Fdtxjr

It is a fundamental result of Sobolve space that

Let $Omega subset mathbbR^d$ or $mathbbR^d_+$, then
$C^infty_0(barOmega)$ 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?

Just like Chris said in his comment, let us suppose we have a bounded Lipschitz domain $Omegasubset mathbbR^d$, $C^infty_0(Omega)$ 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 $uin H^1(Omega)$, 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 $u_nsubset C^infty_0(Omega) subset H^1(Omega)$, such that $u_nto u$ 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
$$
mathscrI: H^1/2(partial Omega)longrightarrow H^1(Omega)
$$
serving like an extension of any function defined on boundary to the interior, and $T(mathscrIg) = g$ for any $gin H^1/2(partial Omega)$, prescribe any $gin H^1/2(partial Omega)$, say $g = 1$ on $partial Omega$, let $u = mathscrIg$.

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

Let $u = 1in H^1(Omega) = W^1,2(Omega)$, 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 $xin Omega$ we have:
$$
|u_n(x) - u_n(0)| = left| int^x_0 u'_n(t),dtright| leq
left| int^1_0 u'_n(t)^2,dtright|^frac12 ;
left| int^1_0 1^2,dtright|^frac12 = |u'_n|_L^2(Omega)
$$
This implies $displaystylesup_xinOmega|u_n(x)| leq |u'_n|_L^2(Omega)$, therefore combining everything we have would lead us to the following:
$$
1-epsilon < | u_n|_L^2(Omega) leq
left| int^1_0 sup_tinOmega|u_n(t)|^2,dtright|^frac12 = sup_xinOmega|u_n(x)| leq |u'_n|_L^2(Omega) < epsilon
$$
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.

Let $f(x) = cosh(x) = tfrace^x+e^-x2$ for $xin (a,b)$. Then obviously $fin C^infty(a,b)subset H^1(a,b)$ with $f'' = f$. Hence, for all $uin C_0^infty(a,b)$ we have
$$
(f,u)_H^1 = (f,u) + (f',u') = (f,u) + [f'overline u]_a^b - (f'',u) = (f-f'',u) = 0.
$$
Therefore $C_0^infty(a,b)$ is not dense in $H^1(a,b)$.