A topology on $D(Ω)$ given by the seminorms $ρ_N (φ) := \max \{|∂^\alpha φ(x)| : x ∈ Ω\}$ is not complete.

Question

Show that the topology on $D(Ω)$ given by the seminorms $ρ_N (φ) := \sup\{ |∂^ αφ(x)| : x ∈ Ω, |\alpha|\leq N\}$ is not complete for any nonempty open set $Ω ⊆ \mathbb R^d$. Where $\Omega\subseteq\mathbb{R}^d$ is open and non-empty and $D(\Omega)$ is the space of all smooth functions whose support lies in $\Omega$.

In general, how does one usually approach questions asking to prove a topology is incomplete? I am only familiar with questions of incompleteness in a metric space with a given metric.

Answer 1

A collection of semi-norms induce a uniform structure on the space. For uniform spaces you can define the notion of completeness via convergence of Cauchy filters. Specifically on a uniform space you define the notion of a Cauchy filter:

Let $\mathcal F$ be a filter on a uniform space. If for every entourage $U$ there exists $A \in \mathcal F$ so that $A\times A \subset U$, then $\mathcal F$ is called Cauchy.

A uniform space is called complete if every Cauchy filter converges.

In the case that the topology and the uniform structure are generated by a family of semi-norms $\{|\cdot|_\alpha\}_{\alpha \in I}$ (which induce pseudo-metrics $d_\alpha(\cdot,\cdot)$), the condition of a filter being Cauchy becomes:

Let $\mathcal F$ be a filter. If for every $\epsilon >0$, $\alpha \in I$, there exists an $A \in \mathcal F$ so that $A \times A \subset d_\alpha^{-1}([0,\epsilon])$, then $\mathcal F$ is Cauchy.

In the case that $\mathcal F$ is the elementary filter of a sequence $(x_n)_{n \in \mathbb N}$, the statement that $\mathcal F$ is Cauchy is equivalent to the sequence $x_n$ being Cauchy wrt every pseudo-metric $d_\alpha$.

So to show that the space is not complete we just give an example of a Cauchy-sequence that does not converge.

Let $f$ be some smooth function with support $[0,1]$. Then define $f_n(x):=f(x\cdot(1-1/n))$. $f_n$ has support $[0,1-1/n]$ and thus lies in $D(-1,1)$ for all $n$. But it converges uniformly to $f$ and all derivatives converge uniformly to the derivatives of $f$, which does not lie in $D(-1,1)$ since the support of $f$ is $[0,1]$.

The uniform convergence of $f_n$ and its derivatives makes it a Cauchy sequence wrt any of the semi-norms.