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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 $\langle \psi, A \phi \rangle \leq||A|| \langle \psi, \phi \rangle$.

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  • $\begingroup$ It's not clear what the "lemma" is that you're referring to. Merely calling $\ \psi\ $ and $\ \phi\ $ "disjoint" if $\ \langle\psi,A,\phi\rangle=0\ $ for all $\ A\ $ is just a definition, not a lemma. And $\ \langle\psi,A\phi\rangle=0\ $ for all $\ A\ $ if and only if $\ \psi=0\ $, so under that definition $\ \psi\ $ and $\ \phi\ $ would be "disjoint" if and only if one of them is zero. $\endgroup$
    – lonza leggiera
    May 28 at 22:35
  • $\begingroup$ The paper is e-periodica.ch/cntmng?pid=hpa-001:1972:45::1204 and the $\langle \psi , A \phi \rangle = 0$ is actually $\langle \psi , \pi(A) \phi \rangle = 0$ only for the $\pi(A), A \in \mathscr{A}$, the lemma is Lemma 1 and in Lemma 7 we arrive that $\langle \psi , \phi \rangle =0$ and this implies in some way that $\langle \psi , \pi(A) \phi \rangle = 0 , \forall A \in \mathscr{A}$, I am trying to understand this last implication in Lemma 7. $\endgroup$
    – Felipe Dilho
    May 29 at 0:58

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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 $$ ||A||\langle e_2,e_1 \rangle=0. $$

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  • $\begingroup$ Do you know any inequality relating $\langle e_1, e_2\rangle $ to $\langle e_1, Ae_2 \rangle$ ? $\endgroup$
    – Felipe Dilho
    May 28 at 22:18

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