I have some doubts about the Riesz theorem. Firstly can you check my proof?
Fa = for all
Fa (H,<,>) a pre Hilbert space
Fa x in H different from the zero vector
Fa F a functional on H
Fa T: (H->R) -> H : T(F) = F(x) x/||x||^2 a map from a functional on H to H
Then F(x) = F(x) < x, x > / ||x||^2 = < F(x) x/||x||^2, x > = < T(F), x >
End of proof, where F(x) = < T(F), x > for any vector x in a pre Hilbert space.
If my proof is correct, it implies that H doesn’t have to be an Hilbert space but only a pre Hilbert one and that the functional doesn’t have to be neither linear and continuous, but only a functional.
Is it true or I’m missing something?
The Riesz Representation theorem is:
$$\forall F \in H^*, \exists y \in H, \forall x \in H, F(x) = \langle y, x\rangle$$
What you have shown is
$$\forall F \in H^*, \forall x \in H, \exists y \in H, F(x) = \langle y, x\rangle$$
$y$ or $T(F)$ in the Riesz Representation theorem is picked knowing only $F$, not $x$. This single value of $y$ must work for every $x$. In your result, you pick $x$ first, and then define $T(F)$ in terms of it. Thus your $y$ changes from one value of $x$ to another.