I need to prove that $B \in \mathcal{L}(X,Y)$ when $D \subset X$ is a dense subspace and $A \in \mathcal{L}(D,Y)$ is some bounded linear operator from $D$ to $Y$ when we assume that $Y$ is a Banach space.
My approach was to show that B is a bounded linear operator, if it satisfies:
i) $A(\alpha x_n)=\alpha A(x_n) \forall \alpha \in \mathbb{K} $
and
ii) $A(x_n + x_m) = A(x_n) + A(x_m)$
I started to feel pretty lost since this is an introduction to real analysis course, so please correct me if I'm wrong in approaching this problem. Thanks
I believe your question is that if $A = B|_D$ then prove that $B$ is bounded. Let $x \in X\setminus D$ and let $\{x_n\}$ be a sequence in $D$ such that $x_n \to x$. Then compute the following limit:
\begin{eqnarray*} \lim_{n\to \infty} ||Bx - Bx_n|| & = & \lim_{n\to\infty} ||Bx - Ax_n|| \\ & = & \lim_{n\to\infty} ||Bx - Bx_n + Bx_n - Ax_n|| \\ & \leq & \lim_{n\to\infty} ||B(x-x_n)|| + ||Bx_n - Ax_n||. \end{eqnarray*}
Notice that the above limit must be zero since $x-x_n \to 0$ and $By = Ay$ for each $y \in D$. This proves that $B$ is continuous, and is therefore bounded.