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Let $\mathcal{D}_K(\Omega)$ denote the smooth functions with compact support, $\mathrm{supp}(u) \subset K \subset \Omega \subset \mathbb{R}^n$ with $K$ compact. When studying this space, many texts provide a family of seminorms, given by: $$ \|\phi\|_{K,j} = \max_{|\alpha| \leq j}\max_{x\in K} \left|\frac{\partial^{\alpha}\phi}{\partial x^\alpha}\right| $$ My question is, why are these seminorms instead of proper norms? If we have $\|\phi\|_{K,j} = 0$, is it not then the case that for $\alpha$ with $|\alpha| = 0$ (i.e $\alpha = 0$ so that we may consider the function itself) that: $$ \max_{x\in K}\left|\frac{\partial^0\phi}{\partial x^0}\right| = \|\phi\|_\infty = 0 $$ which only happens if $\phi$ is identically $0$. Am I missing something here?

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    $\begingroup$ You are right, I think the term semi-norm is used since many times these semi norms define the smooth topology on $C^\infty (\mathbb{R})$, for which then they don't define strict norms. $\endgroup$ – pitariver Mar 11 at 19:15
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    $\begingroup$ If $\phi\in C^\infty_c([2,3])$ then $\|\phi\|_{[-1,1],j} = 0$ even if $\phi \ne 0$ $\endgroup$ – reuns Mar 11 at 20:51
  • $\begingroup$ @reuns: this is not a valid counterexample. The support in $K$ condition is built into the definition of the space. So if a function vanishes on $K$ it vanishes everywhere. $\endgroup$ – Abdelmalek Abdesselam Mar 12 at 10:14
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A norm is a particular case of seminorm. So calling the $||\cdot||_{K,j}$ seminorms is perhaps not optimal in precision but is not meant to deny that these are norms for the space $\mathcal{D}_{K}(\Omega)$ which they are. The reason I think one refers to them as seminorms is because in the general theory of (locally convex) topological vector spaces the basic input for defining the topology is a set of seminorms which may or may not be norms. This distinction comes to the forefront when discussing whether the space is Hausdorff separated or not. For this one needs the collection ofs eminorms as a whole to be able to separate points. Asking one of the seminorms to do this by itself is asking it to be a norm.

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