Ranges square root compact operator Let $A$ and $B$ two self-adjoint positive compact operators defined on a Hilbert space $H$ to itself, such that
$$
\mathcal{R}(A^{1/2})\bigcap\mathcal{R}(B^{1/2})=\{0\}
$$
Let $g$ an eigenfunction of $B$, is it true that $g\not\in\mathcal{R}(A^{1/4})$?
 A: It seems the answer could be yes or no. As I discussed in my previous answer, if $A$ and $B$ are finite rank, for example, then the answer is yes. On the other hand, for $\alpha,\delta>0$, suppose
$$A= \sum_{i=1}^\infty i^{-\alpha} v_i \otimes v_i,$$ for orthonormal $v_i$, and let $B= g \otimes g$, where $g=\sum_{i=1}^\infty i^{(1+\delta)/2} v_i$. As such $g$ is a rank 1 operator with one eigenfunction proportional to $g$. Now so long as $\alpha/2-\delta -1>-1$, i.e. $\alpha> 2\delta$, then
$$g \notin R(A^{1/2}) =\left\{x \in H\; : \; x= \sum_{i=1}^\infty c_i v_i, \; \sum_{i=1}^\infty c_i^2 i^{\alpha/2}<\infty \right\}.  
$$
This follows since the eigenvalues of $A^{1/2}$ are $i^{-\alpha/2}$, and with $c_i = i^{(1+\delta)/2}$, so $g=\sum_{i=1}^\infty c_i v_i$,
$$
\sum_{i=1}^\infty c_i^2i^{\alpha/2}= \sum_{i=1}^\infty  i^{\alpha/2-1-\delta} = \infty. 
$$
So in this case
$$
R(B^{1/2})\cap R(A^{1/2})=0. 
$$
Following the same line of reasoning, since the eigenvalues of $A^{1/4}$ are $i^{-\alpha/4}$, we have if  $\alpha/4-\delta -1>-1$, i.e. $\alpha> 4\delta$, then $g\notin R(A^{1/4})$. If though $4 \delta > \alpha > 2 \delta$, then $g\in R(A^{1/4})$, whilst still $R(B^{1/2})\cap R(A^{1/2})=0$.
