In the previous question, I find that situation is much less favorable than expected…. So I add more details to focus on the specific case I have in mind.

Let us consider the Schwartz space $\mathcal{S}(\mathbb{R}^{mN})$ and denotes its elements as $f(x_1, \dotsc, x_N)$ with $x_1, \dotsc, x_N \in \mathbb{R}^m$. In the original paper Osterwalder and Schrader - Axioms for Euclidean Green's functions on Schwinger functions, the following subspace is introduced: \begin{equation} ^{0}\mathcal{S}(\mathbb{R}^{mN}):= \{ f \in \mathcal{S}(\mathbb{R}^{mN}) \mid f \text{ and all its derivatives vanish at $(x_1, \dotsc, x_N)$ whenever } x_i=x_j \text{ for some } 1 \leq i < j \leq N \} \end{equation}

Let $ ^{0}\mathcal{S}(\mathbb{R}^{mN})'$ denote the space of continuous linear functionals on $ ^{0}\mathcal{S}(\mathbb{R}^{mN})$.

Obviously, each $f \in \mathcal{S}(\mathbb{R}^{mN})$ defines an element $T_f \in {^{0}\mathcal{S}}(\mathbb{R}^{mN})'$ by the formula $T_f (g) := \int_{\mathbb{R}^{mN}} fg$ for $g \in {^{0}\mathcal{S}}(\mathbb{R}^{mN})$.

Now, my question is as follows:

Is the above mapping $f \to T_f$ injective?

Is $\mathcal{S}(\mathbb{R}^{mN})$ sequentially dense in ${^{0}\mathcal{S}}(\mathbb{R}^{mN})'$ w.r.t the weak$^*$ topology in the sense that for any $T \in {^{0}\mathcal{S}}(\mathbb{R}^{mN})'$, we can find a sequence $\{ f_l \} \subset \mathcal{S}(\mathbb{R}^{mN})$ with $T_{f_l}(g) \to T(g)$ as $l \to \infty$ for each $g \in {^{0}\mathcal{S}}(\mathbb{R}^{mN})$?

I hope that this question is more solid than the previous one.

Add) After all, I am looking for a space of "ordinary functions" injectively and densely embedded in ${^{0}\mathcal{S}}(\mathbb{R}^{mN})'$. If Schwartz space doesn't work, I hope to know if there is any other candidate.


1 Answer 1


$\newcommand\R{\Bbb R}\newcommand\S{\mathcal S}$Yes, the linear mapping $f\mapsto T_f$ is injective.

Indeed, suppose that $T_f=0$ for some $f\in\S(\R^{mN})$.

Consider the open set $$X:=\{x=(x_1,\dots,x_N)\in\Bbb R^{mN}\colon x_i\ne x_j \text{ if }i\ne j\}.$$

Take any $x\in X$. Then there is a sequence $(g_n)$ in $^0\S(\R^{mN})$ such that eventually (that is, for each large enough $n$) we have $g_n\ge0$, $g_n=0$ outside the open ball $B_x(1/n)$ of radius $1/n$ centered at $x$, and $\int g_n=1$. Then $0=T_f(g_n)=\int fg_n\to f(x)$.

So, $f=0$ on $X$. Therefore and because $f$ is continuous and because $X$ is dense in $\R^{mN}$, we have $f=0$ on $\R^{mN}$. So, the mapping $f\mapsto T_f$ is injective.

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    $\begingroup$ @Isaac : I think this should also be true, but do not have a complete proof at this point. Since there should be only one question in one post you may want to consider posting the sequential density separately. $\endgroup$ Commented May 8 at 12:09
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    $\begingroup$ It should be remarked that if ${}^0\mathcal{S}(\mathbb{R}^{mN})$ is endowed with the topology induced from $\mathcal{S}(\mathbb{R}^{mN})$, then any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ admits a(n usually non-unique) continuous linear extension to the whole of $\mathcal{S}(\mathbb{R}^{mN})$ by the Hahn-Banach theorem. In other words, any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ is the restriction to ${}^0\mathcal{S}(\mathbb{R}^{mN})$ of some $\widetilde{T}\in\mathcal{S}(\mathbb{R}^{mN})'$. Since $\mathcal{S}$ is weak-* sequentially dense in $\mathcal{S}'$, the result should follow. $\endgroup$ Commented May 8 at 18:28
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    $\begingroup$ I was going to mention the strategy of extension, followed by the usual mollification and truncation to get the approximation in the usual $S'$ but Pedro beat me to it. It's better to say sequentially dense for the strong dual topology, which is a stronger statement anyway, instead of weak-$\ast$ sequentially dense. Note that Hahn-Banach is a cheap solution, the alternative being to note that $T$ has at most algebraic singularities on the big diagonal and therefore admits an extension using resolution of singularities, or Bernstein polynomial techniques. My understanding is that... $\endgroup$ Commented May 8 at 18:34
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    $\begingroup$ ...the big diagonal is as hard as it gets from the point of view of resolution of singularities. BTW see mathoverflow.net/questions/470858/… for a new question about resolution of singularities for the big diagonal. $\endgroup$ Commented May 8 at 18:36
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    $\begingroup$ @AbdelmalekAbdesselam well, the whole subject of renormalization of distributions (as done in perturbative QFT) is a constructive form of the Hahn-Banach theorem, as Klaus Hepp put it long ago.... $\endgroup$ Commented May 8 at 18:42

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