Welcome to mathimatikoi.org forum; Enjoy your visit here.

## An Interesting Exercise

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
Tsakanikas Nickos
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

### An Interesting Exercise

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
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

### Re: An Interesting Exercise

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.
Articles: 0
Posts: 16
Joined: Fri Aug 12, 2016 4:33 pm

### Re: An Interesting Exercise

1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear
Tsakanikas Nickos
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

### Re: An Interesting Exercise

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?
Articles: 0
Posts: 16
Joined: Fri Aug 12, 2016 4:33 pm

### Re: An Interesting Exercise

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.

0