Bounded Operators


by tommy01
Tags: bounded, operators
tommy01
tommy01 is offline
#1
Feb10-10, 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.
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
CompuChip
CompuChip is offline
#2
Feb10-10, 06:20 AM
Sci Advisor
HW Helper
P: 4,301
Clearly, the '<=' implication is true. For suppose that ||Tf|| <= C ||f||. Take N1 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 :)
Fredrik
Fredrik is offline
#3
Feb10-10, 07:16 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 8,997
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.

tommy01
tommy01 is offline
#4
Feb10-10, 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? General Astronomy 0