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Let $H$ be a Hilbert space. Let $A: H\to H$ be a bounded linear operator which is positive definite. Suppose $B \eqsim A^{-1}$, i.e. $$\forall x\in H,\quad m(A^{-1}x,x) \le (Bx,x) \le M(A^{-1}x,x).$$

Do we have the following $$\frac{1}{M}(Ax,x)\le (B^{-1}x,x)\le\frac{1}{m}(Ax,x)?$$

My thoughts:

Since $A$ is positive definite, then $\|x\|_A =(Ax,x)$ defines a norm. So, the above given condition is equivalent to $$m\|x\|_{A^{-1}} \le \|x\|_B \le M\|x\|_{A^{-1}} ~.$$ So, these two norms are equivalent. I guess $\|\cdot\|_A$ and $\|\cdot\|_{B^{-1}}$ are equivalent as well. But I don't know what to do next.

XYZ
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1 Answers1

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Yes. Note that the conditions also imply that $B$ is positive-definite.

In general, if $X,Y$ are positive-definite and $X\leq Y$, then $Y^{-1}\leq X^{-1}$ (the argument there is written for matrices, but everything works the same for operators).

From $mA^{-1}\leq B$, you get $ B^{-1}\leq m^{-1}A$. And from $B\leq MA^{-1}$, you get $ M^{-1}A\leq B^{-1}$. Putting both inequalities together, $$ M^{-1}A\leq B^{-1}\leq m^{-1} A. $$

Martin Argerami
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