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Bounded Operators

  1. Feb 10, 2010 #1

    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. jcsd
  3. Feb 10, 2010 #2


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    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 :)
  4. Feb 10, 2010 #3


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    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.
  5. Feb 10, 2010 #4

    Many thanks to both of you.
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