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Let $I\subseteq \mathbb{R}^n$ open and let $1\le p\le \infty$. Denote with $C_0^\infty(I)$ the set of functions $\varphi\in C^\infty(I)$ with compact support in $I$.

We denote with $W^{1,p}(I)$ the set of functions $u:I\to\mathbb{R}$ with $u\in L^p(I)$ whose weak partial derivatives $\frac{\partial u}{\partial x_i}$ are in $L^p(I)$ for every $i=1,..,n$. We endow this space with the following norm:

$\|u\|_{W^{1,p}}=(\|u\|_{L^p}^p+\|\nabla u\|_{L^p}^p)^{\frac{1}{p}}$, if $1\le p<\infty$ and

$\|u\|_{W^{1,\infty}}=\max\{\|u\|_{L^\infty},\|\nabla u\|_{L^\infty}\}$ if $p=\infty$.

Now, if $1\le p<\infty$, we define $W_0^{1,p}(I)$ to be the $W^{1,p}$-norm closure of $C_0^\infty(I)$.

For $p=\infty$, we set $W_0^{1,\infty}(I)=W^{1,\infty}(I)\cap W^{1,1}(I)$.

Question: Why don't we set $W_0^{1,\infty}(I)$ as the $W^{1,\infty}$-norm closure of $C_0^\infty(I)$, analogously as for $1\le p<\infty$, i.e. why is $W_0^{1,\infty}(I)=W^{1,\infty}(I)\cap W^{1,1}(I)$ the better definition?

Regards

Edit additional question: The above question is already answered now, but now I have a further question related to the answer below: How to see that the $W^{1,\infty}$-norm closure of $C_0^\infty(I)$ doesn't contain functions which are only weakly differentiable, and which are not in $C^1$?

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The $W^{1,\infty}$-closure of $C_0^\infty$ will only contain $C^1$-functions, which does not contain functions with merely weak derivatives:

The convergence in $W^{1,\infty}$-norm is equivalent to convergence in $C^1$-norm for $C^1$-functions: one can replace the essential supremum by the maximum.

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  • $\begingroup$ thank you. And how to see that only $C^1$-functions are in the closure, and not in general those which are only weakly differentiable? $\endgroup$ – Ryan Feb 13 '17 at 7:57

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