# Silly question but is it true that $\langle \psi, A \phi \rangle \leq ||A|| \langle \psi, \phi \rangle$?

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

• 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. May 28 at 22:35
• 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. May 29 at 0:58

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 $$||A||\langle e_2,e_1 \rangle=0.$$
• Do you know any inequality relating $\langle e_1, e_2\rangle$ to $\langle e_1, Ae_2 \rangle$ ? May 28 at 22:18