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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 $H^1$?

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Functions $[0,1]\to \mathbb{R}$ don't have to have bounded variation. And why do you consider then $(0,1)$ rather than $[0,1]$? –  Ilya May 21 '11 at 11:41
    
I consider $(0, 1)$ because Sobolev spaces are (as far as I know) defined over open sets. –  Giuseppe Negro May 21 '11 at 11:57
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Sorry, I don't understand your question: An element of $H^1$ has, by definition, a generalized derivative. This is, by definition, an additional property that elements of $H^1$ need to have, and elements of $L^2$ need not to have. –  Tim van Beek May 22 '11 at 8:39
    
@Tim: It is my fault: my question is not well-posed. The fact is that I'm trying to understand the intuitive meaning of Sobolev norms, and to do so I chose a phenomenon which I thought might be enlightening. Not so enlightening, on second thought! :-) –  Giuseppe Negro May 22 '11 at 9:29
    
Sobolev's norm deals with distances between all possible derivatives (and also the degree of the function in integration depends on the type of the Sobolev's space). $L_2$ norm is a norm in a Sobolev's $W^{0,2}$ space as far as I understand. –  Ilya May 22 '11 at 21:19

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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).

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Great point, thank you! –  Giuseppe Negro Sep 9 '11 at 16:19

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