extending a bounded linear operator

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!

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 $$x_n$$.

General notions in this area are the concept of closable operator and the closed graph theorem. This matter is discussed in Kato's "Perturbation Theory of Linear Operators". Here the boundedness simplifies matters. See Problem 5.17 on p 166 (second corrected printing of second edition) which states that every bounded operator is closable. In the present case this means that $T_0$ indeed extends in a unique way to all of $E$.

For (i) there's only one way you can extend this linear operator. We want to $T$ to be continuous, which in this case means that whenever $\|x_n - x\| \to 0$ then $\|T x_n - Tx\| \to 0$. Uniqueness should follow because sequences can't converge to two different points

For (ii) try showing that $$\sup \{ \|T_0 x \| : \|x\| \leq 1 \mbox{ and } x \in E_0 \} = \sup \{ \|T x \| : \|x\| \leq 1 \mbox{ and } x \in E_0 \}.$$ This should work because the idea behind supremums is that they are already limits.

• I disagree with (i). To say that $T$ is continuous means that $$\|x_n-x\|\to 0 \Rightarrow \|Tx_n-Tx\|\to 0,$$ which is different from what you wrote. Moreover, you are downplaying the difficulty of the problem, since you need to construct $T$, you are not given it a priori. – Giuseppe Negro Oct 24 '14 at 15:31
• I don't really understand why showing the two supremums are the same means the norms are the same? – Victoria Oct 25 '14 at 14:05
• @GiuseppeNegro: Chris Cave wrote "...one way you can extend...". So he's not assuming the extension being given. – C-Star-W-Star Oct 25 '14 at 15:46
• @Freeze_S: Yes, but it is not obvious that such an extension makes sense. (Of course it becomes obvious after you have done this exercise). A point $x$ can be approximated by many sequences $x_n\in E_0$, and in principle each one of them leads to a different value for $Tx$. One needs to prove that this does not happen and this is precisely the main difficulty of the extension process. – Giuseppe Negro Oct 25 '14 at 16:37
• @Victoria, the norm of a linear operator $T \colon E \to F$ is defined as $\|T \|= \sup \{ \|Tx\|_{F} : \|x\| \leq 1\}$. – Chris Cave Oct 27 '14 at 8:01