Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$.

share|improve this question
    
After some thought, I think the answer is yes. First we extend $C$ to its affine span $$L=\lbrace \sum_{i=0}^n \alpha_ix_i: n\ge 0, x_i\in C\rbrace.$$ –  TCL Nov 10 '10 at 20:40
    
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. –  TCL Nov 10 '10 at 20:50
    
$\sum_{i=0}^n\alpha_i=1$ is missing from the definition of L above. –  TCL Nov 10 '10 at 21:04

1 Answer 1

up vote 1 down vote accepted

I believe the answer given in my comments is correct.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.