As far as I know those operators are defined as follows:
- Positive operator is an operator $L: H\rightarrow H$ such that $\langle L\textbf u|\textbf u\rangle \geq0$ for all $\textbf u \in H$ and the equality could be achieved even for nonzero $\textbf u$-s.
- Positive dedinite operator is an operator $L: H\rightarrow H$ such that $\langle L\textbf u|\textbf u\rangle \geq 0$ for all $\textbf u \in H$ where we have equality only if $\textbf u=\textbf 0$. Or equivalently said $L: H\rightarrow H$ is positive definite operator if $\langle L\textbf u|\textbf u\rangle >0$ for all nonzero $\textbf u \in H$.
But according to a book I read (can't link to the book since it is in Bulgarian not English) the definitions are:
- Positive operator is an operator $L: H\rightarrow H$ such that $\langle L\textbf u|\textbf u\rangle \geq 0$ for all $\textbf u \in H$ where we have equality only if $\textbf u=\textbf 0$. Note that this is the definition of positive definite operator as far as my reasoning above is correct.
- The operator $L: H\rightarrow H$ is positive definite if there is a constant $\gamma >0$ such that $\langle L\textbf u|\textbf u\rangle \geq \gamma \langle \textbf u|\textbf u\rangle$ for all $\textbf u \in H$. Note that this definition says nothing about the case $\textbf u=\textbf 0$.
I am confused with the definitions in the book because it is known that every positive definite operator is a positive operator, while the opposite is not true. But according to the books' definitions I interpret that there is no difference between positive definite and positive operator, i.e. they are equivalent.
So which definition is the correct one?
Best Answer
Both definitions are correct and common in certain contexts. Yours is common for the context of finite dimensional vector spaces (in which one may use "positive semidefinite" for your notion of positive).
Your book's definition is one that is strictly applicable to infinite dimensional vector spaces. By the definition it gives, the "positive definite" operators are the positive operators that are not only injective, but also have a bounded inverse.
There are other definitions in use, however. Some books will agree with your definition of positive, even in infinite dimensional spaces (see e.g. Pedersen or Kreyszig). I have never seen the phrase "positive definite" used that way; normally I'd see "positive and bounded away from zero".
Example of difference (between the book's "positive" and "positive definite")
We define $H = \ell^2$ to be the Hiblert space of all square-summable real sequences. In particular, $\ell^2 = \{(x_k)_{k=1}^\infty: \sum_{k}x_k^2 < \infty\}$ is a Hilbert space with the inner product $\langle x,y \rangle = \sum_{k=1}^\infty x_k y_k$.
We define the map $L: H \to H$ by $$ L(x_1,x_2,x_3,\dots) = \left(x_1, \frac 12 x_2, \frac 13 x_3, \dots\right) $$ I will leave it to you to verify that this operator is indeed positive. To show that it fails to be positive definite, it suffices to show that the ratio $\frac{\langle Lu \mid u \rangle}{\langle u \mid u \rangle}$ has no lower bound. To that effect, define the "standard basis element" $e^{(n)}$ by $$ e^{(n)}_k = \begin{cases} 1 & k = n\\ 0 & k \neq n \end{cases} $$ We then find that $$ \frac{\langle Le^{(n)} \mid e^{(n)} \rangle}{\langle e^{(n)} \mid e^{(n)} \rangle} = \frac 1n $$ so, there is no lower-bound $\gamma$ that works for all of $H$.
If you want an example that applies to the space $L^2([0,1])$ (with which you seem to be more familiar), any map of the form $$ L[f(x)] = \int_0^1 k(x,y)f(x)\,dx $$ will also be positive but fail to be positive definite. Another example more akin to ours from above is the map $$ L[f(x)] = -\int_0^x \int_0^t f(\tau)\,d\tau $$