Discontinuous linear functional in $\mathbb{R^{\infty}}$

Question

Let $\mathbb{R^{\infty}}=\{x=(x_n)_{n=1}^\infty:x_n\in\mathbb{R}, \forall n\in\mathbb{N}\}$, with the topology given by the set of neibourhoods of zero of the form $\mathcal{U}(n_1,n_2,...,n_m;\epsilon)=\{x\in\mathbb{R^{\infty}}:|x_{n_j}|<\epsilon, \forall j=1,...,m\}$.

I am trying to find a linear functional $f:\mathbb{R^{\infty}}\longrightarrow\mathbb{R}$ which is not continuous for this topology.

I have tried to use the following result:

In a topological vector space $E$, a linear functional $f:\mathbb{E}\longrightarrow\mathbb{R}$ is continuous if, and only if, $\exists \mathcal{U}$ neighbourhood of zero such that $f(\mathcal{U})$ is bounded in $\mathbb{R}$.

Everything I have tried has been useless because I have been unable to find a functional that is linear and for which $f(\mathcal{U})$ is unbounded for all $\mathcal{U}=\mathcal{U}(n_1,n_2,...,n_m;\epsilon)$.

Any hints would be very helpful. Thanks in advance.

Answer 1

There exists a (Hamel) basis for $\mathbb R^{\infty}$ which includes the sequence $e_1,e_2,...$ (where $e_n$ has $n-th$ coordinate $1$ and all other coordinates $0$). Define $f(e_n)=1$ for all $n$ and $f(x)=0$ for all the basis elements not in $\{e_1,e_2,...\}$. You can extend $f$ to $\mathbb R^{\infty}$ by linearity and it is trivial to check that $f$ is not continuous. [Just use the fact that $f(e_n) = 1$ for all $n$ so we cannot have $|f(x)|<1$ on any basic neighborhood of $0$.

Answer 2

Why can you not "find" such an example? Because the Axiom of Choice is required. (It is consistent with ZF that there is no such functional.)

example
Let vectors $\mathbf e_n \in \mathbb R^\infty$ be defined by $$ \mathbf e_n = (0,0,\dots,0,1,0,0,\dots) $$ with $1$ for coordinate number $n$, the other coordinates all zero.

Let $$ \mathbf u = (1,1,1,1,\dots) $$ all coordinates $1$.

Note that the set $\mathcal A_1 := \{\mathbf u, \mathbf e_1, \mathbf e_2,\dots\}$ is linearly independent. Extend it to a Hamel basis $\mathcal A$ for all of $\mathbb R^\infty$. (Axiom of Choice used!) This means: each $\mathbf x \in \mathbb R^\infty$ can be written uniquely as $$ \mathbf x = \sum_{\mathbf a \in \mathcal A} c_{\mathbf a} \mathbf a $$ where $c_{\mathbf a} \in \mathbb R$ and only finitely are nonzero.

Define a linear functional $F : \mathbb R^\infty \to \mathbb R$ by $$ F\left(\sum_{\mathbf a \in \mathcal A} c_{\mathbf a} \mathbf a\right) = c_{\mathbf u} $$

I claim the functional $F$ is not continuous. Indeed $$ \lim_{n \to \infty} \sum_{k=1}^n \mathbf e_k = \mathbf u $$ in the (product) topology of $\mathbb R^\infty$, but $$ F\left(\sum_{k=1}^n{e_k}\right) = 0 \quad\text{for }n=1,2,3,\dots \\ F(\mathbf u) = 1 . $$