Regarding a homogeneous map on a Vector space

Question

Let $(X,\|.\|)$ be a complex Banach space. Let $S$ be a dense subspace of $X$. Let $f:X\longrightarrow\mathbb{C}$ be a map such that $f$ is linear and bounded on $S$ and also $f$ is homogeneous on $X$ i.e $f(\alpha x)=\alpha f(x)$ for every $\alpha\in \mathbb{C}$ and $x\in X$.

Can we say that $f$ is continuous or linear on $X$?

I know that the Hahn Banach theorem assures the existence of a unique linear and bounded extension to $f$ restricted to $S$. But I ask for the linearity or continuity of $f$ itself on all of $X$.

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. — José Carlos Santos, Feb 15 '18 at 10:13
MO copy of the question: Regarding additivity of a function. — Martin Sleziak, Feb 16 '18 at 12:29
The information about the domain of $f$ seems to be missing. Is "$f:\longrightarrow\mathbb{C}$" supposed to be "$f:X\longrightarrow\mathbb{C}$"? — Martin Sleziak, Feb 16 '18 at 12:30
I believe this answer gives a very reasonable advice about cross-posting - the most important point seems to be that you should like each post to the copy on other sites. Of course, you can also have a look at other discussions about cross-posting on meta. — Martin Sleziak, Feb 16 '18 at 12:32

Alex M. · Accepted Answer · 2018-02-16T13:09:09.303

No, $f$ is not necessarily linear or continuous. Let $X = l^2$ with elements denoted $x=(x_0, x_1, \dots)$. Consider the hemisphere $H = \{ \|x\| = 1, x_0 \ge 0\}$ and let $g : H \to \mathbb C$ be any function. Define $h : X \to \mathbb C$ by $h(x) = \begin{cases} g \left( \frac x {\|x\|} \right) \|x\| , & x_0 \ge 0 \\ g \left( -\frac x {\|x\|} \right) \|x\|, & x_0 < 0 \end{cases}$ and notice that it is a homogeneous, not necessarily continuous functional on $X$. Define $f : X \to \mathbb C$ as $f(x) = \begin{cases} 0, & x \in S \\ h(x), & x \in X \setminus S \end{cases}$. To see that $f$ is homogeneous, notice that $x \in S \iff rx \in S \ \forall r \in \mathbb \setminus \{0\}$. Notice that, in general, $f$ is neither continuous, nor linear.

Martin Sleziak · Answer 2 · 2018-02-16T13:00:29.770

Suppose that you have a Banach space $X$ and a dense subspace $S$ such that $X=S\oplus T$ where $T$ is not finite-dimensional.¹ (By $\oplus$ I mean direct sum as vector space.)

Since $T$ is infinite-dimensional, there is a discontinuous linear functional² $f\colon T\to\mathbb R$. Extending it by choosing $f|_S=0$ you get a linear functional defined on $X$ which is zero on $S$. So it fulfills your conditions. (It is not continuous if you take $X$ as the domain, but the restriction to $S$ is continuous. It is linear on $X$ and, consequently, also on $S$.)

¹It is possible to find such situations. Consider $X=\ell_2$ and $S=c_0$, i.e., $S$ consists of all sequences with finite support. Hamel dimension of $S$ is $\aleph_0$, but $\ell_2$ has uncountable Hamel dimensions, since it is a Banach space. (See also: Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.)

²See also: Discontinuous linear functional.

Regarding a homogeneous map on a Vector space

2 Answers2