Register to reply 
Bounded Operators 
Share this thread: 
#1
Feb1010, 05:08 AM

P: 39

hi.
i'm reading "quantum mechanics in hilbert space" and a don't get a basic point for bounded operators. def. 1 a set S in a normed space [tex]N[/tex] is bounded if there is a constant C such that [tex]\left\ f \right\ \leq C ~~~~~ \forall f \in S[/tex] def. 2 a transformation is called bounded if it maps each bounded set into a bounded set. and now comes the part i don't understand. for linear operators [tex]T: N_1 \rightarrow N_2[/tex] def. 2 is equivalent to: there exists a constant C such that [tex]\left\ T f \right\ \leq C \left\ f \right\ ~~~~~ \forall f \in N_1[/tex] this is stated without a proof. i don't think it's obvious or at least not to me. i'm thinking of a map, for example from the real numbers (normed space) to the real numbers, where the bounded set [tex]N_1=(0,1][/tex] is transformed in a way to another interval say [tex]N_2=(a,b][/tex] now the norm of elements from [tex]N_1[/tex] can get arbitrary small. So there can't exist a constant fulfilling [tex]\left\ T f \right\ \leq C \left\ f \right\ ~~~~~ \forall f \in N_1[/tex] when the norm of all elements of [tex]N_2[/tex] has a lower bound say [tex]m>0[/tex]. Or is such a map forbidden because of the continuity (zero has to be mapped on zero) and bounded linerar operators are continuous and vice verca? i would be glad if someone can show me a proof or a source where a can get one. thanks and greetings. 


#2
Feb1010, 06:20 AM

Sci Advisor
HW Helper
P: 4,300

Clearly, the '<=' implication is true. For suppose that Tf <= C f. Take N_{1} to be a bounded set. That means that for all f, f <= C' for some constant C'. So clearly, Tf <= C C' where the right hand side is again a constant.
As for the direct implication '=>' ... I didn't quite see it right away. How to connect the statement about bounded sets to a statement about any subset of S eludes me right now. If I think of something I will let you know :) 


#3
Feb1010, 07:16 AM

Emeritus
Sci Advisor
PF Gold
P: 9,248

Suppose that T takes bounded sets to bounded sets. The set of all vectors of the form [tex]\frac{f}{\f\}[/tex] is bounded, so there must exist a C such that
[tex]\T\frac{f}{\f\}\\leq C[/tex] This implies [itex]\Tf\\leq C\f\[/itex]. To prove the converse, suppose instead that there exists a C such that [itex]\Tf\\leq C\f\[/itex] for all f, and let B be a bounded set with bound M (i.e. [itex]\g\\leq M[/itex] for all g in B). We need to show that there exists an upper bound for the set of all [itex]\Tg\[/itex] with g in B. [tex]\Tg\\leq C\g\\leq CM[/tex] If you're also wondering why an operator is continuous if and only if it's bounded, the Wikipedia page titled "bounded operator" has a very nice proof of that. 


#4
Feb1010, 07:25 AM

P: 39

Bounded Operators
Great!
Many thanks to both of you. 


Register to reply 
Related Discussions  
Pointwise bounded, uniformly continuous family of functions locally uniformly bounded  Calculus & Beyond Homework  0  
What does it mean for a set to be bounded?  Set Theory, Logic, Probability, Statistics  2  
Every sequence of bounded functions that is uniformly converent is uniformly bounded  Calculus & Beyond Homework  4  
A question on bounded linear operators (Functional Analysis)  Calculus  23  
Do quantum fluctuations bounded above correspond to those bounded below?  Astronomy & Astrophysics  0 