said to belong to the approximate point spectrum of $A$, written as $\sigma_{app} (A),$ if there exists a sequence $\{x_n\} $ of unit vectors such that $ \|(A-\lambda I)x_n \| \longrightarrow 0.$ In the following theorem we show that if $0 \notin \sigma_{app} (A),$ then the notions $T \bot_B A$ and $T \bot_{SB} A$ are equivalent.

\[theorem:hilbert\] Let $\mathbb{H}$ be a Hilbert space, $T,A \in \mathbb{B}(\mathbb{H})$ and $0 \notin \sigma_{app} (A)$. Then $T \bot_B A$ and $T \bot_{SB} A$ are equivalent.

Clearly $T \bot_{SB} A $ implies $T \bot_{B} A.$ On the other hand, if possible suppose that $T \bot_{B} A $ but $T\not \perp_{SB} A$. Then there exists a nonzero scalar $\lambda_0 \in \mathbb{R}$ such that $\|T\|= \|T- \lambda_0 A\|$. Now, for all $\lambda \in \mathbb{R}$, $\|T- \lambda_0 A\| \leq \|T + \lambda A\|= \|(T- \lambda_0 A)+ (\lambda + \lambda_0) A\|$. Hence $(T- \lambda_0 A) \bot_B A$. Then from Paul [@P], it follows that there exists $\{x_n\}\subseteq S_{\mathbb{H}}$ such that $\|(T-\lambda_0 A)x_n\| \rightarrow \|T- \lambda_0A\|$ and $\langle (T- \lambda_0 A)x_n, Ax_n \rangle \rightarrow 0$. Now, using the fact that $\|Ax_n\|\not\rightarrow 0$ (since $ 0 \notin \sigma_{app} (A)$) we have, $\|T- \lambda_0 A\|^2 = lim_{n \rightarrow \infty} \|(T- \lambda_0 A)x_n\|^2= 