Papa Rudin $4.12 $ theorem There are things that we need for the proof of the theorem:
]1

There is the theorem:
If $L$ is a continuous linear functional on $H$, then there is a unique $y$ $\in$ $H$ such that
$Lx$ $=$ $(x,y)$ ($x$ $\in$ $H$).
There is the proof:
If $Lx$ $=$ $0$ for all $x$, take $y$ $=$ $0$.
Otherwise, define
$M$ $=$ ${x: Lx = 0 }$.
The linearity of $L$ shows that $M$ is a subspace. The continuity of $L$ shows that $M$  is closed. Since $Lx$ $\neq$ $0$ for some $x$ $\in$ $H$, Theorem $4.11$ shows that $M^\bot$ does not consist of $0$ alone.
Hence there exists $z$ $\in$ $M^\bot$, with $||z||$ $=$ $1$ . Put
$u$ $=$ $(Lx)z$ $-$ $(Lz)x$.
Since $Lu$ $=$ $(Lx)(Lz)$ $-$ $(Lz)(Lx)$ $=$ $0$, we have $u$ $\in$ $M$.
Thus $(u,z)$ $=$ $0$.
This gives
$Lx$ $=$ $(Lx)$$(z,z)$ $=$ $(Lz)$$(x,z)$.
Thus the theorem holds with $y$ $=$ $\alpha z$, where $\bar \alpha$ $=$ $Lz$.
I don't understand why do we get $y$ $=$ $0$ if $Lx$ $=$ $0$ for all $x$.
I also don't understand why is $Lu$ equal of $(Lx)(Lz)$ $-$ $(Lz)(Lx)$ and how does it give $Lx$ $=$ $(Lx)$$(z,z)$ $=$ $(Lz)$$(x,z)$, and why does theorem hold when $y$ $=$ $\alpha z$, where $\bar \alpha$ $=$ $Lz$ ?
Any help would be appreciated.
 A: Let's go through your questions.

*

*Why can we take $y = 0$ when $Lx = 0$ for all $x$?
Well, suppose $Lx = 0$ for all $x \in H$. What we want is some $y \in H$ such that $$0 = Lx = (x, y)$$ for all $x \in H$. But we know that $(x, 0) = 0$ for all $x \in H$ (see the discussion at the beginning of Ch. 4 of Rudin), so we might as well take $y = 0$. (In fact, if you think about it, $y = 0$ is the only value of $y$ that makes this possible.)


*Why do we have $Lu = (Lx)(Lz) - (Lz)(Lx)?$
Remember that $L$ is a linear functional, which means that $L$ takes an element of $H$, and spits out a scalar. So, $Lx$ and $Lz$ are scalars.
Now, we have $u = (Lx)z - (Lz)x.$ Applying $L$, we have
$$Lu = L((Lx)z) - L((Lz)x).$$ But since $Lx, Lz$ are scalars and $L$ is linear, we can bring them out in front of $L$ to obtain
$$Lu = (Lx)(Lz) - (Lz)(Lx),$$ as claimed.


*How does $Lx = (Lx)(z, z) = (Lz)(x, z)$ follow?
First: do you understand why $(u, z) = 0$? (Hint: as Rudin notes, the previous question shows that $Lu = 0$, hence $u \in M$. How is $z$ defined?)
Given this fact, we have
\begin{align}
0 &= (u, z)\\
&= ((Lx)z - (Lz)x, z)\\
&= ((Lx)z, z) - ((Lz)x, z)\\
&= (Lx)(z, z) - (Lz)(x, z).
\end{align}
So, we have $(Lx)(z, z) = (Lz)(x, z).$ But why do we have $Lx = (Lx)(z, z)$? Well, note that we chose $z$ such that $||z|| = 1.$ What is the relationship between $(z, z)$ and $||z||$?


*Why does the theorem hold when $y = \alpha z$, where $\overline{\alpha} = Lz$?
To see that the theorem holds, we need to show that
$$Lx = (x, y).$$ In other words, we need to show that
$$(x, \alpha z) = Lx.$$ What happens if you bring a scalar out of the second entry of an inner product? Once you bring $\alpha$ out, the result follows from the fact that $\overline{\alpha} = Lz$ and the previous question.
