Embedding: Extension

Question

Problem

Given Banach spaces $E_0$ and $E$

Regard dense domain: $$\overline{\mathcal{D}_0}=E_0\quad\overline{D}=E$$

Consider an embedding: $$\Phi:\mathcal{D}_0\hookrightarrow\mathcal{D}:\quad\|\Phi(\varphi)\|\leq\|\varphi\|_0$$

Then one is tempted: $$\overline{\Phi}:E_0\hookrightarrow E:\quad\|\overline{\Phi}(\varphi)\|\leq\|\varphi\|_0$$

Does this really follow?*

*I mean injectivity.

Reference

I'm trying to check: Embedding

Since the function is uniformly continuous then we can extend it on the comletion metric space to a continuous mapping. It remains to prove that it is linear. A feeling exists that this a routine matter. — Adelafif, Jul 11 '15 at 07:08

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

Thanks alot to T.A.E.!!

Counterexample

Given the Hilbert space $\ell^2(\mathbb{N}_0)$.

Consider the domain: $$\mathcal{D}_0:=\langle\{e_0+\frac{1}{n}e_n:n\in\mathbb{N}\}\rangle$$

It is dense since: $$e_0=\lim_n(e_0+\frac{1}{n}e_n):\quad e_0\in\overline{\mathcal{D}_0}\Rightarrow e_n\in\overline{\mathcal{D}_0}$$

Define the operator:* $$L_0:\mathcal{D}_0\to\ell^2(\mathbb{N}_0):\quad (L\varphi)(k):=\varphi(k+1)$$

Then it is contractive: $$\|L_0\varphi\|=\|L\varphi\|\leq\|\varphi\|$$

And it is injective: $$e_0\notin\mathcal{D}_0\implies\mathcal{N}L_0=(0)$$

But for its closure it is: $$\overline{L_0}e_0=Le_0=0$$

That gives a counterexample!

*Nothing but left-shift!

Reference

See the thread: Construction

1 Answers1