Topology on the space of test functions

Question

I try to read into the theory of distributions and there is one thing which bothers me. I read that a distribution is a linear, continuous functional from the space of test functions, which, depending on the author, is sometimes defined as the Schwartz space and sometimes as $C^\infty_K$ (smooth functions with compact support).

For starters, where does this ambiguity come from and does it matter?

Now for the case $C^\infty_K$, I found two definitions of the associated topology. One as the final topology induced by the inclusion maps from $C^\infty(K)$ (I already understand how the topology on those spaces is defined) and one as the final locally convex topology induced by those maps, i.e. the finest locally convex topology which makes those maps continuous.

I couldn't proof their equivalence, i.e. how do I show that the final topology induced by those maps is already linear and locally convex?

Answer 1

Let $K_n$ be the unit ball of radius $n$, so your $C_K^\infty = \bigcup_n C^\infty(K_n)$ and let $f_n:C^\infty(K_n)\longmapsto C_K^\infty$ be the natural embeddings. The system $$ \bigl\{ \bigcup_n f_n[U_n]\, ;\, U_n \text{ is a $0$-neighbourhood in } C^\infty(K_n) \bigr\} $$ is a filter base in $C_K^\infty$, immediately to verify (using $U_n = f_n^{-1}(f_n[U_n])$) that it is a filter base for a for a fundamental system of $0$-neighbourhoods of a linear topology.

An absolutely convex set U is $0$-neighbourhood in $C_K^\infty$ for the topology described above, iff $f_n^{-1}[U]$ is a $0$-neighbourhood in $C^\infty(K_n)$ for each $n$. But the latter is precisely the characterization for a $0$-neighbourhood in the final topology on $C_K^\infty$.

Answer 2

OK, I think I got a counterexample now, i.e. the final topology w.r.t. to the inclusion maps is not linear. I only worked it through on ${\bf R}$ though.

The only assumption that I need is that for given $\epsilon>0$ and $k,n\in{\bf N}$ we find some $f\in C^\infty([-1,1])$ such that $\|f^{(j)}\|_\infty<\epsilon$ for $j<k$ and $f^{(k)}(0)>n$, which I think is easily constructable.

Now the set $$A:=\{f\in C^\infty_K:\forall j>0.f(j)<1/f^{(j)}(0)\}$$ (where we set $1/0=\infty$) is open in the final topology, i.e. $A\cap C^\infty([-i,i])$ is open for each $i$. Yet, if the final topology were linear, we would find an neighbourhood $B$ of $0$ such that $B+B\subseteq A$. But any such $B$ contains a basic open neighbourhood from $C^\infty([-i,i])$ of the form $$U_i=\{f\in C^\infty([-i,i]):\forall j<k(i).\|f^{(j)}\|_\infty<\epsilon(i)\}$$ for each $i$. Now we can construct $f\in U_1$ such that $f^{(k(i))}(0)$ is large enough so $f+U_{k(i)+1}\subseteq B+B$ is no longer a subset of $A$.

So the majority of books and scripts I read (e.g. Rudin) got it right, when they defined the topology on the space of test functions as the limit topology in the category of locally convex spaces, i.e. the finest locally convex topology which makes the inclusion maps continuous.