Let $E,||.||$ be a Banach space. If $F$ is a closed subspace of $E$.
Consider the quotient space $E_{|F}$ the quotient space of $E$ by $F$ such that:$x,y\in E\:\:\implies x\sim y\implies x-y\in F$ is an equivalence relation.
Prove that function:
$[x]\to||[x]||=\inf_{y\in F}||x-y||$ is a well defined norm on the quotient space.
I do not understand ||[x]||. $||[x]||=\inf_{y\in F}||x-y||=?$
Question:
I am not understanding this exercise. How should I prove the ||.|| is a norm? I know the norm is the distance of x to $F$. How should I write that?
Thanks in advance!
Best Answer
The most important thing here is to understand the definitions. We define an equivalence relation on $E$: $x\sim y\iff x-y\in F$. Then for any $x\in E$ we can define its equivalence class:
$[x]=\{y\in E: x-y\in F\}=\{x+f: f\in F\}=x+F$
And now the quotient space $E/F$ is the space of all equivalence classes with the operations defined in the most natural way: $[x+y]=[x]+[y]$ and $[\lambda x]=\lambda[x]$. The operations are well defined, so $E/F$ is a vector space. And now we can define a norm on it like this:
$||[x]||=\inf_{y\in F}||x-y||=\inf_{y\in F}||x+y||=\inf_{z\in [x]}||z||$
So all you have to prove is that this is really a norm. I'll give an example. Obviously the function we defined is always non negative. Now why $||[x]||=0$ if and only if $[x]=[0]$? Suppose $||[x]||=0$. By the definition of infimum we can take a sequence of vectors $y_n\in F$ such that $||x-y_n||\to ||[x]||=0$. That implies $y_n\to x$ as vectors. But since $F$ is a closed subspace we conclude that $x\in F$, which means $[x]=[0]$. So we have that property. Now try to prove the other properties of norm.