Welcome to our community

Be a part of something great, join today!

[SOLVED] Continuous Extension

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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. :)