Why I can write function in $C([0, 1])$ as such? I do not understand a passage in the notes: I am dealing with a Banach Space $C([0,1])$ defined with the norm $||f|| = \sup_{[0,1]} |f(x)|$
In the exercise, we define the space $X = \{f\in C; f(0) = f(1) = 0\}$ and then we want to prove $X$ is a closed subspace. So far so good, I understand the proof. But then he says: show that $C[0,1]$ can be written as $X + \tilde{X}$ where $$\tilde{X} = \{f\in C; f(x) = ax + b\}$$
where $a, b$ reals.
Now to prove this he starts with "Every function $f\in C[0,1]$ can be written as $f(x) = f_0(x) + ax + b$ where
$$a = f(1) - f(0)$$
$$b = f(0)$$
$$f_0(x) = f(x) - (ax  +b)$$
Here I do not get this: why can we say that EVERY function in $C[0,1]$ can be written in such way? How does he find out $a$ and $b$?
I perhaps got that $b = f(0)$ but still I need some explanation
 A: Intuition is a funny thing... Who knows what one person's intuition is? This proof is somehow so simple, it's hard to guess what any one person's intuition might be for this proof.
But, let me explain my intuition, for what it's worth. Since I'm a geometer, my intuition will be geometric. (I suspect that there are different geometric intuitions for this exercise, and I suspect algebraists might have completely different intuitions, e.g. the comment of @Hyperplane).
Imagine the graph $y=f(x)$ of the continuous function $f : [0,1] \to \mathbb R$ in the ordinary $x,y$ plane; draw whatever squiggly thing you like, although of course it should be continuous, and it should pass the vertical line test so that it is a function. And of course the values of $f(0)$, $f(1) \in \mathbb R$ can be anything whatsoever.
But then, we are all very familiar with superimposing new coordinate systems on top of old ones. So, here's a new $x',y'$ coordinate system. First, put the new origin $(x',y') = (0,0)$ at the point $(x.y) = (0,f(0))$. Next, put the new point $(x',y') = (1,0)$ at the point $(x,y) = (1,f(1))$. Now you have the $0$ and $1$ points on the $x'$ axis; draw the new $x'$ axis by passing through those two points. Finally, draw the new $y'$ axis lying right on top of the old $y$ axis, although with a different position for the origin as already stated. In this new coordinate system, name your new function $y' = f_0(x')$. Now you've got enough to verify that
$$f_0(x') = f(x') - \bigl((f(1)-f(0)) \, x' + f(0)\bigr)
$$
Finally, just substitute $x=x'$.
Well, perhaps there's one last step. Write down the outcome of this geometric intuition as a mathematical statement:

Theorem: For all $f \in C[0,1]$ there exists $a,b \in \mathbb R$ and $f_0 \in C[a,b]$ such that for all $x \in [0,1]$ we have
$$f(x) = f_0(x) + (ax + b)
$$

And then, as it turns out, the proof of this statement is shorter than the intuition. But that's how intuition is often used: for discovering theorems and their proofs. And if the proof is very short, sometimes the author will just write down the proof and discard the intuition.
