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Rudin functional Analysis chapter 10, exercise 13
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This is Rudin's functional Analysis chapter 10, exercise 13.
I am confused about the notation σA (f ), what does that mean?(What role does the subscript A play here). And can
someone illustrate how to solve this question?
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Update: thanks to the comment, and a theorem in Rudin:
is the collection of polynomials of x (|x| ∈ [1, 2] ), whose power ranges from negative to positive. So the only
possible λ (here I use λ to denote the spectral, while x as variable in the function, not the way in the question) is 1.
B
Actually, if we want (λ ∗ e − f )(x) ∗ P (x)
So by the theorem in Rudin, σA
= σB = 1
= 1
when λ is not 1, we can simply choose P (x) to be
1
(λ−1)x
.
.
Is this correct?
banach-algebras
share cite improve this question follow
edited Apr 22 '16 at 20:16
asked Apr 22 '16 at 5:25
Toad Jiang
867
4
17
it seems there is a special notion of spectrum in a Banach algebra : en.wikipedia.org/wiki/Banach_algebra#Spectral_theory
and σA (T ) would be the set such that z ∈ σA (T ) ⇔ (T − zI )−1 ∉ A – reuns Apr 22 '16 at 5:49
If B is a subalgebra, by the definition of algebra(by folland, if A is an algebra, f , g
also in B, which implies A ⊂ B ? – Toad Jiang Apr 22 '16 at 19:04
∈ A, ⟹
f ∗ g ∈ A),can
we say that 1 is
yes if B contains f and 1/f it clearly contains f (1/f ) = 1. in the context of Banach algebras, you should think to sub vector
n×n
spaces of the space of matrices C
closed for the multiplication of matrices, and with the usual matrix norm
(with ∥x∥ the Euclidean norm), then you get the same kind of objects when considering the
algebra of bounded linear operator from an Hilbert space to itself, with again the operator norm, and the sub vectors paces
which are closed for the composition of operators. – reuns Apr 22 '16 at 19:13
∥M ∥ = max∥x∥=1 ∥M x∥
K
and remember that an algebra is the basic structure where we can consider the set of polynomials P (T ) = ∑k=1 ck T k where
T is any element of the algebra, and cn the constants of the underlying field of the vector space – reuns Apr 22 '16 at 19:14
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y
1
g
,
y g
n
p
p
Write σ(f ) for the spectrum of f in C(K). Note that α ∈ σ(f ) ⟺ α − f is not invertible, and this is equivalent to say
that α − λ is not invertible for some λ ∈ K , which means that α ∈ K . Therefore, σ(f ) = K . Now, remember that
σA (f ) ⊃ σB (f ) ⊃ σ(f ) = K , so it's not possible to have σA (f ) or σB (f ) equal to 1. PS: I'm using Rudin's notation,
f (λ) = λ . – Integral Apr 22 '16 at 23:59
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