This is related to a question I answered earlier which raised a question in my mind.
My question is the following,
Suppose we have a vector space $\mathbb{V}$ with real coefficients.
Let $\textbf{T}$ be an operator on this space which has the following two properties:
- $ \textbf{T}( \textit{u} + \textit{v})= \textbf{T}(\textit{u}) + \textbf{ T}( \textit{v}) \qquad (\textit{u},\textit{v} \in \mathbb{V})$
- $\textbf{T}(c v) = c \textbf{T}(\textit{v}) \qquad (\textit{v} \in \mathbb{V}, c \in \mathbb{Q})$
Can we prove that $\textbf{T}(c\textit{v}) = c\textbf{T}(\textit{v}) $ for $c \in \mathbb{R}$ ?
I suspect we can prove this for any particular $r\in \mathbb{R}$ by taking a sequence of rational points $\lbrace q_n \rbrace$ which converge to $r$ and writing the following equality.
$$\textbf{T}(q_n \textit{v}) = q_n \textbf{T}(\textit{v}) \qquad (\forall n)$$
$$\lim_{n\rightarrow \infty} \textbf{T}(q_n \textit{v}) = \lim_{n\rightarrow \infty} q_n \textbf{T}(\textit{v}) $$
$$\lim_{n\rightarrow \infty} \textbf{T}(q_n \textit{v}) = r \textbf{T}(\textit{v}) $$
The statement would be proven if we can pull the limit inside but I'm not enough of an analyst to be able to justify that at this stage.
You can move the limit past the $T$ only if $T$ is continuous (strictly, if the restriction of $T$ to every one-dimensional subspace is continuous).
Every $\mathbb{R}$ vector space can be canonically considered also a $\mathbb{Q}$ vector space, and the additivity of $T$ gives you $\mathbb{Q}$-linearity, but not $\mathbb{R}$-linearity.
We can "construct" operators that are $\mathbb{Q}$-linear but not $\mathbb{R}$-linear if we extend an $\mathbb{R}$-basis of $\mathbb{V}$ to a $\mathbb{Q}$-basis, and define the operator to take some non-zero values on the basis elements belonging to the original $\mathbb{R}$-basis, and $0$ on the added basis elements. Since, however, one cannot explicitly construct a $\mathbb{Q}$-basis of an $\mathbb{R}$ vector space, one cannot obtain an explicit example in that way.