# Norm on continous functions with compact support and distributions.

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 :)

• It would not be complete otherwise. Commented May 11, 2023 at 12:27

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.
• The topology of uniform convergence on $\Omega$ of all partial derivatives $\partial^\alpha f=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n} f$ would make differentiation of distributions well defined. However, this (metrizable) topology is not cpmplete which is a strong sign that it is not appropriate and it would not allow to consider all locally integrable functions as distributions. Commented May 11, 2023 at 7:00