Two parts I cannot understanding on simple proof about Hilbert Space I am currently studying Hilbert Space in Real analysis, and I have a part not understandable. This is a theorem for Hilbert Space $H$.
$Theorem$ : 
If $L$ is a bounded linear functional on $H$, then there exists a unique $y \in H$ such that $L(x) = <x,y>$
$Proof$
Proof for uniqueness of $y$.  If $L(x) = <x,y> = <x,y'>$, then $<x,y-y'> = 0$ for all $x$. For $x = y-y'$, $y-y' = 0$. 
Proof for existence. If $L(x) = 0$ for all $x$, take $y = 0$. Otherwise, let $M = \{x : L(x) = 0 \}$. take $z \neq 0$ in $M^ㅗ$, and let $y = \alpha z$ where $\alpha = \frac{\overline{L(z)}}{<z,z>} $. Notice that $y \in M^ㅗ$, 
$$
L(y) = \frac{\overline{L(z)}}{<z,z>}L(z) = \frac{|L(z)|^2}{<z,z>} = <y,y> 
$$
and $ y \neq 0$
If $x \in H$ and 
$$
w = x - \frac{{L(x)}}{<y,y>}y,
$$
then $L(w) = 0$, so $w \in M$, and hence $<w,y> =0$. Then 
$$
<x,y> = <x-w,y> = L(x)
$$ 
as desired

Then I have two questions. 
First, it seemed we used the fact $L(x) = <x,y>$ during the proof. Is it right? I think we have to prove without using $L(x)$ is an inner product. 
Second, how do we guarantee that $z \in M^ㅗ$ exists?
Thanks in advance. 
 A: If no such $z\neq 0$ in $M^{\perp}$ exists, then since $M$ is a closed subspace, either $M=H$ or $M^{\perp}\neq\{0\}$ (every closed subspace is complemented). 
As for the first question, I think you are misunderstanding the proof. The existence part does not assume $L(x)=\langle x, y\rangle$ for some $y$, you are using the value of $L$ at $z$ to create this $\alpha$ and hence $y$. We are claiming this $y$ will be the one such that $L(x)=\langle x, y\rangle$. The proof then goes on to show that this is indeed the case. Now the uniqueness says that if we have two such $y$'s that do this, they must actually be the same.
A: The existence of vectors in $M^{\perp}$ , i.e., showing it is not empty, can also be shown like this: find a basis for $M$ , which always exists, if you accept choice, i.e., if you're pro-choice*. Then extend this basis into a basis for the whole space.Then do the Gram-Schmidt algorithm , first on the basis for $M$ , and then for the basis for $H-M$ . The "orthogonalized" basis for $H-M$ give you elements of $M^{\perp}$ , and all the linear combinations of these elements give you more elements in $M^{\perp}$ , showing $M^{\perp}$ is non-empty. And I agree with user 43687 on the fact that $L(x)=<x,.>$ is just a definition, and you have to show that there exists a single $y$ with  $L(x)=<., y>$; despite the fact that $99680-43687=55993$, we're not that far apart in our views on this.
EDIT, as suggested: Notice here that $M$ is a closed subset, since the map $L(x)=<x,y>$ is continuous, e.g., by Cauchy-Schwarz, so that $M=L^{-1}(0)$ is closed. 


*

*hope I don't get in trouble for this stupid joke.

A: In your demonstration, you should have $y = \alpha z$  instead of $y = \alpha x$ 
