In Lang's Linear Algebra the following theorem was mentioned:
I am not sure if I truly understand the result. The author here is referring to the Representation theorem, so I wrote the following proof: For any functional $L$ on $V$ there exists a unique $v$ in $V$ such that $L(w)=\langle w,v\rangle$. In particular if we define for a fixed $w$ in $V$ the functional $L_w$ by $L_w(v)=\langle Av,w\rangle$ then there is a unique $w'$ in $V$ such that $L_w(v)=\langle v,w'\rangle$ for each $v$ in $V$.
The failure of the association mentioned in the remark to be an isomorphism is due to the fact that the same vector could be associated to different functionals on $V$ which makes the association not even a linear map on $V$. Is that right? And is my proof fine? Any help is appreciated!
No. The failure of the association to be an isomorphism is due to the fact that the form is hermitian which in particular means that it is linear in one argument but anti-linear in the other, i.e.
$$ \langle \alpha v, w\rangle = \alpha\langle v, w\rangle \\ \langle v, \alpha w\rangle = \overline{\alpha} \langle v, w\rangle. $$
Consequently, upon fixing $w$ the associated $L_w = \langle ., w\rangle$ is linear, but the mapping $w \mapsto L_w$ is antilinear. To see the former, write
$$ L_w(\alpha v) = \langle \alpha v, w\rangle = \alpha\langle v, w\rangle = \alpha L_w(v) $$
and to see the latter write
$$ L_{\alpha w} = \langle ., \alpha w\rangle = \overline{\alpha}\langle ., w\rangle = \overline{\alpha} L_w. $$
Note that the mapping $w \mapsto L_w$ is a bijection and hence an anti-isomorphism, see Riesz representation theorem.