In the middle of another proof (Theorem 3.4, p. 60) in his Functional Analysis book, Rudin says that
"every nonconstant linear functional on $X$ (topological vector space) is an open mapping."
Is this obvious? Could someone please show me how the proof goes?
Let $f : X \to \mathbb{R}$ be a nonconstant linear functional and $H= \text{ker}(f) \subsetneq X$. There exists an linear map $\varphi : X/H \to \mathbb{R}$ such that $f = \varphi \circ \pi$ with $\pi : X \to X/H$ the canonical projection. Using the definition of quotient topology, on can show that $f$ is open (resp. continuous) iff $\varphi$ is open (resp. continuous).
Two cases happen: either $f$ is continuous and $X/H$ is Hausdorff, or $f$ is not continuous and $X/H$ is not Hausdorff. However, any real one dimensional topological vector space is linearly homeomorphic to $\mathbb{R}$ if it is Hausdorff (see for example here), or it has the trivial topology otherwise.
To see this, let $Y$ be a real one dimensional topological vector space, say $Y= \mathbb{R}$, and $U$ be an open neighborhood of $0$. Suppose there exists $x \notin U$. Then, for all $y,z \in Y$ there exists $a \in \mathbb{R} \backslash \{0\}$ such that $y \notin z+aU$. Consequently, $Y$ is $T_1$ and the diagonal of $Y \times Y$ corresponds to $f^{-1}(0)$, with $f : (x,y) \mapsto x-y$, and it is closed for the product topology since $f$ is continuous; therefore, $Y$ is Hausdorff.
Consequently, either $X/H$ is Hausdorff, and $\varphi$ is a linear homeomorphism so $f$ is in fact continuous and open, or $X/H$ is not Hausdorff, and $\varphi$ is open (but not continuous) so $f$ is open (and not continuous).