I am trying to prove the following, but I am not confident in my work.
Let $D$ be a linear subspace of a normed space $X$ that is dense in $X$. Let $Y$ be a Banach space. Show that any bounded linear operator $T : D \to Y$ has a unique extension to a bounded linear operator $X \to Y.$
What I tried. I can prove the lemma
Lemma: Assume that $D$ is a dense subset of $X_1$, that $(X_2, d_2)$ is complete and that $f : D → M_2$ is a uniformly continuous mapping. Then there exists a unique continuous mapping $F : M_1 → M_2$ such that $F|_D = f$.
Bounded linear operators are uniformly continuous the existence and uniqueness of a continuous extension follows from the Lemma. If I am correct, I can use the proof of the lemma to construct the desired extension.