Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds $$ c^- x^\top D x \le x^\top A x \le c^+ x^\top D x $$ for any $x \in \mathbb{R}^n$, where $c^+, c^- > 0.$ The matrices $A$ and $D$ can be diagonalized, that is $$ A = V\Lambda_A V^\top, \quad D = W\Lambda_D W^\top $$ where $V$ and $W$ contain the eigenvectors of $A$ and $D$, and $\Lambda_A$ and $\Lambda_D$ are diagonal matrices containing the respective eigenvalues. Based on the Reighleigh quotient, it should follow that $$ cond(D^{-1}A) \le \frac{c^+}{c^-},$$ thus $c^+$ and $c^-$ upper and lower bounds for the range of the eigenvalues of $D^{-1}A.$
Now my question is: For $0 < \alpha \le 1,$ does $$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$ hold ? Here, $A^\alpha := V\Lambda_A^\alpha V^\top,$ and $D^\alpha := W\Lambda_D^\alpha W^\top,$ where $\Lambda_A^\alpha, \Lambda_D^\alpha$ can be computed by taking the power $\alpha$ of each diagonal entry.