Not understanding the operator $I(f)(x)=\int_0^x f(t)dt$. In my course I have the following operator :
Let define $$I: \mathcal C^0([0,1])\longrightarrow \mathbb R$$
by $$I(f)(x)=\int_0^x f(u)du$$
for all $x\in [0,1]$.
Since $I(f)$ is a number, isn't false to write $I(f)(x)$ ? Shouldn't we have $$I:\mathcal C^0([0,1])\times [0,1]\longrightarrow \mathbb R \ \ ?$$
 A: Why do you say that $I(f)$ is a number? It isn't; it's a function.
A: Your definition of $I$ is wrong; it is a map from $C^0([0,1])$ to $C^1([0,1])$.
A: $I$ is indeed an operator that maps a (real $\mathcal C^0$) function (defined in the unit interval) to a (real $\mathcal C^1$) function (defined in the unit interval).
You have
$$I:\mathcal C^0([0,1])\to \mathcal C^1([0,1]):f\to\int_0f(t)\,dt^*,$$
an operator that maps a function to a function, and
$$I(f)\in\mathcal C^1(\mathbb R):[0,1]\to\mathbb R: x\to \int_0^xf(t)$$
a function that maps a real in the unit interval to a real.
$^*$Anyway, the notation of the integral with no upper bound, to reflect the absence of the argument of $f$ and its antiderivative, does not exist !
A: $f$ is a function; $x$ is a number that is its input; $f(x)$ is a number that is its output.
Similarly $I(f)$ is a function; $x$ is a number that is its input; $I(f)(x)$ is a number that is its output.
The statement that $I : \mathcal C^0([0,1]) \to \mathbb R$ is wrong. If $I$ is defined in the way you give, then it should say $I:\mathcal C^0([0,1]) \to \mathcal C^0([0,1])$ or $I:\mathcal C^0([0,1]) \to \mathcal C^1([0,1]).$
(I would prefer to write it as $(If)(x).$)
A: I think the easiest way to think about this is with a concrete example.
Suppose $f(x) = x^2$.  Let's figure out what $I(f)(x)$ is.  From the definition,
$$I(f)(x) = \int_0^x u^2 du 
=\frac{1}{3} x^3 $$
and in general the operator $I$ takes a function $f(x)$ and maps it to an antiderivative $F(x)$ satisfying the initial condition $F(0) = 0$.
