# Bounded Operators

by tommy01
Tags: bounded, operators
 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 $$N$$ 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.
 HW Helper Sci Advisor P: 4,282 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 :)
 PF Patron Sci Advisor Emeritus P: 8,837 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.
P: 39

## Bounded Operators

Great!

Many thanks to both of you.

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