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 Post subject: An Interesting ExercisePosted: Thu Jun 09, 2016 10:10 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
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 $\displaystyle C \left( \mathbb{R}^{n} , \mathbb{R} \right) / Ker(\Phi) \cong C \left( X , \mathbb{R} \right)$

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 Post subject: Re: An Interesting ExercisePosted: Sat Jun 11, 2016 4:27 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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.

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 Post subject: Re: An Interesting ExercisePosted: Tue Aug 30, 2016 3:02 pm

Joined: Fri Aug 12, 2016 4:33 pm
Posts: 15
1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear

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 Post subject: Re: An Interesting ExercisePosted: Wed Aug 31, 2016 1:45 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
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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 Post subject: Re: An Interesting ExercisePosted: Mon Sep 05, 2016 10:50 pm

Joined: Fri Aug 12, 2016 4:33 pm
Posts: 15
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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