AnswerBun.com

Why is identity map on a separable Hilbert space not compact? False proof.

Mathematics Asked on January 5, 2022

Why is identity map on a separable Hilbert space not compact? False proof.

Let $e_n$ be the orthonormal basis. Then the projection map onto $H$ is defined by $sum(x,e_k)e_k$. What is stopping us from taking a finite part of this series. This gives us a compact operators, that converge to the identity map. Clearly something is wrong here not all Hilbert spaces are finite dimensional.

One Answer

Beacause that sequence of operators does not converge to the identity: if $ninBbb N$,$$leftlVert e_{n+1}-sum_{j=1}^nlangle e_{n+1},e_krangle e_krightrVert=lVert e_{n+1}rVert=1$$and therefore$$leftlVertoperatorname{Id}-sum_{k=1}^nlanglecdot,e_krangle e_krightrVertgeqslant1.$$

Answered by José Carlos Santos on January 5, 2022

Add your own answers!

Related Questions

complex norm inequality

1  Asked on December 6, 2020 by stranger

     

Intersection of a quadratic and a plane is a quadratic?

0  Asked on December 6, 2020 by twosigma

 

Is $x^2$ analytic in $mathbb{R}$

1  Asked on December 5, 2020 by joey

 

Understanding Seifert Van Kampen

0  Asked on December 5, 2020 by moooose

   

Trying to evaluate a complex integral?

4  Asked on December 5, 2020 by gray

   

Ask a Question

Get help from others!

© 2023 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir