\[theorem:rotund\] is not sufficient to characterize Birkhoff-James orthogonality of bounded linear operators.

\[example:not rotund\] Define $T, A : \ell_1 \rightarrow \ell_1$ by $ Te_1 = \frac{1}{2}e_1,~Te_n = (1- \frac{1}{n^4})e_n$, $n\geq 2$, and $ Ae_1 = \frac{1}{2}e_1,~Ae_n = \frac{1}{n^2} (\frac{1}{n^2}-1)e_n$, $n\geq 2$, where $e_n =(0,0, \ldots ,0,1,0, \ldots ),$ with $1$ in the $n$-th position and $0$ elsewhere. Then both $T$ and $A$ are bounded linear operators on $\ell_1$ with $\|T\| =1.$ Also $ T \bot_B A$, since $\|T + \lambda A \| \geq 
\| (T + \lambda A)e_n \| = \| (1 - \frac{1}{n^4})e_n + \lambda \frac{1}{n^2}(\frac{1}{n^2}-1) e_n \| = 
\mid 1 - \frac{1}{n^4} + \frac{\lambda}{n^4} - \frac{\lambda}{n^2} \mid \longrightarrow 1 = \|T\|.$\
Now for each $x = (a_1,a_2,\ldots,\dots) \in S_{\ell_1} $ we have $$\begin{aligned}
\|Tx + Ax\| & = & \| (a_1, (1-\frac{1}{2^2})a_2,\ldots,(1-\frac{1}{n^2})a_n,\ldots ) \| \\
            & = & \mid a_1 \mid + \mid a_2 \mid (1-\frac{1}{2^2}) + \ldots + \mid a_n \mid (1-\frac{1}{n^2}) +                  \ldots \\
                        & = & \mid \frac{a_1}{2} \mid + \mid a_2 \mid (1-\frac{1}{2^4}) + \ldots + \mid a_n \mid (1-\frac{1}{n^4}) +                  \ldots \\
                        &   & + \mid \frac{a_1}{2} \mid + \mid a_2 \mid (\frac{1}{2^4}-\frac{1}{2^2}) + \ldots + \mid a_n \mid                    (\frac{1}{n^4}-\frac{1}{n^2}) +                  \ldots \\
                        