Strong limit from the unitary orbit of $A$

Question

Let $A,X\in L(H)$ with $\Bbb D\subset W_e(A)$ ,where $W_e(A)$ is the essential numerical range of $A$,and $\|X\|\leq 1$. Then there exists a sequence of unitaries $(U_n)_n$ in $L(H)$ such that

wot $lim_{n\to \infty}U_nAU_n^*=X$.

The above conclusion is from Bourin and Lee ' s paper "Pinchings and positive linear maps".

In that paper, it mentions that the above result is not true if we replace the weak convergence by the strong convergence.

For instance, if $A$ is invertible and $\|Xh\|＜\|A^{-1}\|^{-1}$ for some unit vector $h$, then $X$ cannot be a strong limit from the unitary orbit of $A$.

If there exists a sequence of unitaries $(U_n)_n$ in $L(H)$ such that SOT $lim_{n\to \infty}U_nAU_n^*=X$, how to get the contradiction?

Answer 1

I don't immediately see to get a contradiction the way it is stated. But for instance you could take $A$ to be a unitary with $\sigma_e(A)=\mathbb T$, and $X$ any contraction that is not an isometry. Then there exists a unit vector $h$ with $\|Xh\|<1$, while $$\|U_nAU_n^*h\|=\|h\|=1$$ for all $n$. So $U_nAU_n^*$ does not converge SOT to $X$.