# Implying a positive definite operator

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 Theorem is on page 40.

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.

• Thanks for your response. I have edited my post so that you can see what inspired the question. I have attached the beginning of a proof of a Theorem. In the proof it states $E(u) \geq c_{o}^{2}\Vert u \Vert_{V}^{2} - 2\Vert F \Vert_{V^{*}}\Vert u \Vert_{V}$. Can you maybe see where the term $c_{o}^{2}\Vert u \Vert_{V}^{2}$ comes from? Thanks for your assistance.
– user100431
Jul 5, 2014 at 15:59
• My guess is that this author calls "strictly positive" what I called "strongly positive", it should be defined earlier in the book. Jul 7, 2014 at 19:28
• It doesn't state what definition he is using. I have edited to include the whole proof. If you have a chance you can see what is meant. It states that the function is also self-adjoint, but I can't see where this is used?
– user100431
Jul 8, 2014 at 12:31
• It seems that it would follow if that is how the author defines it.
– user100431
Jul 8, 2014 at 17:36
• Can you maybe see how the line $2 \Vert u_{j} \Vert^{2} + \Vert u_{l} \Vert^{2}] - \Vert u_{j} + u_{l} \Vert^{2} = 2 [E(u_{j})+E(u_{l})]-4E(\frac{u_{j}+u_{l}}{2})$ follows on page 40?
– user100431
Jul 8, 2014 at 17:40