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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 $O\subset B$?

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  • $\begingroup$ 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
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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.

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