Reading through Chapter 8 of Brezis. We see that $C_{c}^{\infty}(\mathbb{R})$ is dense in $W^{1,\,p}(I)$. Later on in the chapter we define $W^{1,\,p}_{0}(I)$ to be the closure of $C_{c}^{1}(I)$ in $W^{1,\,p}(I)$. This $W^{1,\,p}_{0}(I)$ space has really confused me.

  1. What happens when look at $C_{c}^{\infty}(I)$ which causes us to lose density in $W^{1,\,p}(I)$?

  2. This question is linked to the first question. Suppose $(u_{n})\in C_{c}^{1}(I)$, then $(u_{n}')\in C_{c}(I)$. Both $C_{c}^{1}(I)$ and $C_{c}(I)$ are dense in $L^{p}(I)$, so $u_{n}\rightarrow u$ and $u_{n}'\rightarrow g$ in $L^{p}(I)$. Hence why cannot we not say that $u_{n}\rightarrow u$ in $W^{1,\,p}$ by taking $u'=g$?

  3. By definition $\overline{C_{c}^{1}(I)}=W^{1,\,p}_{0}(I)$ in the $W^{1,\,p}$ norm. So how does one show density of $C_{c}^{\infty}(I)$ in $W^{1,\,p}_{0}(I)$? Is it sufficient to show that $C_{c}^{\infty}(I)$ is dense in $C_{c}^{1}(I)$ with respect to the supremum norm?


It's not true that $\mathcal C_c^\infty (I)$ is dense in $W^{1,p}(I)$ when $I\neq \mathbb R$. What is true is $\mathcal C_c^\infty (\mathbb R)$ is dense in $W^{1,p}(\mathbb R)$. And indeed, if $I\neq \mathbb R$ is an interval, we define $W_0^{1,p}(I)$ as the closure of $\mathcal C_c^\infty (I)$ in $W^{1,p}(I)$.

  • $\begingroup$ Yes I know this. I am asking why is $C_{c}^{\infty}(I)$ not dense in $W^{1,\,p}(I)$? What do we lose when we go from $C_{c}^{\infty}(\mathbb{R})$ to $C_{c}^{\infty}(I)$ that makes us sacrifice density? $\endgroup$ – Zeta-Squared Jun 5 at 7:40
  • $\begingroup$ Just take any sequence in $\mathcal C_c^\infty (0,1)$ that converges to $f(x)=1\in W^{1,p}(0,1)$. The gradient will explose. (with a draw it's very easy to see). $\endgroup$ – Surb Jun 5 at 7:49
  • $\begingroup$ I see what you are saying. But $C_{c}^{\infty}((0,1))$ is dense in $L^{p}((0,1))$. So if $f_{n}\rightarrow f$ in $L^{p}((0,1))$ and likewise since $(f_{n}')\in C_{c}((0,1))$ then $f_{n}'\rightarrow f'$ in $L^{p}((0,1))$ wont we have that $f_{n}\rightarrow f$ in the $W^{1,\,p}$ norm? $\endgroup$ – Zeta-Squared Jun 5 at 7:59
  • $\begingroup$ The thing is in $W^{1,p}(0,1)$ you cannot define a function on the boundary as you want. In $W^{1,p}(0,1)$, the function $f(x)=1$ must have $f(0)=f(1)=1$. You'll find more information with the trace operator $\endgroup$ – Surb Jun 5 at 8:04
  • $\begingroup$ I agree. I am clearly not understand something properly here. I think I may have identified what I am mistaken about. Please clarify for me. We can show that a function $f$ belongs to $L^{p}$ simply by showing $\|f\|_{p}<\infty$. However, I have been thinking that it is sufficient to show $f$ belongs to $W^{1,\,p}$ if $\|f\|_{W^{1,\,p}}<\infty$. This is incorrect, that is, $\|f\|_{W^{1,\,p}}\nRightarrow f\in W^{1,\,p}$ $\endgroup$ – Zeta-Squared Jun 5 at 8:09

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