Conceptual question about the topology of the test function space $\mathcal{D}(\Omega)$

Question

I am studying distribution theory using Walter Rudin's Functional Analysis and although I seem to have overcome most of the technical details, there is an aspect I cannot fully understand. It is related to the topology of the test funcion space $\mathcal{D}(\Omega)$.

When equipping the space with a topology, Rudin insists on the importance of having the following condition satisfied:

If a sequence $(\phi_k)$ converges in $\mathcal{D}(\Omega)$, then there is a compact set $K \subset \Omega$ such that $\text{supp } \phi_k \subset K$ for all $k\in \mathbb{N}$

As far as I understand, it is the very condition Schwartz imposed when developing the theory; in fact, I have read that he did not even specify at topology at first, but only focused on assuring that bounded sets satisfied this condition.

In fact, Rudin starts with the topology generated by the following norms

$$ \lVert\phi \rVert_N = \{ \text{max}|D^\alpha \phi|: x\in \Omega, |\alpha|\leq N \} $$

but finds and easy counterexample that proves that Cauchy Sequences in this topology are not convergent. In case it is relevant, the example is the following: let $\Omega = \mathbb{R}$, let $\varphi$ be a test function whose support lies in $(0,1)$ and

$$ \phi_k = \sum_{i=0}^k \frac{\varphi(x-i)}{i+1} $$

He then argues that it is necessary to add more open sets (i.e., find a strictly finer topology) such that the above mentioned condition holds.

My question is: Is there a fundamental reason why this boundedness condition is needed? Why are we so interested in having a complete test function space (I understand that it is desirable to have a complete distribution space, but I cannot see why it is a requirement in the test function space)? In the following question,

Motivation for test function topologies

someone argues that this condition is necessary to verify that distribution derivative is a local operator, but it does not seem to make a lot of sense to me. In fact, this finer topology we work with renders the distribution space larger rather than smaller, so everything that works for the distribution space should work for the dual space that would result if we used this coarser topology.

Whether this is the answer or not, it is an answer of this type what I am looking for. Thanks in advance!

Answer 1

An answer that satisfies me is the following: if we do not impose bounded sets to lie inside some $\mathcal{D}_K$ (i.e., for any bounded set $B\subset \mathcal{D}(\Omega)$, there exists a $K\subset \Omega$ such that $\text{supp}\phi \subset K$ for all $\phi$), then, firstly, some functions in $L^1_{\text{loc}}(\Omega)$ would not be in $\mathcal{D}'(\Omega)$. This should raise a first concern on the quality of our extension of the function concept (rather than an extension, we would be suggesting an alternative).

To prove this claim, I will just give an example from which the general argument can follow. I will work in $\Omega = \mathbb{R}$ and with a particular set, but, once again, the argument can be made general easily. Consider the set $B = \{\phi_k\}_{k\in\mathbb{N}} = \{\varphi(t-k+1)\}_{k\in\mathbb{N}}$, where $\varphi$ is a $C_c^\infty(\Omega)$ function such that $\text{supp}\varphi \subset (0,1)$ and $\int_{\mathbb{R}}\varphi \neq 0$. Let now $\Lambda_f$ be the functional associated to the function $f = \sum_{i=1}^\infty i^2 \chi_{(i-1,i)}$, where $\chi_S$ represents the indicator function of the set $S$.

Then,

$$ \Lambda_f(\phi) = \int_{\mathbb{R}}f\phi $$

is not a continuous functional. Indeed, we know from the general topological vector spaces theory that

$$ \text{A linear functional is continuous} \implies \text{it is bounded}$$

and hence

$$ \text{A functional is unbounded} \implies \text{it is not continuous} $$

Let us now come back to the set $B$ I mentioned before. Our hypothesis is that $B$ is bounded, but then

$$ | \Lambda_f(\phi_k)| = |\int_{(k-1,k)}k^2\varphi(x-k+1)dx| = k^2 |\int_{\mathbb{R}}\varphi| $$

Hence, the set $\Lambda_f(B)$ is clearly not bounded, so $\Lambda_f$ cannot be in the dual space.

With little effort, an argument of this kind can be done general: without the boundedness condition,we can always find a function $L^1_{\text{loc}}(\Omega)$ such that $\Lambda_f$ is not in the dual space.

For a particular example, we can show that things can get even worse. If, for example, we use the toplogy that Rudin rejects, it can be shown that all $L^1$ functions are such that the associated distribution is continuous. But then, if we equip the dual space with the weak-$\star$ topology, it can be seen that the sequence of distributions associated to the functions $\sum_{i=1}^n i^2 \chi_{(i-1,i)}$ is a Cauchy sequence. Hence, as we have shown that the distribution associated to $\sum_{i=1}^\infty i^2 \chi_{(i-1,i)}$ would not be in the dual space, we can see that the distribution space would not be complete!