Friday, April 5, 2019

Bounded operator - definition?


As mentioned also in Bounded and Unbounded Operator, an operator $A$ is said to be bounded, if $$\|Af\|\leq k \|f\|,$$ where the constant $k$ does not depend on the choice of $f$ (let us consider a map to the same Banach space).


However, in a mathematical physics text I came across a definition: a symmetric operator $B$ is said to be bounded from below if there $\exists$ a constant $c$ such that $$\langle\psi,B\psi\rangle\geq c\|\psi\|^2$$ for all $\psi$ in the domain of $B$.


Both definitions are logical (in the second one we can imagine $B$ being the Hamiltonian, than the system energy is bounded from below and hence the system is stable).



The only think that bothers me is when we rewrite the first definition into a similar form to the second one (we assume the norm comes from an inner product), namely:


$$\langle Af, Af\rangle \leq k\|f\|^2,$$ we get something quite different on the left-hand side, so the same words (bounded operator) refer to different things. Any hints how I can clarify this to myself?



Answer



TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia.


In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$. It follows that $\langle A \rangle_{\psi}\in\mathbb{R}$ is real.




  1. That $A$ is bounded from below means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\geq ~C. $$





  2. That $A$ is bounded from above means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\leq ~C. $$




  3. That $A$ is bounded means that $$\exists C\geq 0~ \forall \psi\in D\backslash\{0\}: ~~ \frac{||A \psi||}{||\psi||}~\leq ~C, $$ which is equivalent to $A^{\dagger}A$ $(=A^2)$ being bounded from above, which in turn is equivalent to $A$ being bounded from both above and below.




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