If $A,B,A-B$ are all symmetric positive definite and real, how to prove that $\forall 0<\alpha<1, A^{\alpha}-B^{\alpha}$ is also positive?

Question

It is necessary to mention that we difine the $A^{\alpha}$ in this way:
if $A=P^Tdiag\{c_1,c_2,...,c_n\}P$, then $A^{\alpha}:=P^Tdiag\{c_1^{\alpha},c_2^{\alpha},...,c_n^{\alpha}\}P$
($P$ is a unitary matrix)

To solve this question, I tried to calculate it directly like this:
set $A=P_1^Tdiag\{a_1,a_2,...,a_n\}P_1, B=P_2^Tdiag\{b_1,b_2,...,b_n\}P_2$,
then $A^{\alpha}-B^{\alpha}=P_1^Tdiag\{a_1^{\alpha},a_2^{\alpha},...,a_n^{\alpha}\}P_1-P_2^Tdiag\{a_1^{\alpha},a_2^{\alpha},...,a_n^{\alpha}\}P_2$
if we set $P_1\beta:=(c_1,c_2,...,c_n),P_2\beta:=(d_1,d_2,...,d_n)$, then $\beta^T(A^{\alpha}-B^{\alpha})\beta=\sum_{i=1}^na_i^{\alpha}|c_i|^2-\sum_{j=1}^nb_i^{\alpha}|d_i|^2$
On the other hand, since $A-B$ is postive, we have:
$\beta^T(A-B)\beta=\sum_{i=1}^na_i|c_i|^2-\sum_{j=1}^nb_i|d_i|^2>0$

But then I don't know how to proceed the solution. I would appriciate it if somebody could do me a favor!

Answer 1

One possible (admittedly advanced) proof goes as follows; for another one see the reference in my comment to the question. I am presently not aware of a more "elementary" proof.

As a further reference you may see https://encyclopediaofmath.org/wiki/L%C3%B6wner-Heinz_inequality

1. Establish the identity $$ M^\alpha=\frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\lambda(M+\lambda)^{-1}\right)d\lambda $$ valid for any self-adjoint matrix $M>0$ and any $0<\alpha<1$. The integral is understood entry-wise.

2. Use the identity and the assumptions $A>B>0$ to write, for any $v\in\mathbb C^n$, $$ \langle v,A^\alpha v\rangle =\frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\lambda\langle v,(A+\lambda)^{-1}v\rangle\right)d\lambda \\ >\frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\lambda\langle v,(B+\lambda)^{-1}v\rangle\right)d\lambda= \langle v,B^\alpha v\rangle, $$ where $A>B>0$ implies that for all $\lambda>0$ we have $-(A+\lambda)^{-1}\geq -(B+\lambda)^{-1}$.

To establish 1. let indeed $M=P^\top CP$ where $C=diag(c_1,...,c_n)$. The integral is $$ P^\top X P,\qquad X:=\frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\lambda(C+\lambda)^{-1}\right)d\lambda. $$ Using the integral $$ \frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\frac\lambda{x+\lambda}\right)d\lambda=x^\alpha,\qquad 0<\alpha<1, $$ we obtain that the entries of $X$ are $$ X_{ij}=0,\mbox { for }i\not=j,\qquad X_{ii}=\frac{\sin(\alpha\pi)}\pi\int_0^{+\infty}\lambda^{\alpha-1}\left(1-\frac{\lambda}{c_i+\lambda}\right)d\lambda=c_i^\alpha, $$ for all $0<\alpha<1$, and so we get the original definition of $M^\alpha$.