Show that $ \Vert Tx_\varepsilon -x_\varepsilon \Vert \leq \varepsilon $

Question

Exercise:

Let $X$ be a Banach space, $\Vert \quad \Vert$ is a norm on $X$. Let $B_1= \{ x \in X:\Vert x \Vert=1 \} $. If the oprator $T: B_1\rightarrow B_1$ satisfies for all $x, y\in B_1$, there is $\Vert Tx-Ty\Vert \leq \Vert x-y\Vert$. Then for all $\varepsilon \in (0,1)$, there exists $x_\varepsilon\in B_1$ such as $\Vert Tx_\varepsilon-x_\varepsilon \Vert\leq \varepsilon$.

My Attempt:

This exercise looks quite like the Fixed Point Theorem so I want to prove it by the similar way. But I failed because without $0< \theta< 1$ in the Fixed Point Theorem I can't find a Cauchy sequence. So I can't find a $x$ even is dependent on $\varepsilon$ to make $Tx$ and $x$ close enough. I'm not sure if my attempt is on a correct way.

Update:

According to @Brian-Moehring, there is a mistake in this exercise.

Maybe the $B_1$ should be a unit ball rather than a unit sphere.

Answer 1

I think $B_1$ stands for the unit ball instead of the unit sphere. In that case, consider $Ux = \frac12 Tx + \frac12 (1- \epsilon)x$. You can easily prove that the map is a contraction map and so you can find a fixed point $x_\epsilon$. $$Ux_\epsilon = x_\epsilon \implies Tx_\epsilon - x_\epsilon = \frac\epsilon2 x_\epsilon \implies \left\|Tx_\epsilon - x_\epsilon \right\|\le \frac\epsilon2 < \epsilon$$

Answer 2

In fact, if $B_1$ is a unit ball rather than a unit sphere, I can give a proof.

For all $r$ that $0<r<1$, it's clear that $rT$ is a contraction mapping from $B_1$ to $B_1$. So there is a fixed point $y$ for $rT$ such that $rT(y)=y$. Then we have $$\Vert T(y)-y\Vert =\Vert T(y)-rT(y) \Vert=(1-r)\Vert T(y) \Vert \le 1-r.$$ So given $\varepsilon$, let $r$ close enough to 1, we have $\Vert T(y_\varepsilon)-y_\varepsilon\Vert \le 1-r\le \varepsilon$. Here $y_\varepsilon$ is the fixed point of $rT$.