Suppose $0\leq A\leq B$ (and hence $A$ and $B$ are self-adjoint) where $A$ and $B$ are compact on a complex Hilbert space $H$. That is, $$0\leq \langle x,Ax\rangle\leq \langle x,Bx\rangle$$ for all $x\in H$.
Is it true that $0\leq A^{\frac{1}{2}}\leq B^{\frac{1}{2}}$?
Since $A$ and $B$ are positive and self-adjoint, by the spectral theorem there exist orthonormal bases of eigenvectors $(\varphi_n)$ and $(\psi_m)$ with corresponding non-negative eigenvalues $(\lambda_n)$ and $(\mu_m)$ for $A$ and $B$ respectively. That is, $A=\sum_n\lambda_n\langle{\varphi_n,\cdot}\rangle\varphi_n$ and $B=\sum_n\mu_m\langle{\psi_m,\cdot}\rangle\psi_m$.
So for any $x\in H$, $$\langle x,Ax\rangle = \langle x, \sum_n\lambda_n\langle{\varphi_n,x}\rangle\varphi_n\rangle=\sum_n\lambda_n|\langle \varphi_n,x\rangle|^2,$$ and similarly $$\langle x,Bx\rangle =\sum_m\mu_m|\langle \phi_m,x\rangle|^2.$$
The assumption $0 \leq A \leq B$ implies that $$\sum_n\lambda_n|\langle \varphi_n,x\rangle|^2\leq \sum_m\mu_m|\langle \phi_m,x\rangle|^2.$$
But to compare $A^{\frac{1}{2}}$ and $B^{\frac{1}{2}}$ I must consider $$\langle x,A^{\frac{1}{2}}x\rangle = \sum_n\sqrt{\lambda_n}|\langle \varphi_n,x\rangle|^2\quad \text{and} \quad \langle x,B^{\frac{1}{2}}x\rangle=\sum_m\sqrt{\mu_m}|\langle \phi_m,x\rangle|^2$$ and I'm not sure how to do so!