Just a simple construction to illustrate Nate Eldredge's answer about functions with dense graphs. Pick any -vector space E with a norm. On E, choose a non-continuous linear form ; now this can only be done if , of course.
Then, pick y such that , and let be defined by . Then obviously T maps E onto the kernel of L; it is not difficult to prove that must be dense in E for any non-continuous L (the two conditions are even equivalent), and thus the graph of T must be dense in .
No comments:
Post a Comment