Welcome to our community

Be a part of something great, join today!

[SOLVED] inverse operator limit

  • Thread starter
  • Banned
  • #1

Boromir

Banned
Feb 15, 2014
38
Let $T_{n}$ be a sequence of invertible bounded linear operators with limit $T$ Prove that $(T_{n})^{-1}$ tends to $T^{-1}$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Let $T_{n}$ be a sequence of invertible bounded linear operators with limit $T$ Prove that $(T_{n})^{-1}$ tends to $T^{-1}$
This is not true without the additional assumption that the limit operator $T$ is invertible (in general it need not be).

As a hint, notice that $T_n^{-1}-T^{-1} = T_n^{-1}(T-T_n)T^{-1}$.
 
  • Thread starter
  • Banned
  • #3

Boromir

Banned
Feb 15, 2014
38
This is not true without the additional assumption that the limit operator $T$ is invertible (in general it need not be).

As a hint, notice that $T_n^{-1}-T^{-1} = T_n^{-1}(T-T_n)T^{-1}$.
How do you get that equality?

Once I have got that equality, take the norm, then $T-T_{n}$ tends to zero. Though what happens to the $T_{n}^{-1}$? Its not neccesarily bounded even though individually they are.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
... what happens to the $T_{n}^{-1}$? Its not neccesarily bounded even though individually they are.
Good catch, I was being careless there. :eek:

I think what you need to do is something like this. For $n$ large enough, $\|T-T_n\| < \frac12\|T^{-1}\|^{-1}$. It follows that $\|I - T^{-1}T_n\| = \|T^{-1}(T-T_n)\| \leqslant \|T^{-1}\|\|T-T_n\| <\frac12$. It follows from the Neumann series that $T^{-1}T_n$ is invertible, with $\|(T^{-1}T_n)^{-1}\| = \|T_n^{-1}T\| <2.$ Thus $\|T_n^{-1}\| = \|T_n^{-1}TT^{-1}\| \leqslant \|T_n^{-1}T\|\|T^{-1}\| <2\|T^{-1}\|.$
 
  • Thread starter
  • Banned
  • #5

Boromir

Banned
Feb 15, 2014
38
Good catch, I was being careless there. :eek:

I think what you need to do is something like this. For $n$ large enough, $\|T-T_n\| < \frac12\|T^{-1}\|^{-1}$. It follows that $\|I - T^{-1}T_n\| = \|T^{-1}(T-T_n)\| \leqslant \|T^{-1}\|\|T-T_n\| <\frac12$. It follows from the Neumann series that $T^{-1}T_n$ is invertible, with $\|(T^{-1}T_n)^{-1}\| = \|T_n^{-1}T\| <2.$ Thus $\|T_n^{-1}\| = \|T_n^{-1}TT^{-1}\| \leqslant \|T_n^{-1}T\|\|T^{-1}\| <2\|T^{-1}\|.$
I don't understand how this implies $||T_{n}^{-1}-T^{-1}||$->0.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Good catch, I was being careless there. :eek:

I think what you need to do is something like this. For $n$ large enough, $\|T-T_n\| < \frac12\|T^{-1}\|^{-1}$. It follows that $\|I - T^{-1}T_n\| = \|T^{-1}(T-T_n)\| \leqslant \|T^{-1}\|\|T-T_n\| <\frac12$. It follows from the Neumann series that $T^{-1}T_n$ is invertible, with $\|(T^{-1}T_n)^{-1}\| = \|T_n^{-1}T\| <2.$ Thus $\|T_n^{-1}\| = \|T_n^{-1}TT^{-1}\| \leqslant \|T_n^{-1}T\|\|T^{-1}\| <2\|T^{-1}\|.$
I don't understand how this implies $||T_{n}^{-1}-T^{-1}||$->0.
It answers your criticism of my earlier comment by showing that (for $n$ large enough) $\|T_n^{-1}\|$ has a uniform bound $2\|T^{-1}\|$. That earlier comment then gives you the hint for proving that $T_{n}^{-1}-T^{-1} \to0.$
 
  • Thread starter
  • Banned
  • #7

Boromir

Banned
Feb 15, 2014
38
It answers your criticism of my earlier comment by showing that (for $n$ large enough) $\|T_n^{-1}\|$ has a uniform bound $2\|T^{-1}\|$. That earlier comment then gives you the hint for proving that $T_{n}^{-1}-T^{-1} \to0.$
that makes sense now haha