Let $H$ be a Hilbert space and $f \in H^*$. Then there is unique $y \in H$ such that $$ f(x)= \langle x,y \rangle$$ for all $x\in H. $
In the proof of this, first we use projection theorem and express $H$ as direct sum of null space of functional and it's orthogonal complement.
Then to find the value of $y$ we take non zero element say $z$ from orthogonal complement of null space of functional and we can show that the required candidate of $y$ is some scalar multiple of $z$. In particular $y=\frac{f(z)z}{||z||^2}.$
My doubt is here $y$ is dependent on choice of $z$ so if we choose another element say $z^1$ then $y$ also varies according to this, so there are more than one value for fix $f(x)$.
Can anyone explain it.
Thank you.
The uniqueness statement can easily be proven. Let's assume we have elements $y,y'$ such that $f(x)=\langle x,y\rangle$ and $f(x)=\langle x,y'\rangle$ holds true for all $x \in H$. This implies $\langle x,y-y'\rangle=0$ for all $x \in H$.
Then, it particular it must hold for $x :=y-y'$ which implies $y=y'$.
I agree that is not clear in the first place that $y$ is independent of the choice of $z$ but we can show it this way. I am not sure whether there is way to see this differently.