In Schwartz's distribution theory, $T_\nu\rightarrow 0$ in $\mathscr{D}'(\mathbb{R}^n)$ is defined to be $\langle{T_\nu,\varphi\rangle} \rightarrow 0,$ $\forall \varphi\in C_c^\infty(\mathbb{R}^n),$ and $T_\nu(B)\rightarrow 0$ for all bounded set $B$ in $\mathscr{D}.$

Here $B$ is a bounded set in $\mathscr{D}$ means $B$ has the form like: $$ \{\varphi\in C_c^\infty(K)|\sup_x|\partial^\alpha\varphi(x)|\le M_\alpha\} $$ for certain compact set $K,$ and $T(B):=\sup_{\varphi\in B}|\langle T,\varphi\rangle|.$

However, I can't come up with an example that $\langle T_\nu,\varphi\rangle\rightarrow 0,$ $\forall\varphi\in C_c^\infty(\mathbb{R}^n),$ but without $T_\nu(B)\rightarrow 0$ for certain bounded set $B$ in $C_c^\infty(\mathbb{R}^n).$ Is there any obvious example? Otherwise we shall delete the supplement after the word "and".

I think about these when solving my question: $T\mapsto T*\varphi$ is a continuous map . I think $T_\nu\rightarrow 0$ shall add condition that $T_\nu(B)\rightarrow 0.$ After that I can solve it.

  • $\begingroup$ I have only seen the condition $\langle T_\nu, \varphi \rangle \to \langle T, \varphi \rangle,$ not the additional condition $T_\nu(B) \to 0.$ For example, see Definition 7.1 in Lecture notes on Distributions by Hasse Carlsson. $\endgroup$
    – md2perpe
    Dec 13, 2021 at 12:47
  • $\begingroup$ @md2perpe Yes I think this is the most common definiton, simulating $w^*$-convergence in Banach space ($T\rightharpoonup 0$). I regard the definition in my question as a simulation of strong convergence in Banach space ($\|T\|\rightarrow 0$). So my question is: do we really have $T_\nu\rightarrow 0$ in weak sense but not in strong sense? $\endgroup$
    – DreamAR
    Dec 13, 2021 at 12:54
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    $\begingroup$ This about convergence to zero with respect to a topology. The first statement with $\varphi$ corresponds to the weak-$\ast$ topology (the topology for children). The second statement with $B$ corresponds to the strong dual topology (the topology for adults). Most texts on distribution theory use the weak-$\ast$ topology and thay is ok for elementary applications. However, when one go into more advanced topics, starting with things related to the nuclear theorem, the proper topology to use is the strong one... $\endgroup$ Dec 13, 2021 at 21:14
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    $\begingroup$ ...It just happens that this kind of question about continuity of convolution is at the cusp of the transition from elementary to advanced. Here there should not an and because one has to chose one or the other topology. For a proof of hypocontinuity of convolution see my answer to mathoverflow.net/questions/72450/… It's for the simpler case of $\mathscr{S},\mathscr{S}'$ instead of $\mathscr{D},\mathscr{D}'$ but the proof is identical. The multiplier space here should be $\mathscr{E}$ of smooth functions with no support restriction. $\endgroup$ Dec 13, 2021 at 21:19
  • $\begingroup$ @AbdelmalekAbdesselam Thanks! I've learned a lot from your answers. $\endgroup$
    – DreamAR
    Dec 14, 2021 at 3:20


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