Let A and B be positive semidefinite matrices. It is not hard to see that $(A-B)^2 \leq 2A^2 + 2B^2$. In fact, $2A^2 + 2B^2 - (A-B)^2 = (A+B)^2$ is positive semidefinite.
My question is: Is there a constant c (independent of A and B and the dimension) such that
$$(A-B)^2 \leq c (A+B)^2?$$
Thanks.
There is no such $c$ even if we use only $2 \times 2$ matrices. For any $c \geq 1$ let $A,B$ be the positive-semidefinite matrices $$ A = \left( \begin{array}{lc} c^2 & c \cr c & 1 \end{array} \right), \phantom\infty B = \left( \begin{array}{cc} 1 & 0 \cr 0 & 0 \end{array} \right). $$ of rank $1$. Then we calculate that the difference $$ D := c(A+B)^2 - (A-B)^2 $$ has determinant $(c-1)^2 - 4c^3 < 0$, and is thus not positive semidefinite.
In fact this counterexample works for all $c \in \bf R$: looking around $\ker A = {\rm span} \lbrace(-1,c)\rbrace$ we find the negative vector $v = (-c, c^2+1)$. To verify that $\langle v, Dv \rangle < 0$, recall that for any vector $x$ and any symmetric matrix $M$ of the same order we have $$ \langle x, M^2 x \rangle = \langle Mx, Mx \rangle = |Mx|^2. $$ Here we compute $v(A+B) = (0,1)$ and $v(A-B) = (2c,1)$, so $$ \langle v, Dv \rangle = c \phantom. |(0,1)|^2 - |(2c,1)|^2 = -4c^2 + c - 1, $$ which is negative for all real $c$. If $c$ is small enough that $\det(D) > 0$ then $D$ is negative definite.
A weaker version is also not true, that is, there is no constant $c$ such that $$\lambda_j(A-B)^2 \leq c \lambda_j(A+B)^2, ~~ j=1, 2,\ldots$$ where $\lambda_j(A)$ means the $j$-th largest eigenvalue of $A$.
Nevertheless, it is trivial to see that
$$trace(A-B)^2 \leq trace(A+B)^2.$$
