Can smoothness of curves into a convenient locally convex vector space be tested with just a dense subspace of the dual?

Question

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 (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.

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?

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.

Answer 1

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:

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.

Answer 2

$\def\sp{\kern.4mm}\def\bbR{\mathbb R}$A previous idea for getting a counterexample in the reflexive Banach case is here, and a modification of it is in the following: With $1<p<+\infty$ let $c:\bbR\to E=L^p(\bbR)$ be given by $t\mapsto c(t)$ where $c(t)$ is the coset containing the function $\bbR\owns s\mapsto{\rm e}^{\sp s}$ for $s\le t$, and $s\mapsto 0$ for $t<s$. Then $u\circ c$ is smooth for every $u$ in the dense subset of $E'$ formed by the elements given by integration against elements in the space $\mathscr S\,(\bbR)$ of rapidly decreasing functions, but $c$ is not even once differentiable at any point by the argument loc. cit.