norm of bounded linear operator restricted to dense subspace

Question

I have no idea how to do this question I was given in class.

Let $E$ and $F$ be normed spaces and let $T \in \mathcal{L}(E,F)$. Suppose that $E_0 \subseteq E$ is a dense subspace. Show that $\parallel T_{E_0} \parallel = \parallel T \parallel$.

I know we can approximate every $x \in E$ by a sequence in $E_0$ but I can't see how to use it to get the result. Any clues would be a big help

score 3 · Answer 1 · answered Nov 27 '14 at 18:38

3

First note that, because its a subset of $E$,

$$||T||_{E_0} \leq ||T||$$

On the other hand given $\varepsilon >0$, there is some $x \in E$ such that $$||T|| \leq ||Tx|| + \varepsilon $$

And because $E_0$ is dense, and $T$ is continuous, there is an $y \in E_0$ near $x$ such that

$$||Tx|| - ||Ty|| \leq ||||Tx|| - ||Ty|||| \leq ||Tx - Ty|| < \varepsilon$$

Froms this we get

$$||T|| \leq ||Ty|| +2\varepsilon$$

We finish by letting $\varepsilon \rightarrow 0$

answered Nov 27 '14 at 18:38

aram

1,976

I'm not really sure I understand why this last part gives the result? – Victoria Nov 27 '14 at 19:07
@Victoria It means $||T|| = \sup{||Ty|| \colon y \in E_0 \land ||y|| = 1}$ – aram Nov 27 '14 at 19:12
I think I get it now, thanks for your help – Victoria Nov 27 '14 at 19:23

norm of bounded linear operator restricted to dense subspace

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