Let $E$ be a (Hausdorff) locally convex vector space (from now on just "lcs" for short). We say that $E$ is *convenient* (also called *locally complete*, *Mackey-complete* or *$c^\infty$-complete*) if, given any disk (i.e. an absolutely convex and bounded subset) $B\subset E$ which is *closed*, the vector subspace $E\_B$ generated by $B$ $$E\_B=\bigcup\_{\lambda>0}\lambda B=\bigcup^\infty\_{n=1} nB$$ is a *Banach* space when the norm is given by the Minkowski functional of $B$ $$\mu\_B(x)=\inf\{\rho>0\ |\ x\in\rho B\}=\inf\{\rho>0\ |\ \rho^{-1}x\in B\}\ ,\quad x\in E\_B\ .$$ In other words, any closed disk $B\subset E$ is a so-called *Banach* disk. This is a rather weak kind of completeness - for instance, any sequentially complete $E$ is convenient.


It is well known that, if $E$ is a convenient lcs, a curve $\gamma:\mathbb{R}\rightarrow E$ is *smooth* (i.e. the iterated derivatives of all orders of $\gamma$ exist everywhere) if and only if $l\circ\gamma:\mathbb{R}\rightarrow\mathbb{R}$ is smooth (in the usual sense) for all $l\in E'=$ topological dual of $E$ - see e.g. Lemmata 2.1, 2.2 and Corollary 2.3, pp. 14-15 of the book by Andreas Kriegl and Peter Michor, [*The Convenient Setting of Global Analysis*](https://www.mat.univie.ac.at/~michor/apbookh-ams.pdf) (AMS, 1997). This generalizes the usual concept of smoothness of curves with a finite-dimensional target by means of smoothness of each component of the curve.



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> **Question:** suppose $E$ is a *convenient* lcs and $F\subset E'$ is a *dense* vector subspace of $E'$ in the strong topology $\beta(E',E)$. Let $\gamma:\mathbb{R}\rightarrow E$ be such that $l\circ\gamma:\mathbb{R}\rightarrow\mathbb{R}$ is *smooth* for all $l\in F$. *Is* $\gamma$ *smooth*?
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My expectation is that this is true if $E$ is in addition *semi-reflexive* (i.e. $E=(E',\beta(E',E))'$ as a vector space), for in this case $E$ is also the topological dual of any strongly dense vector subspace of $E'$, but I fear I might be missing something and there might still be room for counterexamples.



Can smoothness of curves into a convenient locally convex vector space be tested with just a dense subspace of the dual?
Here is a counter example. Let $E=\ell^2$. Consider the curve $\gamma:\mathbb R \to E$ given by
$$\gamma(t)= \Big(\frac{\sin(2^nt)}{2^n}\Big)\_{n\in \mathbb N}$$
In the dual $\ell^2$ consider the dense linear subspace of all sequences $l=(l\_n)$ with finite support. For such $l$ with $l\_n=0$ for $n\ge N$,
$$(l\circ \gamma)(t) = \sum\_{n=1}^N l\_n.\frac{\sin(2^nt)}{2^n} $$
is smooth, but the prospective derivative $\gamma'(t) = (1)\_{n\in \mathbb N}$ does not lie in $\ell^2$.


Positive results are the following:


(A) If $F\subset E'$ is large enough to recognize bounded sets, then $l\circ \gamma$ smooth for all $l\in F$ implies $\gamma$ smooth. See 5.22 of Convenient Setting.


(B) Theorem.
[Thm 4.1.19 in Frölicher, Kriegl: Linear Spaces and Differentiation Theory, Wiley, 1988]
Let $c:\mathbb R\to E$ be a curve in a convenient vector space $E$,
and let $\mathcal{V}\subseteq E'$ be a subset of bounded linear functionals such that the bornology of $E$ has a basis of $\sigma(E,\mathcal{V})$-closed sets.
Then the following are equivalent:
  

(i) $c$ is smooth
  

(ii) For each $k\in\mathbb N$ there exists a locally bounded curve $c^{k}:\mathbb R\to E$ such that for each $\ell\in\mathcal V$ the function $\ell\circ c$ is smooth $\mathbb R\to \mathbb R$ with $(\ell\circ c)^{(k)}=\ell\circ c^{k}$.
  

If $E=F'$ is the dual of a convenient vector space $F$, then for any point separating subset $\mathcal{V}\subseteq F$ the bornology of $E$ has a basis of
$\sigma(E,\mathcal{V})$-closed subsets, by [4.1.22 in loc.cit.


Theorem (B) is surprisingly strong:
note that $\mathcal V$ does not need to recognize bounded sets.
One can use the theorem in situations where $\mathcal V$ is just the set of all point evaluations on a Sobolev space $H^s(M)$ with $s>\frac{\dim(M)}2$, $M$ a smooth manifold.


Added, and later edited:
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About your question in the comment. My counterexample was wrong. Here is the a proof of the converse: For simplicity's sake I just take the real line:


(C) Corollary:
Let $c:\mathbb R \to C^\infty\_c(\mathbb R)$ be curve such that for each $t$ the the support $\text{supp}(c(t'))\subset K\_t$ for $t'$ neat $t$ and a compact $K\_t\subset \mathbb R$, and such that
$$t\mapsto \int\_{\mathbb R} c(t)(x).f(x)\,dx$$
is smooth for every $f\in C^\infty(\mathbb R)$.
We also need to assume that $t\mapsto c(t)\in C^\infty\_c(\mathbb R)$ is locally bounded. For this it is sufficient, that $t\mapsto c(t)(x)$ is bounded locally in $t$ for each fixed $x$, by 6.2 of Convenient Setting.
  

Then $c:\mathbb R \to C^\infty\_c(\mathbb R)$ is smooth.


For an equivalent formulation of smoothness, see 42.5 of convenient setting.


Proof:
By the support condition and differentiation by parts,
$$t\mapsto \int\_{\mathbb R} c(t)^{(n)}(x).f(x)\,dx = (-1)^n\int\_{\mathbb R} c(t)(x).f^{(n)}(x)\,dx$$ is again smooth. Since $f\mapsto f^{(n)}$ is surjective, half the conditions of theorem (B) are satisfied. It remains to check that $t\mapsto c(t)^{(n)}$ is locally bounded in $t$ as a curve into $C^\infty\_c(\mathbb R)$. But this follows from the condition for $n=0$ since differentiation is a bounded linear operator on $C^\infty\_c(\mathbb R)$. qed.


