Suppose we have self-adjoint operators $A$ and $B$ on a Hilbertspace $H$ and $C$ denotes the commutator $C=AB-BA$. My task is the following:
Show that $iC$ is self-adjoint on $H$ and that $|(Cx,x)| \leq 2*||Ax||*||Bx||$ holds for all $x \in H$. Moreover find out for which $x$ the inequality becomes an equality.
To show that $iC$ is self-adjoint my attempt was to calculate $(iCx,x)$ straightforward using that $(Ax,x)=(x,Ax)$ and $(Bx,x)=(x,Bx)$ by the self-adjointness of A and B. This lead to
$$(iCx,x) \\= (i(AB-BA)x,x) \\= (iABx-iBAx,x) \\ = (iABx,x)-(iBAx,x) \\ = (iAx,Bx)-(iBx,Ax)$$ but now I don't now how to go on since I can't do $(iAx,Bx) = (ix,BAx)$ or am I wrong here?
For the inequality I use that $ (Ax,y) \leq ||Ax||*||y|| \leq ||A||*||x||*||y||$ and $||L(x)|| \leq ||L||*||x||$.
This leads to
$$ |(Cx,x)| = |(ABx-BAx,x)| = |(ABx,x)-(BAx,x)| \\ \leq |(ABx,x)| + |(BAx,x)| \;\;\;\; \operatorname{(Triangle-inequality)} \\ \leq ||A||*||Bx||*||x||+||B||*||Ax||*||x|| \\ \leq ||A||*||B||*||x||*||x|| + ||B||*||A||*||x||*||x|| \\ = 2*||A||*||B||*||x||^2 ,$$ but this is not what I am aiming for.
Some hints where I miscalculated and some (partial) solutions would be really nice.
For the self-adjointness, you might note that (using that the scalar product is linear in one slot and antilinear in the other slot) $$\langle iCx,x\rangle = \langle Cx, -i x\rangle = \langle x, i(-C)^*x\rangle = \langle x, iCx\rangle.$$ The last equality uses $C^*=(AB-BA)^* =B^*A^* -A^*B^* = -C$.
For the norm inequality you have $$\vert \langle Cx, x\rangle \vert \leq \vert \langle ABx, x\rangle \vert + \vert \langle BAx, x\rangle \vert =\vert \langle Bx, Ax\rangle \vert + \vert \langle Ax, Bx\rangle\vert \leq 2\Vert Ax\Vert \cdot \Vert Bx\Vert,$$ where the last inequality is just Cauchy-Schwarz. Cauchy-Schwarz also tells us that we have equality iff there exists a scalar $\lambda \in \mathbb{C}$ such that $Ax=\lambda Bx$.