In the preprint on pages 19−20, first using Hahn−Banach, one proves

Lemma 38.For any closed convex set $U$ in a real Hausdorff locally convex space $E$ and for any Riemann integrable $\gamma:[0,1]\to E$ with ${\rm rng\ }\gamma\subseteq U$ it holds that$\int_0^1\gamma\in U$ .

Then it is remarked that the argument fails if $U$ is neither open nor closed. Further, one defines a set $U$ in a real locally convex space to be *countably convex* if and only if it is stable under taking countable convex combinations, i.e. one requires that $\sum_{i=1}^\infty c_ix_i\in U$ whenever every $x_i\in U$ and $c_i\ge 0$ with $\sum_{i=1}^\infty c_i=1$ . The author there asks (1) whether Lemma 38 still holds if the assumption of $U$ being closed and convex is replaced with it being merely countably convex.

The author then gives an example of a countably convex set $U$ with compact closure $K$ in the Fréchet space $E=\mathbb R^{\mathbb N_0}$ with the zero vector $\boldsymbol 0$ belonging to $K\setminus U$ such that $U$ cannot be separated by a linear functional from the singleton $\lbrace\boldsymbol 0\rbrace$ . Then one asks more particularly (2) whether $\int_0^1\gamma=\boldsymbol 0$ is impossible for these particular $E,U$ and any admissible Riemann integrable $\gamma$ . The set $U$ has as its elements all $\langle s_0 - \sum_{i=1}^\infty s_i , s_1 , s_2 , s_3 ,\ldots \ \rangle$ such that $\langle s_i\rangle\in[0,1]^{\mathbb N_0}$ with $0<\sum_{i=0}^\infty s_i\le 1$ .

**Now I ask**, can anyone give an argument for a positive answer or a counterexample to (1) or (2) above.

**Edit.** In accordance with the suggestion by Pietro Majer in a comment, I add here the definition of Riemann integrability for a function $\gamma:[0,1]\to E$ when $E$ is a real Hausdorff locally convex space. Namely, we say that $\gamma$ is Riemann integrable with integral $y=\int_0^1\gamma$ if and only if $y\in E$ and for every neighbourhood $V$ in $E$ of $y$ there is $\delta>0$ with the property that $\sum_{i=0}^{k-1}(t_{i+1}-t_i)\gamma(s_i)\in V$ whenever with $k\in\mathbb N$ we have $t_0=0$ and $t_k=1$ and $t_i\le s_i\le t_{i+1}$ and $t_{i+1}-t_i<\delta$ for $i=0,\ldots\ k-1$ .

**Edit2.** Concerning (2), it is actually very easy to see that there can not be any Riemann integrable $\gamma$ with ${\rm rng\ }\gamma\subseteq U$ and $\int_0^1\gamma=\boldsymbol 0$ . Indeed, it there were one such, writing $\gamma(t)=\langle\gamma_i(t):i\in\mathbb N_0\rangle$ , each $\gamma_i$ would be Riemann, hence Lebesgue integrable, and for $i\not=0$ and $0\le t\le 1$ we would have $\gamma_i(t)\ge 0$ and $\int_0^1\gamma_i=0$ . It would follow that $\gamma_i(t)=0$ for a.e. $t\in[0,1]$ and all $i\in\mathbb N$ . Since $\gamma(t)\in U$ for all $t\in[0,1]$ , it would follow that $\gamma_0(t)>0$ for $t\in[0,1]$ outside a set of measure zero. This would imply $\int_0^1\gamma_0>0$ , and hence $\int_0^1\gamma\not=\boldsymbol 0$ , a contradiction.

**Edit 3.** A counterexample to (1) is obtained as follows. Put $I=[0,1]$ , and let $E=\ell^2(I)$ , the nonseparable Hilbert space. Define $\gamma:I\to E$ by $\gamma(t)_i=1$ if $i=t$ , and $\gamma(t)_i=0$ otherwise. Then $\gamma$ is everywhere discontinuous, but Riemann integrable with $\int_0^1\gamma=\boldsymbol 0_E$ . The countably convex hull $U$ of ${\rm rng\ }\gamma$ is countably convex and has as its elements all $\sum_{j=0}^\infty c_j\gamma(t_j)$ where $t_j\in I$ and $c_j\ge 0$ and $\sum_{j=0}^\infty c_j=1$ . Hence $\boldsymbol 0_E\not\in U$ . There still remains the possibility that (1) may hold for (say) separable Banach spaces $E$ and continuous $\gamma$ .

Riemann integrable function$\gamma:[a,b]\to E$ used in that paper is indeed given there (page 18), yet perhaps it would help if you recalled it here in a self-contained form. – Pietro Majer Mar 8 '11 at 13:23