F**S
发帖数: 13 | 1 i've been reading pedersen's 'analysis now' (1989) that received almost
unanimously positive reviews
from working analysts. it seems, however, a little too advanced for a dumb ass
like me--i'm having
hard times filling in the steps that have been skipped in the book. take it as
an example, i've been
severely plagued by the proof of prop 3.2.25 on p98 about hte numerical radius
of an operator in B(H).
Prop 3.2.25
If H is a complex Hilbert space and T is a bounded linear operator in B(H).
we defi |
b****t
发帖数: 22 | 2
this will lead to an upper bound = 4|||T|||.
To get 2|||T|||, you need to (and you can) choose a such that
Re(a^2) = || |
n*******l
发帖数: 2911 | 3 Yes. For any complex number z = r e^(i\theta), we have
e^(-i\theta)z = r.
【在 b****t 的大作中提到】
:
: this will lead to an upper bound = 4|||T|||.
: To get 2|||T|||, you need to (and you can) choose a such that
: Re(a^2) = ||
|
b****t
发帖数: 22 | 4 i think there should be no "Re" there.
It must be a typo.
For every x, you can always find a such that
a^2 = ||.
ass
as
radius
【在 F**S 的大作中提到】
: i've been reading pedersen's 'analysis now' (1989) that received almost
: unanimously positive reviews
: from working analysts. it seems, however, a little too advanced for a dumb ass
: like me--i'm having
: hard times filling in the steps that have been skipped in the book. take it as
: an example, i've been
: severely plagued by the proof of prop 3.2.25 on p98 about hte numerical radius
: of an operator in B(H).
: Prop 3.2.25
: If H is a complex Hilbert space and T is a bounded linear operator in B(H).
|
F**S
发帖数: 13 | 5 thanks so much! it was the real part of the whole thing!!!
didn't see that one coming...
【在 n*******l 的大作中提到】
: Yes. For any complex number z = r e^(i\theta), we have
: e^(-i\theta)z = r.
|
n*******l
发帖数: 2911 | 6 Yes. For any complex number z = r e^(i\theta), we have
e^(-i\theta)z = r.
【在 b****t 的大作中提到】
: i think there should be no "Re" there.
: It must be a typo.
: For every x, you can always find a such that
: a^2 = ||.
:
: ass
: as
: radius
|
n*******l
发帖数: 2911 | 7 1. It should be the real part since when you take absolute value on the
left hard side, the real part is always smaller than or equal to
the absolute value. Besides, we can not directly compare two complex
numbers.
2. The norm of the imaginary part of the left hand side is smaller than the
right hand side too, same argument. Or as in that proof, you can choose
a number a with |a|=1 such the the real part of <(aT)^2x | x> is the
imaginary part of .
Hope this clarifies
【在 F**S 的大作中提到】
: i've been reading pedersen's 'analysis now' (1989) that received almost
: unanimously positive reviews
: from working analysts. it seems, however, a little too advanced for a dumb ass
: like me--i'm having
: hard times filling in the steps that have been skipped in the book. take it as
: an example, i've been
: severely plagued by the proof of prop 3.2.25 on p98 about hte numerical radius
: of an operator in B(H).
: Prop 3.2.25
: If H is a complex Hilbert space and T is a bounded linear operator in B(H).
|
F**S
发帖数: 13 | 8 thanks so much! it was the real part of the whole thing!!!
didn't see that one coming...
【在 n*******l 的大作中提到】
: 1. It should be the real part since when you take absolute value on the
: left hard side, the real part is always smaller than or equal to
: the absolute value. Besides, we can not directly compare two complex
: numbers.
: 2. The norm of the imaginary part of the left hand side is smaller than the
: right hand side too, same argument. Or as in that proof, you can choose
: a number a with |a|=1 such the the real part of <(aT)^2x | x> is the
: imaginary part of .
: Hope this clarifies
|
