So I have a homework question which I have no idea how to start.
Let $E_0$ be a dense subspace of the normed space $E$. Let $T_0:E_0 \rightarrow F$ be a bounded linear operator into the Banach space $F$.
(i) Show that $T_0$ can be uniquely extended to a bounded linear operator $T:E \rightarrow F$.
(ii) Prove that $\|T\| = \|T_0\|$.
Any hints would be much appreciated!
Best Answer
HINT: Imagine for a moment that an extension $T$ of $T_0$ exists and take $x\in E\setminus E_0$. Since $E_0$ is dense, you can approximate $x$ with a sequence $x_n\in E_0$. Since our "imaginary" operator $T$ is continuous, it must hold that $$\tag{1}Tx=\lim_{n\to \infty} T_0 x_n.$$ Now go back to reality, where $T$ does not exist yet. You need to construct it. The formula (1) gives you an obvious candidate, but you have to check that it makes sense at all points $x\in E$ and that it is independent of the choice of an approximating sequence $x_n$.
EDIT. Никита Васильев asked for more details in the comment section. Ok, here they are; we want (1) to be a consistent definition of an operator $T\colon E\to F$. For that, we need two things; first, the limit must exist, and second, if we choose another sequence $x_n'\in E_0$ such that $x_n'\to x$, then $$\tag{2} \lim_{n\to \infty} T_0x_n'= \lim_{n\to \infty} T_0 x_n.$$ To prove the first thing, we observe that the boundedness of $T_0$ gives $$ \lVert T_0x_n-T_0 x_m\lVert \le C\lVert x_n-x_m\rVert.$$ From this it immediately follows that $(T_0 x_n)$ is Cauchy, because $x_n$ is. And since $F$ is complete, this assures that the limit exists. To prove the second thing, we note that $$ \lVert T_0 x_n- T_0x_n'\lVert\le C\lVert x_n-x_n'\rVert; $$ now, since $x_n$ and $x_n'$ converge to the same limit, $\lVert x_n-x_n'\rVert\to 0$. We conclude that $\lVert T_0x_n-T_0x_n'\rVert\to 0$, which immediately implies (2).