# Tempered distributions at non-coinciding points and density of Schwartz functions

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: $$$$^{0}\mathcal{S}(\mathbb{R}^{mN}):= \{ f \in \mathcal{S}(\mathbb{R}^{mN}) \mid f \text{ and all its derivatives vanish at (x_1, \dotsc, x_N) whenever } x_i=x_j \text{ for some } 1 \leq i < j \leq N \}$$$$

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.

$$\newcommand\R{\Bbb R}\newcommand\S{\mathcal S}$$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.

• @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. Commented May 8 at 12:09
• 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. Commented May 8 at 18:28
• 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... Commented May 8 at 18:34
• ...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. Commented May 8 at 18:36
• @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.... Commented May 8 at 18:42