Prove an operator is injective

Question

Denote $E'$ the topological dual of a space $E$, and $C[0,1]$ the space of continuous functions $[0,1] \to \Bbb R$.

Consider the operator $T: [0,1] \to (C[0,1])'$ such that:

$T(x): C[0,1] \to \Bbb R$

$\ \ \ \ \ \ \ \ \ \ \ \ \ f\to f(x)$

Prove that T is injective.

My attempt:

Let $x,x'\in [0,1]: T(x)=T(x')=f\in (C[0,1])$

we have $f(x)=f(x') \Rightarrow f(x-x')=0$.

At this point I would like to deduce that $x-x'=0$ but I guess It's not correct and I don't know if I can use somehow the continuity of $f$?

Thank you for your help.

Answer 1

$Tx=Ty$ means $T(x)(f)=T(y)(f)$ for all $f \in C[0,1]$. By the definition of $T$ this means $f(x)=f(y)$ for all $f \in C[0,1]$. In particular this must hold for the function $f(t)=t \,\forall t$. Hence $x=y$.

Answer 2

I think that you are messing a bit with notations. If $f \in (C[0,1])'$ then you should not evaluate $f$ in a scalar ($ x \in [0,1]$ for instance).

I would suggest to consider $x \neq x'$ and prove that $T(x) \neq T(x')$ by finding a function $f \in C[0,1]$ (I emphasize the fact that $f$ is a continuous function and not an element of the dual space) such that $f(x) \neq f(x')$.