Theorem 3 on page 80 of kolmogorov and fomin's volume 1 Hi All: I will write out exactly what is in the book in case people don't have the book. I understand the first part of the theorem but not the second part. What follows is almost word for word from Theorem 3 starting on page 80 of Kolomogorov and Fomin, volume 1.
Theorem 3: let $f(x) \neq 0 $ be a given functional. The subspace $L_f$ ( defined as a subspace where any point in it is such that $f(x) = 0$ ) has index equal to unity. i.e. an arbitrary element $y \in R$ can be represented in the following form:
$$y = \lambda x_{0} + x $$ where $x \in L_{f}$ and $x_{0} \not\in L_{f}$
Proof: Since $x_{0} \not \in L_{f}$, we have $f(x_{0}) \neq 0 $. If we set $\lambda = 
\frac{f(y)}{f(x_{0}}$ and $x = y - \frac{f(y)}{f(x_{0})}$, then $y = \lambda x_{0} + x$, where $f(x) = f(y) - \frac{f(y)}{f(x_{0})}f(x_{0}) = 0$.
If the element $x_{0}$ is fixed, then the element $y$ can be represented in the form $(5)$ uniquely. This is easily proved by proving the contrary.
In fact, let
$$ y = \lambda x_{0} + x $$
$$y = \lambda^{\prime} x_{0} + x^{\prime} $$
then $$ (\lambda - \lambda^{\prime})x_{0} = (x^{\prime} - x)$$
Now, if $(\lambda - \lambda^{\prime}) = 0$, then $(x^{\prime} - x) = 0$. On the other hand, if $(\lambda - \lambda^{\prime} \neq 0$, then
$x_{0} = \frac{(x^{\prime} - x)}{(\lambda - \lambda^{\prime})} \in L_{f}$ which contradicts the condition that $x_{0} \not \in L_{f}$
Therefore, $(x^{\prime} - x) = 0$ which means that the representation for $y$ is unique.
Conversely, given a subspace $L$ of $R$ of index 1,$L$ defines a continuous linear functional $f$ which vanishes precisely on $L$. Indeed, let $x_{0} \not\in L$. Then, for any $x \in R$, $x = y + \lambda x_{0}$ with $y \in L$, $x_{0} \not\in L$. Let $f(x) = \lambda$. It is easily seen that $f$ satisfies the above requirements. If $f$ and $g$ are two such linear functionals defined by $L$, then $f(x) = \alpha g(x)$, for all $x \in R$, $\alpha$ a scalar. This follows because the index of $L$ in $R$ is 1.
I follow the theorem above up until the part that starts with "Conversely". I'm not even clear what is being proven in the converse nor the steps of the proof. If anyone could explain it, it's appreciated. They also, switched the places $x$ and $y$, compared to the first part of the proof, when representing a point in $R$ and I'm not sure why they did that. Also, I'm not clear on what requirements are being satisfied ? Thanks a lot for any enlightenment.
 A: The theorem you have quoted establishes an equivalence between functionals $f$ and subspaces of index $1$, namely $L_f$.  So the theorem has two parts to it: one which says that given a functional $f$ there is a subspace of index $1$ arising from it (which is the first part, which you say you understand).  The second part says that given a subspace of index $1$ called $L$ you can induce a functional $f$ from it.  "Conversely" introduces the second part of the proof.
The proof of the second part assumes that a subspace $L$ of index $1$ has been given, which means that there is at least one point $x_0$ lying in $R$ but not $L$ (since the index of $L$ in $R$ is $1$).  Then we note that we can write an arbitrary point of $R$ as $y+\lambda x_0$ (where $\lambda$ is possibly $0$) and from that we construct a linear functional $f(x) = \lambda$, and so we can rewrite our equation for $x$ as $x = y + f(x)x_0$ where $y\in L$ and $x_0 \not\in L$.  This is what is meant by meeting the 'requirements' -- this was given in the statement of the theorem as the representation of an arbitrary element of $R$.
This might seem very basic -- it is.  This is essentially the 'trivial' side of the claim and pretty much proceeds by writing down what you know and seeing that it matches the definition.
As for the notation switching -- this happens.  It would be slightly nicer if the notation remained consistent for both parts of the proof, but it's not essential and it doesn't mean much more than the writer prefers to use certain letters for certain things.
