In the Lagrangian formalism, $L$ itself is NOT a functional and $q$ and $\dot q$ are not functions of $t$, but as soon as we write the action, it becomes a functional.

Before: $L(q, \dot q, t)$

After: $S = \int_{t_1}^{t_2} L(q(t), \dot q(t), t)$

What is it that allows us to decide that: "ah, now, we can now start saying $q$ is a function of $t$ and $\dot q$ then is automatically the derivative of it ? Why can we do that ?" Is there some solid explanation about this ? I mean before we wrote an action, $L$ was thought to be the function of $q, \dot q$ which didn't depend on t, but in actio…

@naturallyInconsistent I was asleep, sorry. I think I didn't understand your answer. Before moving further, best would be just to agree on these.

1. I'm saying that in the L step, $q$ and $\dot q$ are not functions of $t$. This is what's said here - https://physics.stackexchange.com/a/2895/366606 . so L is just a function of two variables and is not functional.

2. If (1) is correct, which I believe it is, then at that step, $\dot q$ can NOT be a derivative of $q$. So we can just say to have $v$ instead of $\dot q$ to avoid confusion. If they are not functione of $t$, then you can't just…

@ACuriousMind

In the Lagrangian formalism, $L$ itself is NOT a functional and $q$ and $\dot q$ are not functions of $t$ and $\dot q$ is not a derivative of $q$, but as soon as we write the action, it becomes a functional.

Before: $L(q, \dot q, t)$

After: $S = \int_{t_1}^{t_2} L(q(t), \dot q(t), t)$

What is it that allows us to decide that: "ah, now, we can now start saying $q$ is a function of $t$ and $\dot q$ then is automatically the derivative of it ? Why can we do that ?" Is there some solid explanation about this ? I mean before we wrote an action, $L$ was thought to be the functio…

@ACuriousMind

first, we got L which is function of $q, \dot q$ ($q$ and $\dot q$ are not functions of $t$ and $\dot q$ is not derivative of $q$).

This L is just single value, but we need its values at every point in time, and if so, its value at any moment in time would always depend velocity and position. but position and velocity also depend on time.

My question is, that I get why in the action, q and $\dot q$ are dependent on $t$, but then I don't get why it's so wrong to say that in the Lagrangian only(without action), they can't depend on $t$. We coud just say that even in $L$, q an…

Not sure why it's confusing.

function maps a number to number. functional maps a function to a number.

f(x) = 2x+3. when x=3, f(x) = 9, so it mapped 3 to 9.

but functional, maps a function to a number. So we got some main function that accepts its input/argument as a function.

L(q(t)) = 10q(t)

when q(t) is 2t, L(q(t)) = 20t so it didn't map the function to a number, but it mapped function to a function which is not functional.

but in action, since we got definite integral, it maps a function to a number.

1. Our goal is to get to the action, a functional $S : [\mathbb{R},\mathbb{R}^{2n}]\to \mathbb{R}$, where by $[-,-]$ I mean the space of functions between left and right argument in the brackets.
2. We have a function $L : \mathbb{R}^{2n}\mapsto \mathbb{R}, (q,v),\mapsto L(q,v)$
3. Function composition is itself a map $L_\ast: [\mathbb{R},\mathbb{R}^{2n}]\mapsto [\mathbb{R},\mathbb{R}], q\mapsto L\circ q$.
4. Integration is a map $I : [\mathbb{R},\mathbb{R}]\to \mathbb{R}, f\mapsto \int f$.
5. The action is the composition $S = I\circ L_\ast$.