$\|Tx_n+ \lambda Ax_n\|\geq \|Tx_n\|$. Therefore, letting $n\rightarrow \infty$, we have, $\|Tx+ \lambda Ax\|\geq \|Tx\|$. This completes the proof of the fact that $Ax\in (Tx)^+$.\
Similarly, considering the compact operators $T-\frac{1}{n} A,$ it is now easy to see that there exists $y\in M_T$ such that $Ay\in (Ty)^-$. This completes the proof.

Theorem 2.2 of [@S] now follows easily as a simple consequence of the above theorem since every finite dimensional normed linear space is reflexive and every bounded linear operator there is a compact linear operator.

For bounded linear operators defined on a normed linear space, the situation is far more complicated since in this case the norm attainment set may be empty. In the next proposition, we give a sufficient condition for Birkhoff-James orthogonality of bounded linear operators.

Let $\mathbb{X}$ and $\mathbb{Y}$ be normed linear spaces. Let $T$, $A\in \mathbb{B}(\mathbb{X}, \mathbb{Y})$. Suppose there exists two sequences $\{x_n\}$ and $\{y_n\}$ in $S_\mathbb{X}$ satisfying the following two conditions:\
$(i)\| Tx_n\| \rightarrow \|T\|$ and $\|Ty_n\|\rightarrow\|T\|$, as $n \rightarrow \infty $\
$(ii)Ax_n\in(Tx_n)^+$ and $Ay_n\in(Ty_n)^-$ for all $n\in \mathbb{N}$.\
Then $T\bot_B A$.

Since $Ax_n\in (Tx_n)^+$, for any $\lambda \geq 0$ we have, $\|T+ \lambda A\| \geq \|Tx_n+ \lambda Ax_n\| \geq \|Tx_n\|$ for all $n \in 