Extending Continuous Sublinear maps on dense subsets of a Banach space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:25:00Z http://mathoverflow.net/feeds/question/62180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62180/extending-continuous-sublinear-maps-on-dense-subsets-of-a-banach-space Extending Continuous Sublinear maps on dense subsets of a Banach space Jeffrey 2011-04-18T22:47:32Z 2011-06-13T02:22:13Z <p>Suppose X' and Y are Banach spaces and X is a linear subspace dense in X'. Let T be a continuous map of X to Y satisfying:</p> <p>(1) ||T(x+y)|| is less than or equal to ||T(x)||+||T(y)||.</p> <p>Please prove whether or not in general it is true that T has a continuous extension to X'.</p> <p>Please also answer the same question for the case where we assume in addition that</p> <p>(2) ||T(cx)||=|c|*||T(x)|| for all c real.</p> <p>If so far, the answer has been "no," then assume that Y is an L^p space on some sigma finite measure space, with p>1. (Let me know if it's true for p=1 too.) If instead of using the p norm in the definitions (1) of subadditivity or (1) and (2) of sublinearity, we have pointwise almost everywhere inequalities of absolute values, then can T be extended?</p> <p>To be completely explicit, in this case we would have</p> <p>|T(x+y)| is less than or equal to |T(x)|+|T(y)| and |T(cx)|=|c|*|T(x)| for all c real pointwise almost everywhere on the measure space.</p> http://mathoverflow.net/questions/62180/extending-continuous-sublinear-maps-on-dense-subsets-of-a-banach-space/63663#63663 Answer by Ady for Extending Continuous Sublinear maps on dense subsets of a Banach space Ady 2011-05-02T01:08:34Z 2011-05-02T01:08:34Z <p>To 1) and 2). Let $X^{'}=Y=c_{0}$ , and let $X=c_{00}$. Take some $p\in X^{'}\smallsetminus X$, and define $T:X\rightarrow Y$ by $Tx:=\left\Vert x\right\Vert \cdot\left(\sin\left(\left\Vert x-p\right\Vert ^{-1}\right),1,0,0,0,...\right)$.</p> <p>Then $T$ satisfies your conditions, yet it is not continuously extendable.</p>