Let $\Omega \subset \mathbb{R}^n$ be open, then $H_0^m(\Omega) := \overline{C^{\infty}_c(\Omega)}$ with respect to the Sobolev norm $\|\|_{H^m(\Omega)}$. The problem is that I don't really see what kind of functions are in this space. Apparently for $m=0$ we just have $H_0^0 (\Omega)=L^2(\Omega).$

Despite, from the definition I don't see a defining property for all the functions in $H_0^m$?- I mean it is intuitively clear to me what $H^m$ is (weakly differentiable functions that are square integrable in every derivative) , but this closure definition is pretty unclear to me in the sense that I don't understand what these functions have in common.

What is so special about this space?

In particular, what is the difference between $H^m(\Omega)$ and $H_0^m(\Omega)$? It looks to me as if they are quite often the same, as $C_c^{\infty}$ should be dense in $H^m$

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    $\begingroup$ $C_c^\infty$ is not dense in $H^m(\Omega)$ unless $\Omega$ is all or almost all of $\mathbb{R}^n $. $\endgroup$ – user147263 Feb 25 '15 at 20:26

It may help to observe that in general, $C_c^\infty(\Omega)$ is not dense in $H^m(\Omega)$, so the closure $\overline{C_c^\infty(\Omega)}$ is a proper subspace, and this definition actually says something nontrivial.

Take for example $n=1$ and $\Omega = (0,1)$, the open unit interval. Observe that the constant function 2 is in $H^1(\Omega)$. However, for any $f \in C_c^\infty(\Omega)$, $\|2-f\|_{H^1} \ge 1$.

Consider two cases. If $f \le 1 $ everywhere, then clearly $\|(2-f)\|_{L^2} \ge 1$. Otherwise, suppose there exists $x$ such that $f(x) \ge 1$. Since $f$ is compactly supported inside $(0,1)$, we have $f(0)=0$, and so $$\begin{align*} 1 \le |f(x)| &= \left|\int_0^x f'(t)\,dt\right| && \text{FTC} \\ &\le \int_0^x |f'(t)|\,dt && \text{triangle inequality} \\ &\le \int_0^1 |f'(t)|\,dt && \text{integrating over a larger set}\\ &\le \left(\int_0^1 |f'(t)|^2\,dt\right)^{1/2} && \text{Jensen or Cauchy--Schwarz} \end{align*}$$ and in particular $\|(2-f)'\|_{L^2} \ge 1$. In either case we have $\|2-f\|_{H^1} \ge 1$. So $2 \notin H^1_0(\Omega)$.

By a similar argument, if you know that $H^1((0,1))$ consists (up to almost everywhere equivalence) of the absolutely continuous functions on $(0,1)$ whose derivatives are in $L^2$, then you can show that $H^1_0((0,1))$ consists precisely of those $g \in H^1((0,1))$ satisfying $\lim_{x \to 0^+} g(x) = \lim_{x \to 1^-} g(x) = 0$. In other words, all those $g$ satisfying Dirichlet boundary conditions.

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  • $\begingroup$ can we generalize this Dirichlet boundary condition property or is this a special case in 1-dimension? $\endgroup$ – Jacobian Feb 25 '15 at 20:50
  • $\begingroup$ @Jacobian: You can generalize it, but it gets more complicated since in higher dimensions, $H^m$ functions do not generally have continuous representatives. The appropriate generalization is the trace theorem mentioned by Ian: if $T : H^1(\Omega) \to L^2(\partial \Omega)$ is the operator he describes, you are supposed to think of $Tf$ intuitively as "take the continuous extension of $f$ to $\overline{\Omega}$ and restrict to $\partial \Omega$". (For some $f$ there may not actually be such a continuous extension so you have to interpret this in a sort of "almost everywhere" sense.) $\endgroup$ – Nate Eldredge Feb 25 '15 at 20:55
  • $\begingroup$ @Jacobian: Then $H^1_0(\Omega)$ is those functions whose trace is 0; i.e. intuitively, those which vanish on the boundary $\partial \Omega$. $\endgroup$ – Nate Eldredge Feb 25 '15 at 20:56
  • $\begingroup$ @Jacobian: But don't take this too seriously; in some cases the boundary can be "too small" to impose Dirichlet boundary conditions. For instance, if $\Omega = \mathbb{R}^3 \setminus \{0\}$ is $\mathbb{R}^3$ minus the origin, you can check that in fact $H^1_0(\Omega) = H^1(\Omega) = H^1(\mathbb{R}^3)$, and in particular contains the constants. This has to do with the fact that points in $\mathbb{R}^3$ have zero capacity. $\endgroup$ – Nate Eldredge Feb 25 '15 at 20:59

First of all, $C^\infty_c$ is not dense in all of $W^{k,p}$ ($H^k=W^{k,2}$) in general. For instance, take a bounded domain. Then a constant function is in $W^{1,p}$. But you typically can't approximate it in $W^{1,p}$ by compactly supported functions. The function values will converge easily enough, but the constant function has a zero gradient, while the gradients of the compactly supported functions are bounded away from zero, in the sense of $L^p$. Since constant functions are smooth, we've actually shown that $C^\infty_c$ isn't even dense in $C^\infty$, if we equip both with the $W^{1,p}$ metric.

I think to properly understand this phenomenon, you need the trace theorem. For functions which are smooth up to the boundary, we have a well-defined linear operator which restricts them to the boundary. The trace theorem says that this operator has an extension to all of $W^{1,p}$, and that the extension is continuous as a function from $ W^{1, p}(\Omega) $ into $ L^p (\partial \Omega)$. This operator is called the trace, and commonly denoted by $T$. (Note that the trace theorem requires a certain amount of boundary regularity.)

Where the trace theorem applies, $ W^{1, p}_0$ consists of those Sobolev functions whose trace is zero. (This is easy to prove, because the restriction of a compactly supported function to the boundary is zero, and then you use continuity to conclude.)

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  • $\begingroup$ so it has nothing to do with Dirichlet boundary conditions, let's say? $\endgroup$ – Jacobian Feb 25 '15 at 20:28
  • $\begingroup$ @Jacobian Yes it does. A smooth function in $W^{1,p}_0$ satisfies a homogeneous Dirichlet boundary condition. A nonsmooth function in $W^{1,p}_0$ satisfies the appropriate generalization of this idea (since it has no pointwise restriction). $\endgroup$ – Ian Feb 25 '15 at 21:27

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