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!


3 Answers 3


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).


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.

  • $\begingroup$ 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. $\endgroup$ Oct 24, 2014 at 15:31
  • $\begingroup$ I don't really understand why showing the two supremums are the same means the norms are the same? $\endgroup$
    – Victoria
    Oct 25, 2014 at 14:05
  • $\begingroup$ @GiuseppeNegro: Chris Cave wrote "...one way you can extend...". So he's not assuming the extension being given. $\endgroup$ Oct 25, 2014 at 15:46
  • $\begingroup$ @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. $\endgroup$ Oct 25, 2014 at 16:37
  • $\begingroup$ @Victoria, the norm of a linear operator $T \colon E \to F$ is defined as $\|T \|= \sup \{ \|Tx\|_{F} : \|x\| \leq 1\}$. $\endgroup$
    – Chris Cave
    Oct 27, 2014 at 8:01

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