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An Interesting Exercise

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
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Tsakanikas Nickos
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An Interesting Exercise

#1

Post by Tsakanikas Nickos » Thu Jun 09, 2016 10:10 pm

Let \( \displaystyle X \) be a compact subspace of \( \displaystyle \mathbb{R}^{n} \). Consider the rings \[ \displaystyle C \left( X , \mathbb{R} \right) \, , \, C \left( \mathbb{R}^{n} , \mathbb{R} \right) \] of continuous functions and define the mapping \[ \displaystyle \Phi : C \left( \mathbb{R}^{n} , \mathbb{R} \right) \longrightarrow C \left( X , \mathbb{R} \right) \, , \, \Phi(f) = f|_{X} \] Show that \[ \displaystyle C \left( \mathbb{R}^{n} , \mathbb{R} \right) / Ker(\Phi) \cong C \left( X , \mathbb{R} \right) \]
Papapetros Vaggelis
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Re: An Interesting Exercise

#2

Post by Papapetros Vaggelis » Sat Jun 11, 2016 4:27 pm

Firstly, the mapping \(\displaystyle{\Phi}\) is well defined because, if \(\displaystyle{f\in C(\mathbb{R}^n,\mathbb{R})}\),

then \(\displaystyle{f|_{X}}\) is continuous, that is \(\displaystyle{f|_{X}\in C(X,\mathbb{R})}\).

If \(\displaystyle{f\,,g\in C(\mathbb{R}^n,\mathbb{R})}\) and \(\displaystyle{a\in\mathbb{R}}\), then,

\(\displaystyle{\Phi(f+a\,g)=(f+a\,g)|_{X}=f|_{X}+a\,g|_{X}=\Phi(f)+a\,\Phi(g)}\)

\(\displaystyle{\Phi(f\,g)=(f\,g)|_{X}=f|_{X}\,g|_{X}=\Phi(f)\,\Phi(g)}\)

and \(\displaystyle{\Phi(\mathbb{1})=\mathbb{1}|_{X}=\mathbb{1}}\).

So, the mapping \(\displaystyle{\Phi}\) is an \(\displaystyle{\mathbb{R}}\)- algebra homomorphism.

Let \(\displaystyle{g\in C(X,\mathbb{R})}\). Since, \(\displaystyle{X}\) is subspace of \(\displaystyle{\mathbb{R}^n}\),

and according to \(\displaystyle{\rm{Hahn-Banach}}\) - theorem, there exists \(\displaystyle{f\in C(\mathbb{R}^n,\mathbb{R})}\)

such that \(\displaystyle{f|_{X}=g\iff \Phi(f)=g}\). Therefore, the mapping \(\displaystyle{\Phi}\)

is onto.

According to the 1st Isomorphism Theorem we have that

\(\displaystyle{C(\mathbb{R}^n,\mathbb{R})/\rm{Ker}(\Phi)\cong C(X,\mathbb{R})}\) as \(\displaystyle{\mathbb{R}}\) - algebras.
S.F.Papadopoulos
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Re: An Interesting Exercise

#3

Post by S.F.Papadopoulos » Tue Aug 30, 2016 3:02 pm

1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear
Tsakanikas Nickos
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Re: An Interesting Exercise

#4

Post by Tsakanikas Nickos » Wed Aug 31, 2016 1:45 pm

S.F.Papadopoulos wrote: 1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear
1) As $X$ is a subset of $ \mathbb{R}^{n} $, it is naturally a subspace; We additionally require it to be compact.

2)Hahn Banach Theorem/ Formulation


What can you say about the solution of the exercise? How is your post related to it? Could you write a few more things and be more specific?
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Re: An Interesting Exercise

#5

Post by S.F.Papadopoulos » Mon Sep 05, 2016 10:50 pm

1) subspace of a vector space for me( Rudin Functional Analysis and other) mean linear subspace.
2)This is a simple version of Tietze Extension Theorem.










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