[SOLVED] Continuous Extension

[Sudharaka]

Hi everyone,

I have a little trouble understanding a question. Hope you can figure this out for me.

Question: Let \((Y,\,d')\) be a complete metric space and \((X,d)\) be a metric space. Let \(A\subset X\) be a dense subset of \(X\) and \(f:A\rightarrow Y\) be a uniformly continuous mapping. Prove that there is a unique continuous extension \(g:X\rightarrow Y\), such that \(g|_{A}=f\) and \(g\) is uniformly continuous on \(X\).

So first of all I want to understand the precise meaning of what a continuous extension is. Is it that we extend the continuous mapping \(f\) to the whole space \(X\) such that \(g(s)=f(s)\) for all \(s\in A\) and call this new mapping \(g\) the continuous extension of \(f\)? Am I correct?

 

[caffeinemachine]

Hi everyone,

I have a little trouble understanding a question. Hope you can figure this out for me.

Question: Let \((Y,\,d')\) be a complete metric space and \((X,d)\) be a metric space. Let \(A\subset X\) be a dense subset of \(X\) and \(f:A\rightarrow Y\) be a uniformly continuous mapping. Prove that there is a unique continuous extension \(g:X\rightarrow Y\), such that \(g|_{A}=f\) and \(g\) is uniformly continuous on \(X\).

So first of all I want to understand the precise meaning of what a continuous extension is. Is it that we extend the continuous mapping \(f\) to the whole space \(X\) such that \(g(s)=f(s)\) for all \(s\in A\) and call this new mapping \(g\) the continuous extension of \(f\)? Am I correct?

You are pretty much correct but I think the following will make things clearer.

First let us see what an extension of a function is.

Let $A\subseteq X$ and $f:A\to Y$ be any function.
An extension of $f$ is a function $g:B\to Y$, where $A\subseteq B\subseteq X$, such that $g(x)=f(x)$ for all $x\in A$.

A continuous extension of a function $f$ is an extension of $f$ which also happens to be continuous.

In our problem the domain of the continuous extension is mentioned, namely, the whole space.

 

[Sudharaka]

You are pretty much correct but I think the following will make things clearer.

First let us see what an extension of a function is.

Let $A\subseteq X$ and $f:A\to Y$ be any function.
An extension of $f$ is a function $g:B\to Y$, where $A\subseteq B\subseteq X$, such that $g(x)=f(x)$ for all $x\in A$.

A continuous extension of a function $f$ is an extension of $f$ which also happens to be continuous.

In our problem the domain of the continuous extension is mentioned, namely, the whole space.

Thank you very much for the reply. So the notation \(g|_{A}=f\) means \(f(s)=g(s)\) for all \(s\in A\). Am I correct?

 

[caffeinemachine]

Thank you very much for the reply. So the notation \(g|_{A}=f\) means \(f(s)=g(s)\) for all \(s\in A\). Am I correct?

Yes.

In fact, writing $g|_A=f$ also says that $f$ has $A$ as it's domain.

More generally, $g|_A=f|_A$ means $g(s)=f(s)$ for all $s\in A$.

 

[Sudharaka]

Yes.

In fact, writing $g|_A=f$ also says that $f$ has $A$ as it's domain.

More generally, $g|_A=f|_A$ means $g(s)=f(s)$ for all $s\in A$.

Thank you very much for your help. After hours of trying I finally solved this problem.