If we are given that $A:V \rightarrow V$ is an operator where $V$ is a real Hilbert space. If we are given that $A$ is bounded, strictly positive $\big(\langle Au,u \rangle > 0$ for all $u \neq 0\big)$ and a self-adjoint linear operator then does it also follow that $A$ positive definite $\big(\langle Au,u \rangle \geq \alpha\Vert u \Vert$ where $\alpha > 0\big)$?
Also, it’s a bit confusing as to which is the standard definition for a positive definite operator, since sometimes the definition that I gave for a strictly positive operator is also used for positive definite.
Thanks.
Question inspired by the beginning of the proof of the following theorem:
The link to the document is Document link
The Theorem is on page 40.
Best Answer
No, let $V$ be separable and $e_n$ form an orthonormal basis in $V$, set $Ae_n=\frac1n e_n$. Then $\langle Au,u \rangle>0$ since every non-zero vector has at least one non-zero coefficient in its expansion, but there is no $\alpha$ since otherwise $\|Ae_n\|\geq\alpha$ for all $n$. The first condition is sometimes called "strictly positive definite" and the second "strongly positive definite", but it varies from author to author.
In finite dimensional spaces these are equivalent because the unit sphere is compact and $\langle Au,u \rangle$ must attain a (positive) minimum on it, the $\alpha$. But in infinite dimensions strong positive definiteness is strictly stronger.