- #1
tommy01
- 40
- 0
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.
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.