# $C^\infty_{0}(\bar\Omega)$ dense in $W^{k,p}(\Omega)$: the closure is necessary?

It is a fundamental result of Sobolve space that

Let $\Omega = \mathbb{R}^d$ or $\mathbb{R}^d_+$, then $C^\infty_{0}(\bar\Omega)$ is dense in $W^{k,p}(\Omega)$.

However, in some literatures, the fact is pointed out that $C^\infty_{0}(\Omega)$ fails to dense in $W^{k,p}(\Omega)$ in some cases.

Some I would like to try some counterexamples.

My guess is take $d = 1, k = 1, p = 2$ and $\Omega = (0,1)$, i.e. claim that

$C^\infty_{0}((0,1))$ fails to dense in $W^{1,2}((0,1))$.

But how should I proceed such an argument in a precise way?

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The basic issue can be seen with constant functions, where the derivative is zero but the derivative of $C_0^\infty$ functions that approximate constant functions are highly singular near the boundary. –  Chris Janjigian May 5 '12 at 3:26

Just like Chris said in his comment, let us suppose we have a bounded Lipschitz domain $\Omega\subset \mathbb{R}^d$, $C^{\infty}_0(\Omega)$ fails to approximate an arbitrary $W^{k,p}(\Omega)$-function unless you set the boundary value to be 0.

A possible argument could be constructed using the trace theorem. Roughly speaking, denote $H^1(\Omega) := W^{1,2}(\Omega)$, let $$T: H^1(\Omega) \longrightarrow H^{1/2}(\partial \Omega)$$ be the trace operator. Suppose we prescribe a $u\in H^{1}(\Omega)$, which has a non-zero trace, leading us to the trace inequality $$\|Tu \|_{H^{1/2}(\partial \Omega)} \leq c\| u \|_{H^1(\Omega)}$$ Now assume we have a sequence $\{u_n\}\subset C^{\infty}_0(\Omega) \subset H^1(\Omega)$, such that $u_n\to u$ in the $H^1(\Omega)$-norm, then use trace inequality we will see the contradiction in which zero is greater than or equal to a positive number.

If you are not satisfying with the existence of the "non-zero trace" part, we could circumvent this by using the surjectivity of the trace operator, define its right inverse as $$\mathscr{I}: H^{1/2}(\partial \Omega)\longrightarrow H^1(\Omega)$$ serving like an extension of any function defined on boundary to the interior, and $T(\mathscr{I}g) = g$ for any $g\in H^{1/2}(\partial \Omega)$, prescribe any $g\in H^{1/2}(\partial \Omega)$, say $g = 1$ on $\partial \Omega$, let $u = \mathscr{I}g$.

Counter-example on $\Omega = (0,1)$:

Let $u = 1\in H^1(\Omega) = W^{1,2}(\Omega)$, suppose we have a sequence $\{u_n\} \subset C^{\infty}_0(\Omega)$ such that $$\| u - u_n\|^2_{H^1(\Omega)} = \| u - u_n\|^2_{L^2(\Omega)} + \| u'_n\|^2_{L^2(\Omega)} \longrightarrow 0$$ so for any $\epsilon>0$ we could find $N>0$ for all $n>N$: $$\| u - u_n\|_{L^2(\Omega)} < \epsilon, \text{ and }\; \|u'_n\|_{L^2(\Omega)} < \epsilon$$ By triangle inequality: $$\left|\| u\|_{L^2(\Omega)} - \| u_n\|_{L^2(\Omega)}\right| \leq \| u - u_n\|_{L^2(\Omega)} < \epsilon$$ due to $\| u\|_{L^2(\Omega)} = 1$, above implies $$1-\epsilon < \| u_n\|_{L^2(\Omega)} < 1+\epsilon$$ Now we would like to argue like the proof for Poincaré inequality to reach the contradiction, for 1 dimensional case it is very straightforward, for any $x\in \Omega$ we have: $$|u_n(x) - u_n(0)| = \left| \int^x_0 u'_n(t)\,dt\right| \leq \left| \int^1_0 u'_n(t)^2\,dt\right|^{\frac{1}{2}} \; \left| \int^1_0 1^2\,dt\right|^{\frac{1}{2}} = \|u'_n\|_{L^2(\Omega)}$$ This implies $\displaystyle\sup_{x\in\Omega}|u_n(x)| \leq \|u'_n\|_{L^2(\Omega)}$, therefore combining everything we have would lead us to the following: $$1-\epsilon < \| u_n\|_{L^2(\Omega)} \leq \left| \int^1_0 \sup_{t\in\Omega}|u_n(t)|^2\,dt\right|^{\frac{1}{2}} = \sup_{x\in\Omega}|u_n(x)| \leq \|u'_n\|_{L^2(\Omega)} < \epsilon$$ which is a contradiction, hence such sequence $\{u_n\} \subset C^{\infty}_0(\Omega)$ does not exist.

A last remark: Why would the density argument of $C^{\infty}_0(\Omega)$ is true for the whole space is because of the decaying property of both the function itself and the derivative of the $W^{k,p}$-functions, for something to be $L^p$-integrable on the whole space, its integration must be small outside a ball of certain size, however for bounded domain, decaying property does not hold any more unless you add the compactly-supported condition.

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thanks for your reply. I'm wondering if it is possible to argue the specific case mentioned above($C^\infty_0((0,1))$) without trace? Preferably, a counter-example would be selfstanding. –  newbie May 7 '12 at 5:10
@newbie Edited the counter-example into my answer. –  Shuhao Cao May 7 '12 at 6:32
It is crystal clear. Thanks a lot! –  newbie May 7 '12 at 6:38
Hey, sorry I got one more question. In the estimation of sup, is $u^n(0) = 0$ from definition of $C^\infty_0((0,1))$? –  newbie May 7 '12 at 6:57
@newbie Yes, $C^{\infty}_0((0,1))$ normally denotes the compactly supported infinitely differentiable functions on $(0,1)$, the compact supportedness means it is zero on some small neighborhood of the interval's end points. –  Shuhao Cao May 7 '12 at 7:05