# Extensions of $\mathbb{Q}$-linear operators

Let $X$ be a real Banach space and let $Y$ be a dense $\mathbb{Q}$-vector subspace of $X$. Suppose moreover that $T\colon Y\to X$ is a bounded, $\mathbb{Q}$-linear operator. Can we extend $T$ to a bounded linear operator on $X$?

I think we can as boundedness of $T$ implies uniform continuity so we can extend $T$ to a continuous map. Is it clear that it will be linear?

Yes. Let $C > 0$ be a Lipschitz constant for $T$, and let $u, v \in X$. Then there are sequences $(u_k), (v_k) \in Y^{\mathbb{N}}$ converging to $u, v$, respectively. Similarly, let $\lambda \in \mathbb{R}$, where $(\lambda_k) \in \mathbb{Q}^{\mathbb{N}}$ is a sequence converging to $\lambda$. Now setting $S(u) = \lim_{k \to \infty} T(u_k)$, check that the sequence $T(u_k)$ is Cauchy and that $S(u)$ is well-defined (i.e. independent of which $(u_k)$ you choose). Finally, check \begin{align*} \| S(u + v) - S(u) - S(v) \| & = 0 \\ \| \lambda S(u) - S(\lambda u) \| & = 0 . \end{align*} Confirming those two equalities would be tantamount to proving linearity.
Yes. Note that "$T$ is bounded $\iff$ $T$ is continous" holds for both rational and real vectorspaces alike.
$T$ is continuous and is defined on a dense subspace of $X$. This is enough to define the (unique!) extension extension to the whole $X$ which will again be continuous. (just using metric topology. Not linearity neccessary at all)
You are right, linearity of the extended $T$ is not automatic by such a purely topological extension. But it is easy enough to verify:
Let $x,y\in X$ and $a\in\mathbb{R}$. Then there are sequences $x_n,y_n\in Y$ which converge to $x,y$ and $a_n\in\mathbb{Q}$ converging to $a$. Then you can simply calculate $$T(x+ay) = \lim_n T(x_n+a_ny_n) = \lim_n(Tx_n + a_nTy_n) =Tx + aTy$$ where we used that $\lim$ is additive and multiplicative, $T$ is continuous and $T$ is linear on $Y$ and $\mathbb{Q}$.