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topology question: bounded linear operator

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Poirot

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Feb 15, 2012
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Let T be a bounded linear operator on X and x be in X. Assume ||x||>||T||=1 It follows that ||Tx||>1. Can anyone explain why 0 is not in the closure of T(B), where B is the ball centered at x, with radius 1. Thanks
 

Ackbach

Indicium Physicus
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Jan 26, 2012
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Re: topology question

Counterexample: let $X=\mathbb{R}^{2}$, and let $x= \langle 2,0 \rangle$, and let $T$ be the operator that rotates through an angle $\theta$. That is,
$$T= \begin{bmatrix} \cos( \theta) &-\sin( \theta) \\ \sin( \theta) & \cos( \theta) \end{bmatrix}.$$
Then $T$ is bounded and linear, with norm 1, and maps the ball $B_{1}(x)$ to another ball of radius 1, but it will not include zero. The rotated ball will always be a distance of 1 away from zero.
 
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Poirot

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Feb 15, 2012
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Re: topology question

Please read the question again.
 

MarkFL

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Feb 24, 2012
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Re: topology question

Please read the question again.
Instead of directing Ackbach to reread the question, can you demonstrate why his counterexample does not address your question?
 

Klaas van Aarsen

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Mar 5, 2012
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Re: topology question

Let T be a bounded linear operator on X and x be in X. Assume ||x||>||T||=1 It follows that ||Tx||>1. Can anyone explain why 0 is not in the closure of T(B), where B is the ball centered at x, with radius 1. Thanks
$||T||=1$ means that $||Tv|| \le ||v||$ for all $v$.

So with $||x||>||T||=1$, what follows, is that $||Tx|| \le ||x||$.
It does not follow that $||Tx|| > 1$.


In particular that means that it is possible for 0 to be in the closure of T(B).
Pick for instance $T: \mathbb R^2 \to \mathbb R^2, (x,y) \mapsto (x,0)$.
It is not necessarily the case that 0 is in the closure of T(B) as Ackbach has shown.


With the additional condition that $||Tx||>1$, it becomes a different matter.
For instance, the closest point to the origin is \(\displaystyle \frac{||x|| - 1}{||x||} x\).
Its image is:
$$T\left(\frac{||x|| - 1}{||x||} x\right) = \frac{||x|| - 1}{||x||} Tx$$
This point is still on the same side of the origin.