Positive definiteness for a bounded linear operator on $L_2(j\mathbb{R})$ I am reading the textbook A Course in Robust Control Theory: A Convex Approach (Ch 3.4, operator)

Consider the following (comes from the author of this book) 


*

*$M_{\hat{G}} \succ 0 \ \ $ if and only if $$\exists \ \ \epsilon>0 \text{ such that } \hat{G}(j\omega)\succeq \epsilon I \quad \forall \omega\in \mathbb{R}$$   

*$\langle \hat{u},M_{\hat{G}}  \hat{u}\rangle >0 \ \ $ for all $\hat{u}\neq 0\in L_2(j\mathbb{R}) \ \ $ if and only if $$\hat{G}(j\omega)\succeq 0 \quad \forall \omega \in \mathbb{R} \ \ \text{ and } \hat{G}\neq 0$$   


The author of this book said that $M_{\hat{G}} \succ 0 $ if 1. holds, which is not the same as 2.   

My question: in matrices, there are two definitions for positive definiteness, i.e., 1. and 2.    


*

*focuses on all eigenvalues are positive

*focuses on the inner products are positive for all vector not equal to zero.   


However, it seems that both conditions are not quite the same. What is the main difference and why they are not equivalent?     
Also I am confused that in the condition 2., why not just saying $\hat{G}(j\omega)\succ 0$?
 A: In $\mathbb{C}^n$, the typical definition of positive definiteness of a matrix $M$ is $\langle v, Mv\rangle>0$ for all $v$. This implies that $M$ is self-adjoint and that all eigenvalues are strictly positive. If we let $\epsilon>0$ denote the smallest eigenvalue of $M$, then we have $\langle v, M v\rangle \geqslant \epsilon > 0$. In $\mathbb{R}^n$, one usually imposes the extra condition that $M$ is symmetric.
If we require only that all eigenvalues of a matrix $M$ are strictly positive, then it may still happen that $\langle v, M v\rangle < 0$ for some $v$. Consider the matrix
$$
M = \left[\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}\right]
$$
and take $v=(1,-2)^T$, then $\langle v, M v\rangle = -1<0$. On the other hand, $M$ only has one eigenvalue, namely $1>0$. The problem is that $M$ is not self-adjoint.
The two conditions you give in $L_2(j\mathbb{R})$ are not equivalent because there is a difference between $\hat G\geqslant \epsilon >0$ and $\hat G>0$. Consider the function $\hat G(j\omega)=e^{-\omega^2}$, which satisfies $\hat G>0$ but not $\hat G>\epsilon$ for any $\epsilon>0$.
Concerning your final question, I do not know. Maybe the notation $\hat{G}(j\omega)\succ 0$ has not been defined previously.
Clarification: In $\mathbb{R}^n$, if $M$ is a symmetric matrix, then $\langle v, M v\rangle>0$ if and only if all eigenvalues of $M$ are strictly positive, and all eigenvalues are strictly positive if and only if $\langle v, Mv\rangle \geqslant \epsilon$ for some $\epsilon>0$. Therefore $\langle v, M v\rangle>0$ if and only if $\langle v, Mv\rangle \geqslant \epsilon$ for some $\epsilon>0$. The difference in your setting, i.e. in $L_2(j\mathbb{R})$, is that these two condition are no longer equivalent. That is, it may happen that $\hat G>0$ but there is no $\epsilon>0$ such that $\hat G\geqslant \epsilon$.
A: A linear operator $T$ is positive definite if $x^TTx>0$ for any $x$ in the vector space. This means that all of the eigenvalues must be positive.If we have an eigenvalue that is zero or negative, then there will be an eigenvector that gets mapped to zero or reversed. The inner product would then be 0 or negative.
$<x,Tx> =<x,\lambda x> =\bar{\lambda}||x||^2$
