why is it a compact operator let $B(H)$ the $*-$algebra of bounded linear functions on a separable $\mathbb{C}$-Hilbert space and $\tau:B(H)\to B(H)$ a bounded linear multiplicative map with $\tau(x^*)=\tau(x)^*$ for all $x\in B(H)$.
Given $S\in B(H)$ such that: $S\tau(x)-\tau(x)S$, $(S^2-id_H)\tau(x)$, and $(S^*-S)\tau(x)$ are compaxt for all $x\in B(H)$.

Is $((\frac{S+S^*}{2})^2-id_H)\tau(x)$ compact for all $x\in B(H)$? 

I think so, but I'm not sure how to proof it.
It can be possible that there are redundant assumptions, I'm not sure what will be needed. It is $((\frac{S+S^*}{2})^2-id_H)\tau(x)=(\frac{S^2+(S^*)^2+2SS^*}{4}-id_H)\tau(x)$ and it follows from the assumptions that $S^*\tau(x)-\tau(x)S^*$, $((S^*)^2-id_H)\tau(x)$, and $(S-S^*)\tau(x)$ are compaxt for all $x\in B(H)$.
But I don't know how to proceed. I was thinking about to use spectral theorems or some functional calculus arguments, but still haven't solved it.
 A: The proposed operator, which I will call $\DeclareMathOperator{\id}{id} M$, is compact iff $H$ is finite-dimensional.  
To see this, first note that the adjoint of a compact operator is compact, and vice-versa, so the conditions on $S$ are preserved if we swap $S\mapsto S^*$.  That is, $S*\tau(x)−\tau(x)S^*$, $((S^*)^2−\id_H)\tau(x)$, and $(S−S^*)τ(x)$ are all compact.  
Now consider the operator $\tilde{M}=\left(\frac{(S-S^*)^2}{2}-\id_H\right)\tau(x)$.  We may rewrite $$\tilde{M}=\left(\frac{S-S^*}{2}\right)(S-S^*)\tau(x)-\id_H\tau(x)$$ and recall that the product of a compact operator and a continuous operator is compact.  (Why?  Take $K$ the compact operator and $F$ the continuous one.  To show $FK$ is compact: the image of $K$ is compact, and $F$, being continuous, maps (pre)compact spaces to (pre)compact spaces.  To show $KF$ is compact: $\DeclareMathOperator{\im}{im} \im{(F)}\subseteq H$, so $\im{(KF)}\subseteq\im{(K)}$, which is precompact.)  Given this recollection, the first term in our expansion is compact, as it is the product of continuous $\left(\frac{S-S^*}{2}\right)$ with compact $(S-S^*)\tau(x)$.  The second term is compact iff $H$ is finite-dimensional.  So $\tilde{M}$ is compact iff $H$ is finite-dimensional.  
Now expand $\tilde{M}$ another way: \begin{align*}
\tilde{M}&=\left(\frac{(S-S^*)(S-S^*)}{2}-\id_H\right)\tau(x) \\
&=\left(\frac{S(S-S^*)-S^*(S-S^*)}{2}-\id_H\right)\tau(x) \\
&=\left(\frac{S^2-SS^*-S^*S+(S^*)^2)}{2}-\id_H\right)\tau(x) \\
&=\left(\frac{(S^2-\id_H)-SS^*-S^*S+((S^*)^2-\id_H)}{2}\right)\tau(x) \\
&=\left(\frac{(S^2-\id_H)\tau(x)}{2}\right)+\left(\frac{((S^*)^2-\id_H)\tau(x)}{2}\right)-\left(\frac{(SS^*+S^*S)\tau(x)}{2}\right)
\end{align*}  The first two terms of this sum are compact, but $\tilde{M}$ is not and compact operators are closed under addition.  So the remaining term is a compact operator iff $H$ is finite-dimensional.  
Now return to $M$.  We can expand $M$ similarly: \begin{align*}
\tilde{M}&=\left(\frac{(S+S^*)(S+S^*)}{2}-\id_H\right)\tau(x) \\
&=\left(\frac{S^2+SS^*+S^*S+(S^*)^2)}{2}-\id_H\right)\tau(x) \\
&=\left(\frac{(S^2-\id_H)+SS^*+S^*S+((S^*)^2-\id_H)}{2}\right)\tau(x) \\
&=\left(\frac{(S^2-\id_H)\tau(x)}{2}\right)+\left(\frac{((S^*)^2-\id_H)\tau(x)}{2}\right)+\left(\frac{(SS^*+S^*S)\tau(x)}{2}\right)
\end{align*}  Again, the first two terms must be compact, so $M$ is compact iff the last term is compact.  But we just showed that that last terms is compact iff $H$ is finite-dimensional, so we've proven our claim.  
