Lower bound of a positive self-adjoint operator

Question

Let $A$ be a positive self-adjoint operator in a Hilbert space. Is there a way to show that $$\inf_{\|x\|=1}\|Ax\|=\inf_{\|x\|=1}|\langle x,Ax\rangle|$$ without using the spectral theorem and in a way similar to the proof of $\displaystyle \|A\|=\sup_{\|x\|=1}|\langle x,Ax\rangle|$ ? A book reference will also be welcome.

Answer 1

My proof below is incomplete: for operators, positivity does not mean invertibility. I believe, however, that that case in which the infimum is zero can be handled separately.

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I'm not sure how you proved the statement about the sup without the spectral theorem, but that'll be enough for the opposite statement.

Note that since $A$ is positive, it is invertible, and $A^{-1}$ is positive. We have $$ \sup_{\|x\| = 1} \|A^{-1}x\| = \sup_{\|x\| = 1}|\langle x,A^{-1}x\rangle| $$ from there, we note that $$ \inf_{\|x\| = 1} \|Ax\| = \left[\sup_{\|x\| = 1} \frac{1}{\|Ax\|}\right]^{-1} = \left[\sup_{x \neq 0} \frac{x^*x}{x^*AAx}\right]^{-1/2} =(\text{set }y = Ax)\\ \left[\sup_{x \neq 0} \frac{y^*A^{-1}A^{-1}y}{y^*y}\right]^{-1/2} = \left[\sup_{\|y\|=1}\|A^{-1}y\|\right]^{-1} $$ Similarly, show that $$ \inf_{\|x\| = 1}|\langle x,A x\rangle| = \left[\sup_{\|y\| = 1}|\langle y,A^{-1}y \rangle| \right]^{-1} $$ the statement follows.