Norm of AB+BA when A, B are well understood Here is the situation:
I have the operator/matrix expression AB+BA and want to know the norm.
1.) B is compact, so I believe AB+BA is compact (as are AB and BA, so any finite-dimensional matrix answer should be fine, or at least informative)
2.) A is both unitary and self-adjoint (and therefore A^2 = I)
3.) AB+BA itself is self-adjoint, because I also know that AB and BA are self-adjoint. 
4.) I know the norm, eigenvalues, basically everything about both A and B (and A* and B*), AND I know all of this about AB and BA (and their adjoints) as well. An expression in terms of the norms/eigenvalues of those four operators would give me everything I want. 
5.) A sharp inequality would be okay but really I need to find the exact expression for the norm or the max eigenvalue. 

In finite dimension $n$.
According to 2), $A$ is an orthogonal symmetry and is orthogonally similar to $diag(I_p,-I_{n-p})$.
We assume that $A=diag(I,-I)$ and $B=\begin{pmatrix}P&Q\\R&S\end{pmatrix}$. 
Then $AB+BA=2diag(P,-S)$.
Since $AB$ is self adjoint, $P=P^*,S=S^*,Q^*=-R$. 
We obtain, $||AB+BA||_2=\rho(AB+BA)=2\max(\rho(P),\rho(S))$.
EDIT 1. Let $E_1=\ker(A-I_n),E_{-1}=\ker(A+I_n)$.
Clearly, $M_1=\max_{x\in E_1,||x||_2=1}|x^*Bx|=\rho(P)$, $M_{-1}=\max_{x\in E_{-1},||x||_2=1}|x^*Bx|=\rho(S)$.
Finally, $||AB+BA||_2= 2\max(M_1,M_{-1})$.
EDIT 2. We can also use $\rho(P)=1/2\rho((A+I)B)$ and $\rho(S)=1/2\rho((A-I)B)$.
