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?

  • $\begingroup$ OK, according to the script Distribution Theory by Ivan F Wilde, the difference between linear, continuous functionals from the Schwartz space and from $C^\infty_K$ is the difference between tempered distribution und just distributions. So I'd like to put more emphasis on the second part of my question, because that's what really puzzled me. $\endgroup$
    – fweth
    Dec 5, 2014 at 18:30

2 Answers 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.


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 $C_K^\infty$.

  • $\begingroup$ Do you consider neighbourhoods to be always open? If not, I'm pretty sure you cannot say in general for a final topology generated by $\{f_i:X_i\rightarrow X\}_i$ that $U$ is $0$-neighbourhood iff all $f^{-1}_i(U)$ are $0$-neighbourhoods in $X_i$, I think I can give a counterexample if you're interested. $\endgroup$
    – fweth
    Dec 6, 2014 at 14:20
  • $\begingroup$ BTW, there might be an answer to my question in cmouhot.files.wordpress.com/2010/02/main.pdf p.14-15 however I haven't been able to verify their solution yet but I'm working on it. $\endgroup$
    – fweth
    Dec 6, 2014 at 15:07
  • $\begingroup$ As long as the $f_n$ are linear, I am sure that the characterization for the final topology is correct. $\endgroup$
    – Vobo
    Dec 6, 2014 at 17:18
  • $\begingroup$ OK, maybe I'm wrong, but I thought of the space ${\bf R}^{<\omega}\subseteq{\bf R}^{\omega}$ of finite sequences and the final topology induced by the canonical $f_i:{\bf R}^i\rightarrow{\bf R}^{<\omega}$. For this space, I'm pretty sure the sets $\{x\in{\bf R}^{<\omega}:|x_i|<\epsilon_i\}$ for $(\epsilon_i)_i\in{\bf R}^{\omega}_+$ form a neighbourhood base around $0$. But for the set $A:=\cup_i\{x\in{\bf R}^i:\|x\|_\infty<1/i\}$ we have $A\cap{\bf R}^i$ is always a $0$-neighbourhood yet no basic neighbourhood fits into $A$. $\endgroup$
    – fweth
    Dec 6, 2014 at 17:38
  • $\begingroup$ Should have said '$A\cap{\bf R}^i$ is always $0$-neighbourhood in ${\bf R}^i$', just to be clear $\endgroup$
    – fweth
    Dec 6, 2014 at 17:45

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