Application of the open function theorem (BANACH) with $f\not\equiv 0$. THE PROBLEM STATEMENT IS AS FOLLOWS :
Let $X$ be a Banach space over $\mathbb{R}$ and $f:X\rightarrow \mathbb{R}$ a continuous linear functional and $f\not\equiv 0$ then $f$ is open.
DEMONSTRATION:
it suffices to prove that f is overjective and apply the open function theorem (BANACH) as follows:
let $y\in\mathbb{R}$, as $f\not\equiv 0$ then $\exists x\in X: f(x)\neq 0 $   (i.e)
Then $f$ is overjective and by the open function theorem $f$ is open (i.e.) $f$ sends open in open
MY QUESTION IS THIS
Having the same conditions of the statement, except the continuity of f, you have that f is open?
My suspicion is no, but I have not been able to find a function that complies. Could someone please guide me to find a non-continuous linear function that is not open?
 A: Since $f$ is linear, we have $$ f((1-t)x+ty)=(1-t)f(x)+tf(y),\\0\le t\le 1,\ x,y\in X$$ Thus the image of any open ball is a convex subset  of $\mathbb{R}.$ If $f$ is unbounded the image is an unbounded symmetric convex subset of the line. Thus it  is equal to the entire line.
In the complex valued case the image of the unit ball is an unbounded convex and rotationally invariant subset of $\mathbb{R}^2.$ So it is equal $\mathbb{R}^2.$
A: Discontinuous functions are also open maps, but for a different reason: they map every open ball to the entire scalar field. This is, of course, an open set.
The reason is, unbounded linear functionals $f$ have a dense kernel $f^{-1}\{0\}$. For other scalars $z$, the set $f^{-1}\{z\}$ is simply a translate of $f^{-1}\{0\}$; given some $v$ such that $f(v) = 1$ (such a vector must exist if $f \neq 0$), we then have $f^{-1}\{z\} = zv + f^{-1}\{0\}$. As such, $f^{-1}\{z\}$ is dense, for all scalars $z$. This means that, in every open set, there is some vector $w$ such that $f(w) = z$. So, every open set maps to the entire scalar field.
