is Sobolev norm more restricting, so that a "bad" $L^2$ functional turns out to be a "good" one on $H^1$?
 A: I would not say that $F(u)$ is not even defined everywhere, but more precisely that is not defined at all on $L^2(0,1)$: how would you define the value a function in $L^2(0,1)$ takes in a single point (which has measure zero?).
On the other side, for functions in $H^{1,2}(0,1)$, there is a precise way to define what is their "value" on the boundary of $(0,1)$: in fact this result hold more in general for a "well-behaving" open set $\Omega$ and functions in $H^{1,2}(\Omega)$. For these functions, we can associate a function in $L^2(\partial \Omega)$, which is a continuous extension of the "restriction to $\partial \Omega$ operator" of $C^1(\overline\Omega)$ functions. This is called the trace operator.
The fact that you can define the trace operator for $H^1$ and not for $L^2$ is one of the reasons Sobolev spaces are useful for solving PDEs, and can be seen as one of the motivations for introducing them (at least, this can be one point of view for you, if you are trying to understand the rationale beyond Sobolev spaces).
