Why is $C_c^\infty(\Omega)$ not a normed space?

Question

I am watching a Coursera video on Théorie des Distributions and I am trying to understand one of the slides.

Let $\Omega \subset \mathbb{R}^N$ be an open set and $C_K^\infty(\Omega) = \{ \phi \in C^\infty(\Omega) : \mathrm{supp}(\phi)\subset K \}$.

This space has has an infinite family of norms for each $p \in \mathbb{N}$, and compact $K \subset \Omega$ - the maximum of the partial derivative:

$$ ||\phi||_{p,K} = \max_{|\alpha|\leq p} \max_{x \in K} |\partial^\alpha \phi(x)|$$

Why is $C_c^\infty(\Omega)$ not a normed space, if we can compute a norm of a function in any compact subset?

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Here "compact" means closed and bounded since we are dealing with $\mathbb{R}^n$.

Answer 1

I believe that something more precise is meant here, which is the following:

$C_c^{\infty}(\Omega)$ cannot be given a norm $\|~\|$ that will, on its own, generate the same topology as the family of seminorms $\{\|~\|_{p, K}\}$.

The intuition here is that if you had a finite family of good enough seminorms, you could add them to get one norm that generates the same topology. But in this case, the family being infinite, one cannot just add them all.

Of course, this is not a proof; to actually prove that this is not possible would involve finding some property that all normed spaces satisfy but this one doesn't (with this topology). I can't think of one right now.

Answer 2

The key point is that $C^\infty_c(\mathbb R^n)$ violates the conclusion of the Baire Category Theorem, so cannot be a complete metric space. Namely, it is the countable union of $C^\infty_c(B_n)$ where $B_n$ is the open ball of radius $n$, as $n$ runs through positive integers. All these spaces are nowhere-dense, and the union is the whole.

(So, not only is the space of test functions not Banach, it is not Frechet, either. It is "LF", meaning "(direct) limit of Frechet".)

EDIT: following @AwfullyWeeBilly's comment consider normability without completeness, it suffices to prove that there is no local basis at 0 consisting of bounded sets in the TVS sense. That there is no such local basis follows from the fact that bounds on derivatives of order $\le k$ do not give bounds on higher derivatives.

(Still, I would also claim that we are by-far-most-interested in the complete or quasi-complete TVSs, to the extent that sometimes people say "normable" implicitly suggesting completeness... etc. That is, spaces of functions have natural topologies such that taking limits preserves some desired property.)

Answer 3

At first, you need to understand the topology (the way how sequences converges) on the space of test functions, i.e. what a distribution is: A linear map $T:C_c^\infty(\Omega)\to\mathbb{C}$ is said to be continuous, iff it is continuous on each of the spaces $C_K^\infty(\Omega)$ for $K\subset\subset\Omega$ compact. Unfortunately these are not normed spaces either (as there are still infinitely many norms $\{ ||.||_{p,K}|\,p\in\mathbb{N}\}$), but at least you may imagine the convergence here: A sequence $(\phi_n)_n$ converges in $C_K^\infty(\Omega)$ to $\phi$, iff $(||\phi_n-\phi||_{p,K})_n$ converges to $0$ for each $p\in\mathbb{N}$.

Putting all this together, you may say a linear map $T:C_c^\infty(\Omega)\to\mathbb{C}$ is a distribution, iff for any sequence of test functions $(\phi_n)_n$ such that

• there is a compact set $K\subset\subset\Omega$ with support$(\phi_n)\subseteq K$ for all $n$ and

• for this $K$ and for every $p\in\mathbb{N}$, $||\phi_n||_{p,K}\to 0$ as $n\to\infty$

you have $(T(\phi_n))_n \to 0$ in $\mathbb{C}$.

There is neither a norm nor a metric inducing this type of convergence on the space of test functions. You can find a simple proof of this here. However and even though this type of convergence is not very intuitive, it has proven to be useful for many applications of distributions.

Answer 4

Let me try to add one more answer using your language directly. It does not matter whether or not we can simply compute the norm of any function $\varphi \in C^{\infty}_c(\Omega)$ on any compact set $K$. What matters is this: Is there a single norm $\| \cdot \|$ such that $$ \|\varphi_l\|_{p.K} \to 0\ \text{for all}\ p,\ K \quad \Leftrightarrow \quad \|\varphi_l\| \to 0. $$ It can be shown that no such norm exists:

(There is however a metric $d$ that induces the topology in the sense that $$ \|\varphi_l - \varphi\|_{p.K} \to 0\ \text{for all}\ p,\ K \quad \Leftrightarrow \quad d(\varphi_l,\varphi) \to 0 $$ But any metric that does this is not homogeneous, i.e. it does not satisfy $$ d(\lambda \varphi_l,\lambda\varphi) = |\lambda|d(\varphi_l,\varphi). $$

Edit: Maybe this works:

Take a sequence of functions $\varphi_l \in C_c^{\infty}(\Omega)$ that, roughly speaking take the value 1 in a small strip $$ \Lambda_l := \{x \in \Omega\ :\ \frac{1}{2n} < \mathrm{dist}(x,\partial\Omega) < \frac{1}{n}\} $$ and are zero elsewhere. They are all compactly supported but clearly for any compact $K \Subset \Omega$ and any constants $c_l$ we eventually have $\mathrm{supp}\; c_l\varphi_l \subset \Omega\setminus K$ and hence for any $p \in \mathbb{N}$ we have $\|c_l\varphi_l\|_{p,K} = 0$ for sufficiently large $l$. Therefore $d(c_l\varphi_l,0) \to 0$. However if there is a norm then we can take $c_l$ large enough such that $\|c_l\varphi_l\| \geq 1$ for every $l$ so clearly this sequence does not tend to zero in the norm.

Answer 5

Is it not true that, for any fixed $p\in\mathbb{N}_0$, $\|f\|:=\sup\limits_{\text{compact }K\subseteq\Omega}\|f\|_{p,K}$ defines a norm $\|\cdot\|$ on $C^\infty_c(\Omega)$, making the latter a normed space? Do you mean under a stronger topology (form of convergence/open-ness), e.g., that a sequence converges in this topology if and only if all its $\|\cdot\|_{p,K}$-norms converge? In this case, it is not a normed space, and an explanation can be found here.