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Having had no (proper) answer to this question, I formulate the remaining case as a new question as follows. With $I=[0,1]$, let $E$ be a separable (real) Banach space, and let $\gamma:I\to E$ be continuous. Do there then exist sequences $\boldsymbol c,\boldsymbol t\in I^{\ \mathbb N_0}$, $\boldsymbol c(i)=c_i$ and $\boldsymbol t(i)=t_i$, with

(1) $\quad\mathbb R\ \text{-}\ \lim_{\ k\to\infty\ }\sum_{i=0}^kc_i=1 \quad$ and
(2) $\quad E\ \text{-}\ \lim_{\ k\to\infty\ }\big(k^{-1}\sum_{i=0}^{k-1}\gamma(k^{-1}i)\big) = E\ \text{-}\ \lim_{\ k\to\infty\ }\sum_{i=0}^k(c_i\gamma(t_i)) \quad$ ?

Either a (sketch of a) proof of the positive case or a counterexample is welcome. Countable or σ−convexity has also been considered in this question.

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There is a related concept of "measure convex" ... here it is in MR0808401: $C$ is measure convex if each inner regular probability measure on C has a barycenter and that barycenter belongs to $C$. .... here it is in MR1009196: A subset of a Banach space is said to be ... measure-convex if it contains the closed convex hull of each of its compact subsets, –  Gerald Edgar May 5 '11 at 13:22
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up vote 4 down vote accepted

Counterexample. $E = L^2[0,1]$ and $\gamma \colon [0,1] \to E$ defined by $\gamma(x) = 1_{[0,x]}$, the characteristic function of interval $[0,x]$. Then $\gamma$ is continuous, in fact $\|\gamma(x) - \gamma(y)\| = \sqrt{|y-x|}$. Now suppose $c_i$ and $t_i$ are as given. Let $$u := E\ \text{-}\ \lim_{\ k\to\infty\ }\big(k^{-1}\sum_{i=0}^{k-1}\gamma(k^{-1}i)\big) $$ and $$ v := E\ \text{-}\ \lim_{\ k\to\infty\ }\sum_{i=0}^k(c_i\gamma(t_i)) $$ Then $u(t) = 1-t$ for $t \in [0,1]$. Note $u$ is continous. But (assuming the $t_i$ are distinct) $v$ has a jump of size $c_i$ at $t_i$ for all $i$, so $v$ is certainly not continuous. Not even equal a.e. to a continuous function.

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That is a nice answer which I accept. The example obviously also works with $E=L^p([0,1])$ for $1\le p < +\infty$ . Saying that a set $U$ in a sequentially complete Hausdorff locally convex space $E$ is integral closed iff $\int_0^1\gamma\in U$ whenever $\gamma:[0,1]\to E$ is continuous with ${\rm rng\ }\gamma\subseteq U$ , we see that not even for separable Banach spaces does σ−convexity imply being integral closed. On the other hand, Hahn−Banach implies that (in sequentially complete LCS) open and closed convex sets are integral closed. –  TaQ May 5 '11 at 7:33
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