In the book Distribution theory by Duistermmat/Kolk convergence in the space of test functions is defined as:

$\phi_j \to \phi$ in $C_0^{\infty}(\Omega) $ (where $\Omega$ is an open subset of $\mathbb{R}^n$ )if:

$i)$ There is a compact set $K: \text{supp} \phi_j \subset K$ for every $j$.

$ ii)$ For every multi-index $\alpha$ the sequence $\partial^\alpha \phi_j \to \partial \phi$ uniformly.

I'm not really sure how to interpret this. By "uniformly" in criterion $ii)$ is it meant uniformly on every compact subset of $\Omega$?

It is not a criterion that for every multi-index $\alpha$ there is a compact $K \subset \Omega$ such that $\text{supp} \partial^\alpha \phi_j$? If no is it possible to $\phi_j \to \phi$ in $C_0^{\infty}(\Omega)$ but the support of its derivatives "diverging"?

  • 2
    $\begingroup$ Uniformly on $K$ which is the same for each $\alpha$ since it contains the support of each $\phi_j$ (and of the limit). When discarding the compact support condition for the convergence, you obtain the Schwartz space, which is probably easier to understand (it is generated by the translates of $e^{-n x^2}$) but has a smaller dual (the tempered distributions). $\endgroup$
    – reuns
    Jun 20, 2017 at 12:19
  • $\begingroup$ Ok, I think I get that, but what about the derivatives? Does each derivative have to "converge in the space of test functions" as well? $\endgroup$
    – user202542
    Jun 20, 2017 at 12:21
  • $\begingroup$ The definition guarantees that $\partial^\alpha\phi_j \to \partial^\alpha\phi$ which is why its dual has those nice properties. $\endgroup$
    – reuns
    Jun 20, 2017 at 12:23
  • $\begingroup$ @user202542 I have a silly question, the book you mentioned, is that a good book for distribution theory, or you can suggest better in this matter! $\endgroup$
    – MAN-MADE
    Jun 20, 2017 at 12:39
  • $\begingroup$ @MANMAID I've only just started with it but the very first chapters seems good. I'd recommend taking a look in Rudin's Functional Analysis as a complement though. $\endgroup$
    – user202542
    Jun 25, 2017 at 21:13

1 Answer 1


(ii) requires "uniformly on $\Omega$", i.e., in the sup norm. (Though it is enough to have "uniformly on $K$ and pointwise on $\Omega\setminus K$", as then $\phi\equiv0\equiv\phi_j$ on $\Omega\setminus K$.)

A. It follows that $\phi(x)=\lim_j \phi_j(x)=0$ outside $K$; hence $\mathrm {supp}\, \phi\subset K$.

B. If you assume $\mathrm {supp}\, \phi\subset K$, then "uniformly in $K$" suffices in (ii).

C. It also follows that $\mathrm {supp}\, \partial^\alpha \phi_j\subset K$ $\forall \alpha\in\mathbb N^n$, $j\in\mathbb N$.

Proof: Let $f\in C^\infty_0(\Omega)$. Assume $\mathrm {supp}\, f\subset K$. As $f\equiv0$ on $K^c$, we have $\partial^\alpha f\equiv0$ on $K^c$, so $\mathrm {supp}\, \partial^\alpha f\subset K$, $\forall \alpha\in\mathbb N^n$, QED.

N.B. As $\phi_j\to\phi$ iff $\phi_j-\phi\to0$, we have $\phi_j\to\phi$ iff

$i')$ There is a compact set $K: \text{supp} (\phi_j-\phi) \subset K$ for every $j$.

$ ii)$ For every multi-index $\alpha$ the sequence $\partial^\alpha (\phi_j-\phi) \to 0$ uniformly on $\Omega$.

Although $ii)$ is clearly equivalent to your $ii)$, condition $i')$ is formally different. However, then $\text{supp}\, \phi_j\subset K\cup K'$, where $K':=\text{supp}\, \phi$, so these two common definitions are clearly equivalent (the union of two compact sets is clearly compact).

N.B. Using either definition, it suffices to assume $\phi\in C^\infty(\Omega)$, by "A." (or that $\phi$ has weak derivatives of all orders, as (ii') guarantees that they are actual derivatives).


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