In a book I'm reading (functional analysis of Stein and Shakarshi), they say that a distribution $F$ is not be given by assigning value of $F$ at most point, but will instead be determinated by its average taken with smooth function. So if we think a function $f$ as a distribution $F$, we determine $F$ by the quantity $$F(\varphi )=\int_{\mathbb R}f(x)\varphi (x)dx,$$ where $\varphi $'s range over an appropriate test function.
1) Something I don't get, when they say : "if we think a function $f$ as a distribution $F$", does this mean that $f$ and $F$ are the same ? How can we think $f$ as $F$ ? (or $F$ as $f$ ). But $f$ and $F$ looks very different...
2) A little bit farther for example, it says that if $f$ is smooth, then $$\int f'\varphi =-\int f\varphi '.$$ Therefore, if $f\in L^1(\mathbb R)$ locally, we can define the derivative of $f$ as the distribution as $F'(\varphi )=-\int f\varphi '$. So $F'$ is considered as the dérivative of $f$ or not ? I don't understand. It's the same ?
3) Also, it says that $F$ is not a function. For me it's a function from $$\mathcal C_0^\infty(\Omega )\to \mathbb R.$$ So in what isn't it a function ? For example, let $$F(\varphi )=\int_{\mathbb R}f\varphi ,$$ where $f\in L^1$ and $\varphi $ smooth. In what isn't it a function ? $F(\varphi )$ is well defined (I guess). May be someone could explain ?
The book is pretty misleading, a distribution $F$ is indeed a linear function $F: C^{\infty}_0 (\Omega) \to \mathbb R$.
Given any $f \in C_0^{\infty}(\Omega)$ we can define a corresponding distribution $F_f$ by
$$ F_f(\varphi) = \int f \varphi \; \text{d} \varphi $$
This is common enough that people often abuse notation and write $f$ for both the function $f \in C_0^{\infty}(\Omega)$ and the distribution $F_f: C_0^{\infty}(\Omega) \to \mathbb R$. $f$ and $F_f$ are indeed very different objects, but people will use $f$ to refer to both. Normally this won't cause much confusion as it will be quite clear which they are referring to by what they're acting on.
This all gets at the motivation behind defining what a distribution is. The classic example of a distribution is the Dirac Delta $\delta_0$. You've probably already met this at some point in your studies but in a very informal way, where you've said it's a function such that $$ \int \varphi(x) \delta_0(x)\; \text{d}x = \varphi(0) $$ Of course this is actually nonsense, no such function exists. But we really want to pretend like such a function exists.
Instead we can formalise this idea as a distribution. Considering $\delta_0$ as a distribution $C^{\infty}_0(\mathbb R) \to \mathbb R$ defined by $\delta_0(\varphi) = \varphi(0)$ makes perfect sense. However we still want to keep the intuition that "$\delta_0$ is a function such that $\int \varphi(x) \delta_0(x)\; \text{d}x = \varphi(0)$", despite that fact that such a function does not exist.
For example, how would we define the derivative of $\delta_0$? Let's pretend that $\delta_0$ is in fact a $C^{\infty}_0(\mathbb R)$ function. Then using integration by parts we have $$ \int \varphi(x) \delta'_0(x) \; \text{d}x = - \int \varphi'(x) \delta(x)\; \text{d}x = - \varphi'(0) $$ (since the supports are compact, all boundary terms disappear). Now $\delta_0'(f) = -\varphi'(0) = -\delta_0(\varphi')$ is a perfectly legitimate distribution, and this leads us to an idea to define the derivative of a distribution $F$ as $F': C^{\infty}_{0}(\Omega) \to \mathbb R$ defined by $$ F'(\varphi) = - F(\varphi') $$ This is an idea that permeates distribution theory. Whenever we want to manipulate a distribution, we pretend as if the distribution is actually a $C^{\infty}_0(\Omega)$ function and try and get to something which we can define via the distributions.
One thing I haven't mentioned is that not every linear function $F: C_0^{\infty}(\Omega) \to \mathbb R$ is a distribution, they have to satisfy a continuity condition. With this condition, as reuns mentions below, every distribution $F$ can actually be written as a limit of $C^{\infty}_0(\Omega)$ functions $f_n$ "understood" as distributions $F_{f_n}$. Hence pretending a distribution is a $C^{\infty}_0(\Omega)$ function is not as silly as it might sound.