Please consider the image below.
I have a couple of questions from example 10.8.6.
The example mentions that the mapping $\phi: ~f \rightarrow f_U$ is bijective. It's understandable that the function is subjective. I wanted to prove it's injective as well.
Suppose $\exists~~ f_1, f_2~\in~X^*~|~f_{1|U}= f_{2|U} $
We know that in a normed space the ball $b[a ;r) = a+ r b[0;1)= a+r U$ where $U$ is the open unit ball
How do I conclude that the above conditions imply that $f$ is injective i.e $f_1=f_2$?
In the arguments presented, We have that if $a,b \in U$ then $a+b \in U$ as given in the arguments. This might not be always true? Why does the author assume the same?
Thanks for reading!
The open subset contains a basis of $X$ and two linear functions which are equal on a basis are equal.
there exists $b$ in $U$ and a real $c$ such that $a=cb$, this implies that $f_1(a)=f_1(cb)=cf_1(b)=cf_2(b)=f_2(cb)=f_2(a)$.
Let $x\in a+rU$, $x=a+ry,y\in U$, $f_1(x)=f_1(a+ry)=f_1(a)+rf_1(y)=f_2(a)+rf_2(y)=f_2(y)$.
Now $X$ is the union of balls.