there exists $\{x_n\}, \{y_n\}$ in $ S_\mathbb{X}$ such that\
$(i)|f(x_n)| \rightarrow \|f\|$ and $ |f(y_n)| \rightarrow \|f\|$ as $n \rightarrow \infty$\
$(ii)f(x_n).g(x_n) \geq 0$ and $f(y_n).g(y_n) \leq 0$ for all $n\in \mathbb{N}$.

If $ \mathbb{X} $ is reflexive then every bounded linear functional on $ \mathbb{X}, $ by virtue of being compact, attains norm. In that case, considering $ \mathbb{Y} = \mathbb{R} $ in Theorem  \[theorem:compact\] we get a complete description of the Birkhoff-James orthogonality of bounded linear functionals on $ \mathbb{X}. $ In fact, the following reformulation of Theorem  \[theorem:compact\] is worth mentioning in this context:

Let $\mathbb{X}$ be a reflexive Banach space. Then for any $f, g \in \mathbb{X}^*$, $f\bot_B g$ if and only if there exists $ x, y \in M_f $ such that $ f(x).g(x) \geq 0 $ and $ f(y).g(y) \leq 0. $ In addition, if $ \mathbb{X} $ is strictly convex then $f\bot_B g $ if and only if there exists $ x \in M_f $ such that $ f(x).g(x) = 0. $

However, if $ \mathbb{X} $ is not reflexive then the norm attaining set of a bounded linear functional may be possibly empty and Theorem  \[theorem:compact\] is no longer applicable. In fact, 