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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?

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3 Answers 3

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).

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Let $f$ be any non-constant functional on $X$ and $A$ to be any open subset in $X$.

$f$ is an open mapping if and only if for any $x\in A$, there exists an open subset $U$ of $0$ such that $x+U \in A$ and $f(x+U)=(f(x)-\epsilon, f(x)+\epsilon)$, for some $\epsilon>0$.

Since $f$ is a non-constant functional, according to Theorem 1.18 in Rudin's book, $f$ is bounded in some neighborhood of $0$. Take $W$ to be a neighborhood of $0$ such that $x+W \in A$. By Theorem 1.14, we can take a balanced neighborhood $U'$ of $0$ such that $U' \in W\cap V$. Since $f(U')$ is also a balanced neighborhood of $0$ in $\mathbb{R}$, $f(U')$ is a finite interval whose interior contains $0$. Take $\epsilon>0$ such that $(-\epsilon, +\epsilon)$ is contained in $f(U')$. $f^{-1}(-\epsilon,+\epsilon)= U$ is a neighborhood of $0$ in $X$ and $f(U)=(-\epsilon,+\epsilon)$. $U$ is the desired neighborhood.

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You are assuming $f$ is continuous, which is not part of Rudin's statement. –  Cantor Apr 16 '13 at 17:49
@Cantor as for the proof of Theorem 3.4, the functional \Lambda (following Rudin's symbol)we concern is continuous. –  Salmon Apr 20 '13 at 0:45

I think this is indeed easy to check. Let $f$ be a non zero linear functional from the topological vector space $X$ to $\mathbb{C}$.

  • It is enough to show that $f$ maps a balanced neighborhood of $0$ to an open ball around $0\in \mathbb{C}$ or the whole space $\mathbb{C}$.
  • The former case corresponds to a continuous map $f$, the later to an unbounded map.

Let $V$ be a balanced neighborhood of $0\in X$. There is a point $u\in V$ such that $f(u)\neq0$. Then $$\{f(\theta u):\theta\in\mathbb{C},\, |\theta|\leq1\}=\{z\in\mathbb{C}:|z|\leq|f(u)|\}.$$ It follows from the assumption on $V$ that either

  • $f$ is unbounded and $f(V)=\mathbb{C}$,
  • or $f$ is continuous and either $f(V)=\{z\in \mathbb{C}:|z|<r\}$ or $f(V)=\{z\in \mathbb{C}:|z|\leq r\}$ for some $r>0$.

At this point one can use a big gun such as the open map theorem. I however think that one can still proceed to complete the proof using simple ideas. Here is a sketch.

In the continuous case we now show that $f(V)\neq \{z\in\mathbb{C}:|z|\leq r\}$. If it were the case, then there would be $w\in V$ such that $f(w)=r$. The continuity of the scalar product implies that there is a real number $t>1$ such that $t w\in V$ but then $f(tw)=tf(w)>r$ which is a contradiction.

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