hi.(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# Bounded Operators

Have something to add?

- Similar discussions for: Bounded Operators

Loading...

**Physics Forums - The Fusion of Science and Community**