Assume we have 2 injective continuous operators with dense images $A$ and $B$ on a Hilbert space $\mathbb H$ and $B$ is self adjoint. Further let there be constants $a_1$ and $a_2$ such that-
$a_1\|Bu\| \leq \|Au\| \leq a_2\|Bu\|$ for all $u \in \mathbb H$.
Can we show that for $A^{*}$, the adjoint of $A$, we have -
$a_1\|Bu\| \leq \|A^{*}u\| \leq a_2\|Bu\|$ for all $u \in \mathbb H$.
This is obviously true when $A$ is normal. Is it true for non normal operators?
Consider in $\Bbb{R}^2$,$B(v_1)=3v_1$ and $B(v_2)= 12 v_2$. Let $A(v_1) = v_2$ and $A(v_2)= 4v_1$ where $v_1 = (1,0)$ and $v_2=(0,1)$ Note that $$|A(v)|=\frac{1}{3}|B(v)|$$
So we conclude that $$a_1\|Bu\| \leq \|Au\| \leq a_2\|Bu\| $$
with $a_1 = \frac{1}{3}$ and $a_2 = \frac{1}{3}$
Now note that $A*(v_1)= 4v_2$
$$|A^*(v_1)|\nleq \frac{1}{3} B(v_1)$$
Hence it does not follow that $$a_1\|Bu\| \leq \|A^{*}u\| \leq a_2\|Bu\| $$
Note: and $AA^* \neq A^*A$. $A$ is not a normal operator.
