My question is about the following function space:
Let $\Omega \subset \mathbb{R}^n$ be an open set. Now
$C_0^\infty(\Omega)$ denotes the set of $C^\infty(\Omega)$ functions with compact support in $\Omega$.
One could define a norm on this function space. For example:
$||u|| = sup_{x \in \Omega}|u(x)|$ for $u \in C_0^\infty(\Omega)$
But i think there are many other options to define a norm on $C_0^\infty(\Omega)$.
Is there any (obvious) reason that if we look at distributions
$T: C_0^\infty(\Omega) \rightarrow \mathbb{R}$ with $T$ linear and continous
we take $C_0^\infty(\Omega)$ as a topological vector space $\mathfrak D (\Omega) = (C_0^\infty(\Omega),\tau)$ and not as a normed space (for example with the norm above).
Thank you for your help 🙂
Best Answer
The reason is that we want the derivative of a distribution to be a distribution. An important motivation for this requirement is the possibility to define and manipulate weak solutions of partial differential equations.
Recall that, for $i \in \{ 1, \dotsc, n \}$, the derivative $\partial_i T$ of a distribution $T$ is defined as $$ \langle \partial_i T, \varphi \rangle = - \langle T, \partial_i \varphi \rangle. $$ If, as you suggest, you use consider $C^\infty_0(\Omega)$ as a normed vector spaced equipped with the $\|\cdot\|_{C^0}$ norm, then $\partial_i T$ is not continuous. (You can construct counter-examples quite easily). Also, note that using any $C^k$ norm does not solve the problem either.
This is what leads to the definition of $\mathfrak{D}(\Omega)$ using an inductive limit.
You can also read this question for a very similar question with a slightly different viewpoint.