After reading many interesting posts here about Darboux functions and Conway's base 13 function (http://en.wikipedia.org/wiki/Conway_base_13_function) I have some questions that I don't seem to answer about functions with range equal to whole reals on every open set.

Say, that a function $f$ has the latter property, then it must be unbounded. What about the converse? Let $B$ be a Banach space and consider an unbounded linear functional $f\colon B\to\mathbb{R}$. Is $f(O)=\mathbb{R}$ for any (nonempty) open set $O\subset B$?

  • $\begingroup$ What exactly does it mean to you for $f$ to be "unbounded" in this case? If you merely require that the range of $f$ is unbounded in $\mathbb R$, then take as a counterexample $B=\mathbb R$ and $f(x)=x$. There must be something more to your question? $\endgroup$ – Henning Makholm Jan 13 '15 at 0:00
  • $\begingroup$ @Henning: a linear operator is bounded if the norm of all the vectors on the unit ball is bounded. Equivalently, for Banach spaces, the operator is continuous. $\endgroup$ – Asaf Karagila Jan 13 '15 at 0:02
  • $\begingroup$ @Asaf: Yes, but with that reading the property is not really a "converse" of the Darboux property -- functions such as the base-13 function aren't even linear operators. $\endgroup$ – Henning Makholm Jan 13 '15 at 0:03
  • $\begingroup$ Well $\| f\|$ is not bounded (as an operator). That's why I wanted to point out linear functional (I think it makes it more interesting, but I don't know if it is necessary to have Banach; perhaps it is enough with a normed linear space) $\endgroup$ – PeterA Jan 13 '15 at 0:04
  • $\begingroup$ @HenningMakholm I agree it is perhaps not the converse, but the thought originated from those problems and I've been thinking about it for a while. $\endgroup$ – PeterA Jan 13 '15 at 0:06

Step one: prove it suffices to check the unit ball.

Step two: prove that it is true for the unit ball, since we can shrink each value from the unit sphere itself, and those are arbitrarily large.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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