Questions about the proof of the Riesz representation theorem Let $H$ be Hilbert space, $f:H \rightarrow \Bbb F$ linear and bounded map.
I'm trying to prove that there exists only one $z_0 \in H$ such that:
 $ \forall_{x \in H} : f(x)=\langle x,z_0\rangle$
Proof: 
If $f \equiv 0 $ then we have $z_0 = 0$.  
If $f \not\equiv 0$ then:
Let $K:= \ker f = \{x\in H | f(x)=0 \}  $
$K$ closed, because $f$ is continuous, so $f^{-1}(\{ 0\} )$ is closed, as $\{ 0 \} $ is closed. (- thank you, TooOldForMath)  
$H = K \oplus _{\bot} K^{\bot}$ 
This implies that $H / K \cong K^{\bot}$.
Then $ \Bbb F \cong H/K $.
From this I should assume that $ \dim K^{\bot} = 1$.
$ 0 \neq z \in K^{\bot}$.
$z_0 := \frac{ \overline{f(z)}}{\langle z,z\rangle} \cdot z$
Is it true, that $\forall_x : f(x) = \langle x,z_0\rangle$?
$x= \overline{x} + y$, $\overline{x} \in K$, $y\in K^{\bot}$.
It implies $f(x)=f(y)$, because $f(\overline{x}) = 0$.
$y=rz \Rightarrow f(y)=r f(y)$
So now:
$\langle x,z_0\rangle = \langle \overline{x} + y,z_0\rangle = \langle \overline{x},z_0\rangle + \langle y,z_0\rangle = \langle y,z_0\rangle = \langle y,\frac{ \overline{f(z)}}{\langle z,z\rangle} \cdot z \rangle = \frac{ \overline{f(z)}}{\langle z,z\rangle} \cdot \langle y,z \rangle = \frac{ \overline{f(z)}}{\langle z,z\rangle} \cdot \langle rz,z \rangle = rf(z)$
$\square$
Could you help me with those things I don't understand? Why:
1) $H = K \oplus _{\bot} K^{\bot}$;
2) $H / K \cong K^{\bot}$;
3) $ \Bbb F \cong H/K $;
4) $ \dim K^{\bot} = 1$ ?
 A: (1): The magical ingredient that makes Hilbert spaces behave so much nicer than general Banach spaces is the parallelogram identity:

Lemma: Let $H$ be a Hilbert space and $x,y\in H$. Then
  $$\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$$

Proof: Verify using $\|x\|^2=\langle x,x\rangle$ and straightforward calculation. $\square$

Theorem: Let $H$ be a Hilbert space and $K\subset H$ a closed subspace. Then
  $$H=K\oplus K^\perp$$
  Proof: Let $x\in H$ be fixed. We want to find $y\in K$ and $z\in K^\perp$ such that $x=y+z$. The idea to get $y$ is to project $x$ orthogonally onto $K$. Geometrically this means finding a $y\in K$ with minimal distance to $x$ (imagine that $H$ is the plane and $K$ is a line!). Let $d=\inf\{\|x-y\|\,:\, y\in K\}$. Now choose a sequence $(y_n)_n$ with $y_n\in K$ such that $\|x-y_n\|\rightarrow d$ as $n\rightarrow\infty$ (exists by definition of the infimum). Plugging in $x-y_n$ and $x-y_m$ into the parallelogram identity shows that $(y_n)_n$ is a Cauchy sequence. Because $K$ is closed, $(y_n)_n$ converges to some $y\in K$. Also, $\|x-y\|=d$, that is, $y$ assumes the minimal distance to $x$.

Set $z=x-y$. We still have to show $z\in K^\perp$. So let $w\in K$. The real-valued function $h(t)=\|z+tw\|^2$ with $t\in\mathbb{R}$ is differentiable and has a minimum at $t=0$, because 
$$\|z+tw\|=\|x-\underbrace{(y-tw)}_{\in K}\|\ge \inf_{u\in K}\{\|x-u\|\}=d=\|x-y\|=\|z+0\cdot w\|$$
There the derivative has a zero at $t=0$: $0=h^\prime(0)=2\langle z,w\rangle$. So $\langle z,w\rangle=0$. Because $w\in K$ was arbitrary, we showed $z\in K^\perp$.
Together, we have shown $H=K+K^\perp$. We still have to prove that the sum is direct, i.e. $K\cap K^\perp=\{0\}$. So let $x\in K\cap K^\perp$. Then $\|x\|^2=\langle x,x\rangle=0$, so $x=0$. $\square$

(2): What we did in the theorem above was to construct the orthogonal projection operator on a closed subspace:
$$\pi_K:H\rightarrow K$$
which takes some $x\in H$ onto the unique $y\in K$ such that $x=y+z$ for some $z\in K^\perp$. This is a linear operator with kernel $K^\perp$ and image $K$. 
Instead of $\pi_K$ we can also look at $\pi_{K^\perp}$ ($K^\perp$ is also a closed subspace). This one consequently has kernel $(K^\perp)^\perp=K$ and image $K^\perp$. 
So the first isomorphism theorem for vector spaces shows
$$H/K=H/\mathrm{ker}\,\pi_{K^\perp}\cong \mathrm{im}\, \pi_{K^\perp} = K^\perp$$
Because $K^\perp$ is also a closed 

(3): Because also $f:H\rightarrow\mathbb{F}$ is a linear map, we can apply the isomorphism theorem also to $f$. The kernel of $f$ is $K$, by definition. The image of $f$ is $\mathbb{F}$ because $f\not\equiv 0$. Therefore
$$\mathbb{F}=\mathrm{im}\,f\cong H/\mathrm{ker}\,f=H/K$$

(4): Combining (3) and (4) we get
$$K^\perp \cong H/K \cong \mathbb{F}$$
Therefore the $\mathbb{F}$-dimension of the vector space $K^\perp$ is $1$.

I almost forgot the credible and/or official sources:


*

*Stein, Shakarchi - Princeton Lectures in Analysis - Real Analysis - Vol. 3, Chapter 4, Sections 4 and 5

*Dirk Werner - Functional Analysis, Chapter V

*Rudin - Real and Complex Analysis, Chapter 4
A: $f$ is continuous, because its bounded, so $K=f^{-1}(\{0\})$ is closed as preimage of a closed set under a continuous map.
