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Let $C$ be a convex subset of a Banach space $X$ and $T:C\to C$ a (norm) continuous affine map, i.e. $$T(tx+(1-t)y)=tT(x)+(1-t)T(y)$$ for $0\le t\le 1$. Is $T$ weakly continuous, i.e continuous as a map from $(C,\tau)$ to $(C,\tau)$ where $\tau$ is the topology induced by the weak topology of $X$.

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    After some thought, I think the answer is yes. First we extend $C$ to its affine span $$L=\lbrace \sum_{i=0}^n \alpha_ix_i: n\ge 0, x_i\in C\rbrace.$$2010-11-10
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    After some thought, I think the answer is yes. First we extend $C$ to its affine span $$L=\lbrace \sum_{i=0}^n \alpha_ix_i: n\ge 0, x_i\in C\rbrace.$$ We can also assume these $x_i$ come from a maximal affinely independent subset $S$ of $C$. Then using $S$, we can extend $T$ to an affine map on $L$. After translation, we can assume $T$ is a continuous linear map on a linear subspace. Since every continuous linear map is weakly continuous, we are done.2010-11-10
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    $\sum_{i=0}^n\alpha_i=1$ is missing from the definition of L above.2010-11-10

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I believe the answer given in my comments is correct.