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- Jan 26, 2012

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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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Instead of directingPlease read the question again.

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$||T||=1$ means that $||Tv|| \le ||v||$ for all $v$.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

So with $||x||>||T||=1$, what follows, is that $||Tx|| \le ||x||$.

It does

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

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.