Understanding Bounded Operators in Quantum Mechanics

[tommy01]

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 $$N$$ is bounded if there is a constant C such that $$\left\| f \right\| \leq C ~~~~~ \forall f \in S$$

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 $$T: N_1 \rightarrow N_2$$ def. 2 is equivalent to:
there exists a constant C such that $$\left\| T f \right\| \leq C \left\| f \right\| ~~~~~ \forall f \in N_1$$

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 $$N_1=(0,1]$$ is transformed in a way to another interval say $$N_2=(a,b]$$ now the norm of elements from $$N_1$$ can get arbitrary small. So there can't exist a constant fulfilling $$\left\| T f \right\| \leq C \left\| f \right\| ~~~~~ \forall f \in N_1$$ when the norm of all elements of $$N_2$$ has a lower bound say $$m>0$$.

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.

 

[CompuChip]

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]

Suppose that T takes bounded sets to bounded sets. The set of all vectors of the form $$\frac{f}{\|f\|}$$ is bounded, so there must exist a C such that

$$\|T\frac{f}{\|f\|}\|\leq C$$

This implies $\|Tf\|\leq C\|f\|$.

To prove the converse, suppose instead that there exists a C such that $\|Tf\|\leq C\|f\|$ for all f, and let B be a bounded set with bound M (i.e. $\|g\|\leq M$ for all g in B). We need to show that there exists an upper bound for the set of all $\|Tg\|$ with g in B.

$$\|Tg\|\leq C\|g\|\leq CM$$

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]

Great!

Many thanks to both of you.