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My original question is from this ME post but I think I need a broader understanding for this.

The Schwartz space $\mathcal{S}$ and its subspaces are examples of nuclear spaces. In fact, any closed subspace of the Schwartz space is a nuclear Fréchet space.

Also, for every closed subspace $V$ of $\mathcal{S}$, we know that an element of $V$ may be regarded as an element of $V'$, the space of continuous linear functionals on $V$ by the mapping \begin{equation} f \to T_f \text{ where } T_f(g):= \int fg \text{ for all } g \in V. \end{equation}

Here are my questions now:

Is the above mapping always injective?

Is $V$ sequentially dense in $V'$ w.r.t. the weak$^*$ topology via the above mapping? That is, for each $T \in V'$, can we find a sequence $\{ f_n \} \subset V$ such that $\int f_n g \to T(g)$ as $n \to \infty$ for all $g \in V$?

Add : Following advice below, I restrict my question to closed subspaces of $\mathcal{S}$.

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    $\begingroup$ "Is any closed subspace . . . embedded . . .?" can be taken to mean "Is there any closed subspace that is...embedded...?" Or it could be taken to mean "Is it the case that any closed subspace at all, no matter which one, is...embedded...?" Any ambiguity would be eliminated by writing "every" instead of "any", if that is what is meant. English-speaking mathematicians use the word "any" too much. $\endgroup$ Commented May 8 at 15:35
  • $\begingroup$ @MichaelHardy thank you. $\endgroup$
    – Isaac
    Commented May 8 at 16:30
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    $\begingroup$ You changed "Is any closed subspace …?" to "Is any every subspace …?". I removed 'any'. Did you really mean to remove 'closed' (despite imposing the assumption in your body)? $\endgroup$
    – LSpice
    Commented May 8 at 16:33
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    $\begingroup$ Re, I am not sure what your answer means. If you do not intend to consider only closed subspaces, then I would encourage you to remove the statement from your body that "I restrict my question to closed subspaces of $\mathcal S$." If you do intend to consider only such subspaces, then I would encourage you to restore 'closed' to the title. $\endgroup$
    – LSpice
    Commented May 8 at 17:29
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    $\begingroup$ Ok, I will restore "closed" in that case. Thank you and sorry for confusion.. $\endgroup$
    – Isaac
    Commented May 8 at 17:40

2 Answers 2

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No to both questions. Consider $V=\{ g\in\mathcal S: g(-x)=ig(x) \textrm{ for }x\ge 0\}$. This space is closed and $V\not= 0$ because we can start with a compactly supported function on $[0,\infty)$, with support at some distance from $x=0$ and then define $g$ on $(-\infty, 0)$ as required.

Then $T_f=0\in V'$ for every $f\in V$ because $fg$ is odd when $f,g\in V$.

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    $\begingroup$ I think you can also simply fix one function $f \in V$, or more generarily one function $f \in S$ with $\int f^2=0$ and pick $V$ to be the span of this function. $\endgroup$
    – Nick S
    Commented May 8 at 1:08
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    $\begingroup$ @ChristianRemling Is this correct? $R$ is an involutive linear map, so it can only have $\pm 1$ as its eigenvalues. As such, we must have $V=\{0\}$ if $V$ is as defined above. $\endgroup$ Commented May 8 at 4:15
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    $\begingroup$ @NickS your last suggestion should work - to find such an $f$, let it have the form $f=g+zh$ with $g,h\in\mathcal{S}$ real valued, linearly independent and not identically zero (so that particularly $\int g^2,\,\int h^2>0$) and $z\in\mathbb{C}$. In that case, there must be $z_0\neq 0$ such that the second-order polynomial $P(z)=\int f^2=\int g^2+2z\int fg+z^2\int h^2$ vanishes at $z_0$ but we still have $f\not\equiv 0$. In that case, by the discriminant formula and the Cauchy-Schwarz inequality, $z_0\not\in\mathbb{R}$. $\endgroup$ Commented May 8 at 4:56
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    $\begingroup$ @PedroLauridsenRibeiro: You are right of course, it's obviously wrong as stated. I had actually thought about this issue (existence of such functions), though, but there ended up being a mismatch between what was in my head and on the paper. Fixed now I hope. $\endgroup$ Commented May 8 at 13:12
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    $\begingroup$ @PedroLauridsenRibeiro I was thinking about picking some infinitely many times differentiable functions $g$ supported inside $[1,2]$ and $h$ supported inside $[-2,-1]$, or any other disjoint interval, and defining $f(x)= g(x)+ih(x)$. As long as $g,h$ have the same integral, it works. $\endgroup$
    – Nick S
    Commented May 8 at 13:46
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This is more of a long(ish) comment than an actual answer... The question(s) asked above are not the same as the one posed in the title. In the former you are referring only to closed subspaces of $\mathcal{S}$, whereas in the latter you seem to ask whether any (abstract) nuclear Fréchet space $V$ (i.e. not necessarily topologically embedded into $\mathcal{S}$) has a positive answer to the two questions posed above.

This is important because a.) any closed subspace of $\mathcal{S}$ can be identified with a closed subspace of the nuclear Fréchet space $s$ of sequences of rapid decrease (use e.g. the sequence of Hermite coefficients of elements of $\mathcal{S}$), and b.) a nuclear Fréchet space $V$ is (topologically isomorphic to) a closed subspace of $s$ if and only if it has the so-called dominating norm (DN) property: there is a system $\|\cdot\|_k$, $k\in\mathbb{N}$ of seminorms defining the topology of $V$ and a $p\in\mathbb{N}$ such that for all $k\in\mathbb{N}$ there are $C_k>0$, $n_k\in\mathbb{N}$ such that $$\|x\|^2_k\leq C_k\|x\|_p\|x\|_{n_k}\ .$$ See e.g. Proposition 31.5, pp. 395 of the book by R. Meise and D. Vogt, Introduction to Functional Analysis (Oxford University Press, 1997) for a proof. Not all nuclear Fréchet spaces possess this property - for instance, $V=C^\infty(\Omega)$, where $\varnothing\neq\Omega\subset\mathbb{R}^n$ is open, does not (see e.g. Corollary 31.13, pp. 401-402 of R. Meise and D. Vogt, ibid.). In view of that, it would perhaps be better to clarify either the question or at least its title, for the answer may depend on that.

Edit: with the updated title, the answer to both questions is generally no as pointed by Nick S's comment to Christian Remling's answer. I will leave it to Nick whether he wants to turn his comment into an actual answer.

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    $\begingroup$ I didn't know that such an important issue remains. Thank you for your clarification and I have restricted my question to Schwartz spaces. Could you provide an answer now? Or do I need more improvement? $\endgroup$
    – Isaac
    Commented May 8 at 0:14
  • $\begingroup$ I have refined my question with more context in this MO post. Could you please help me once more? $\endgroup$
    – Isaac
    Commented May 8 at 9:03

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