Correction in Example 2.9. Tayyab Kamran, Quaid-i-Azam University 20 August 2015 Replace$\|A\|=\max\{|a_1|,|a_2|,|a_3|,|a_4|\}$ by $\| A \| =\left(\sum_{i=1}^4 |a_i|^2\right)^{\frac{1}{2}}$. Replace the function $T$ by the following function: \[ T(x) = \begin{cases} \frac{x}{4} & \text{if } x \geq 0 \\ 1 & \text{if } x < 0. \end{cases}\] Then\[d(Ty,T^2y)=a^{*}d(y,Ty)a\] where \[a=\begin{bmatrix} \frac{1}{2} & 0 \\[0.3em] 0 & \frac{1}{2} \\[0.3em] \end{bmatrix} \textrm{ and } \|a\|=\frac{1}{\sqrt{2}}.\] Competing interests No competing interests.
Correction in Example 2.9.
20 August 2015
Replace
$\|A\|=\max\{|a_1|,|a_2|,|a_3|,|a_4|\}$ by $\| A \| =\left(\sum_{i=1}^4 |a_i|^2\right)^{\frac{1}{2}}$.
Replace the function $T$ by the following function:
\[ T(x) = \begin{cases} \frac{x}{4} & \text{if } x \geq 0 \\ 1 & \text{if } x < 0. \end{cases}\]
Then
\[d(Ty,T^2y)=a^{*}d(y,Ty)a\]
where
\[a=\begin{bmatrix} \frac{1}{2} & 0 \\[0.3em] 0 & \frac{1}{2} \\[0.3em] \end{bmatrix} \textrm{ and } \|a\|=\frac{1}{\sqrt{2}}.\]
Competing interests
No competing interests.