Let $X$ be a normed linear space, then the zeros of a discontinuous linear functional is dense in $X$.
If $Z_f$ is set of zeros then $Z_f$ will not be closed i.e. $Z_f\subsetneq \overline{Z_f}$, but don't know what to do after that.
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Suppose $f$ is a linear functional such that $Z_f$ is not dense. Then there is an $x\in X$ and an $\varepsilon > 0$ such that $B_\varepsilon(x) = x + B_\varepsilon(0)$ doesn't intersect $Z_f$. What follows for $f(B_\varepsilon(0))$? How does that imply that $f$ is continuous? – Daniel Fischer Feb 25 '15 at 14:26
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this is a duplicate ... – Learnmore Feb 25 '15 at 14:27
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@DanielFischer Is $f(B_\varepsilon(0))$ dense? How? – Mathronaut Feb 25 '15 at 14:38
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No, it's not dense. And that implies continuity because it is balanced. – Daniel Fischer Feb 25 '15 at 15:51
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I think if I could show that $f(B_\varepsilon(0))=\Bbb{C}$ then I'm done, but how can I show that? – Mathronaut Feb 25 '15 at 17:21