I am not sure if the bidual map $ V^{*} —> K $ (which is a map x: f—> x(f)) is numerically the same than the image of the dual map, i.e, V—>K, where V is a vector space over a commutative field K and the * indicates V is the dual space.
All the books I have read represent the images of the dual and bidual maps as f(x). However, I found one author which wrote x(f), so I guess these values can be different. Correct me if my reasoning is not right:
The dual maps an element x to f(x), given for example x=1 f(x)= 5 x then f(1)= 5. However, for the bidual map I understand that you keep fixed a certain value of x say 5 and you then choose any f you want, such as f= 5x , f=3x, etc. The the images are the values of these f for x=5. Is this correct? Or does f(x) = x(f)?
There is an abuse of notation happening. If $x \in V$, one defines $\hat{x}\colon V^* \to \Bbb K$ by $\hat{x}(f)\doteq f(x)$. If $\dim_{\Bbb K}V<+\infty$, then $V \in x \mapsto \hat{x} \in V^{**}$ is an isomorphism, so we can identify $x$ with $\hat{x}$ and write $x(f) = f(x)$. The idea behind this is to take $x \in V$, $f \in V^*$, and produce an element of $\Bbb K$... there is one sensible way to do it: $f(x)$.