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Please consider the image below.enter image description here

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!

MathMan
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1 Answers1

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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.

Tsemo Aristide
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  • I am sorry but the book which I am reading doesn't explain using basis. On the whole, I haven't dealt basis while going through metric spaces . Would it be possible for you to explain using maybe an another argument? Thanks for your understanding! – MathMan Jan 31 '18 at 12:23
  • I know that every open ball $b[a;r)$with center $a$and radius $r$ can be expressed in terms of the unit ball as as $a+rb[0;1)$ – MathMan Jan 31 '18 at 12:28
  • The above being valid only in a normed linear space – MathMan Jan 31 '18 at 12:36
  • Thank you. Could you also please look into my second query.. the author says that if $a,b \in U, $ then $a+b \in U$. Is this necessary true? – MathMan Jan 31 '18 at 13:07
  • The author does not says that if $a,b\in U$ then $a+b\in U$. He chooses $a,b\in U$ such that $a+b\in U$. – Tsemo Aristide Jan 31 '18 at 13:37