I need help understanding an argument.
Here is a brief
"In reply to:
>>Let B be the unit ball in l^2. Then
T(B) = B (surjective!). As any open neighbourhood
>>U of 0 contains
some rB for a scalar r, T(U) contains an sB for some scalar s.
>>locally compact normed spaces are finite dimensional, T cannot be
>OK, T(B) = B and as any open neighbourhood U of 0
contains some rB for a scalar r, T(U) contains an sB for some scalar
>Now, l2 is not locally compact. How does it follow that T is not
T(B) = B is obvious. An operator is compact iff
the image of the unit ball is pre-compact.
The closure of the unit ball is
not compact hense T is not compact. q.e.d.
I'd like to understand your
topological argument though.
Locally compact means that every point of the
space has a compact neighbourhood.
What you are saying is that every
neighbourhood of 0 has an image which contains sB for some real s.
does not mean locally compact. A more detailed explanation would be appreciated.
For more info check
Apreciate the help