I want that if $H$ Hilbert space where $A$, $B$ are positive operators on $H$ Hilbert space, $0 \leq (Ax|x) \leq (Bx | x)$ $\forall x$, does this mean $(A^2x|x) \leq (B^2x|x)$?
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You said that the operators were positive in the title, but not in the body of the question. In the future, repeat details like this in the body of the question. – Ben Grossmann May 04 '15 at 03:44
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sorry, I thought $A$ positive iff $0 \leq (Ax | x)$ $\forall x$? – dominiquebartone1 May 04 '15 at 03:48
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Yes, that is correct. Read my comment, see the change I made to your question. – Ben Grossmann May 04 '15 at 03:49
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@Omnomnomnom The OP's original statement included that $(Bx|x)\ge(Ax|x)\ge0$, so certainly $A$ and $B$ are positive. – Jason May 04 '15 at 05:18
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@Jason only over complex Hilbert spaces – Ben Grossmann May 04 '15 at 11:48
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Consider $H = \mathbf K^2$ and the operators $$ A = \def\p#1#2#3#4{\begin{pmatrix} #1 & #2 \\ #3 & #4 \end{pmatrix}}\p 1111, \quad B = \p 2111 $$ Then \begin{align*} \def\<#1>{\left<#1\right>}\<Ax,x> &= |x_1 + x_2|^2\\ \<Bx,x> &= (2x_1 + x_2)\bar x_1 + (x_1 + x_2)\bar x_2\\ &= |x_1|^2 + |x_1 + x_2|^2 \end{align*} Hence $0 \le A \le B$. Taking squares, we have $$ A^2 = \p 2222, \qquad B^2 = \p 5332 $$ Hence, $$ B^2 - A^2 = \p 3110 $$ Now $$ (B^2 - A^2)\binom{1}{-3} = \binom{0}{1} $$ And hence $\<(B^2 - A^2)\binom{1}{-3},\binom{1}{-3}> < 0$, so $A^2\not\le B^2$.
martini
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