Let $A,B,C$ be selfadjoint operators of $L^2(\mathbb{R}^n)$ ($A$ and $B$ unbounded), $A\geq 0$, $B \geq 0$, with $\sqrt{A} C$ and $\sqrt{B} C$ bounded. Is the following inequality true for some constant $c \geq 0$, where $\left \! \left \cdot \right \! \right$ is the operator norm, \begin{align*} \left \! \left \sqrt{A+B} C \right \! \right \leq c \left \! \left \sqrt{A} C \right \! \right + c \left \! \left \sqrt{B} C \right \! \right ? \end{align*}

$\begingroup$ Should $A$ be $B$ in the last term? $\endgroup$– Nate EldredgeJul 8, 2020 at 13:34

$\begingroup$ Yes thanks (and sorry) $\endgroup$– user140442Jul 8, 2020 at 13:36

1$\begingroup$ Also, do you need some more assumptions to ensure that $A+B$ is self adjoint? $\endgroup$– Nate EldredgeJul 8, 2020 at 13:37

$\begingroup$ Yes you can assume that $\endgroup$– user140442Jul 8, 2020 at 13:38
2 Answers
$$ \\sqrt{A+B}Cx\^2=(\sqrt{A+B}Cx,\sqrt{A+B}Cx)=\\ ((A+B)Cx,Cx)=(ACx,Cx)+(BCx,Cx)=\\sqrt{A}Cx\^2+\\sqrt{B}Cx\^2, $$ taking the supremum over unit vectors $x$ we get $$ \\sqrt{A+B}C\^2\leqslant \\sqrt{A}C\^2+\\sqrt{B}C\^2\leqslant (\\sqrt{A}C\+\\sqrt{B}C\)^2. $$
For any unit vector $x\in L^2(\mathbb{R}^n)$, $$\\sqrt{A+B}\,Cx\^2=(\sqrt{A+B}\,Cx,\sqrt{A+B}\,Cx) =(ACx,Cx)+(BCx,Cx)=\\sqrt A\,Cx\^2+\\sqrt B\,Cx\^2 \le(\\sqrt A\,Cx\^2+\\sqrt B\,Cx\)^2 \le(\\sqrt A\,C\+\\sqrt B\,C\)^2.$$ So, $$\\sqrt{A+B}\,C\\le\\sqrt A\,C\+\\sqrt B\,C\.$$