Obviously $\langle \psi, A \phi \rangle \leq ||A|| \hspace{1pt} \|\psi\| \|\phi\|$ but is it true that $\langle \psi, A \phi \rangle \leq ||A|| \langle \psi, \phi \rangle$ ?
Since $||A|| = \sup\limits_{x \in \mathscr{H}\setminus \lbrace 0 \rbrace } \frac{|Ax|}{|x|} $ one might think that $\langle \psi, A \phi \rangle \leq ||A|| \langle \psi, \phi \rangle$, but what if $A$ takes $\phi$ to a nonorthogonal vector to $\psi$, but $\phi$ and $\psi$ are orthogonal ? then we would have $0 < \langle \psi, A \phi \rangle \leq ||A|| \cdot 0 = 0$ which is absurd, no?
This question arose because of a article I was reading that proved something for $\langle \psi, \phi \rangle$ and another lemma that if $\langle \psi, A \phi \rangle =0 $ was true for all $A$ then $\psi$ and $\phi$ could be called "disjoint", then it uses this lemma to say that since $\langle \psi, \phi \rangle=0$ then they are "disjoint", this would imply $\langle \psi, A \phi \rangle =0 $ only if $\langle \psi, A \phi \rangle \leq||A|| \langle \psi, \phi \rangle$.
No. Let $A:\mathbb{R}^2\to \mathbb{R}^2$ be a transformation taking $e_1$ to $e_2$. Then $$ \langle e_2, Ae_1\rangle =\langle e_2, e_2 \rangle=1 $$ while $$ ||A||\langle e_2,e_1 \rangle=0. $$