Let M be a Von Neumann Algebra, an automorphism is said to be inner if it is of the form $Ad(u): x \to uxu^*$ for some unitary $ u \in M$ When M is a factor, show that
$\alpha \in Aut(M)$ is inner iff $\exists 0 \neq y \in M$ such that $y \alpha(x)=xy$ for every $x \in M$.
Any help would be appreciated.
Suppose that $\alpha$ is inner. Then $\alpha=u\cdot u^*$ for some unitary $u$, and $$ u^*\alpha(x)=xu^* $$ for all $x\in M$.
Conversely, if there exists $y\in M$ such that $y\alpha(x)=xy$ for all $x\in M$, let $y=vr$ be the polar decomposition, with $r=(y^*y)^{1/2}$. We have $$ y^*y\alpha(x)=y^*xy=(x^*y)^*y=(y\alpha(x^*))^*y=\alpha(x)y^*y. $$ As this works for all $x\in M$ and $\alpha$ is an automorphism, we get that $y^*y\in\mathcal Z(M)=\mathbb C I$. It follows that $r\in\mathbb C I$ too. So now we have that $$ v^*\alpha(x)=xv,\ \ \ x\in M. $$ Using the same trick as before we find that $vv^*\in\mathbb C I$; as it is a projection and $y\ne 0$, we have $vv^*=I$. Then $$ \alpha(x)=vxv^*,\ \ \ x\in M. $$ Also, $$ \alpha(v^*v)=vv^*vv^*=vv^*=I. $$ As $\alpha$ is an automorphism, $v^*v=I$ and so $v$ is a unitary.