field theory - Is the Lagrangian density a functional or a function?

Weinberg at page 300 of The Quantum Theory of Fields - Volume I says:

$L$ itself should be a space integral of an ordinary scalar function of $\Psi(x)$ and $\partial \Psi(x)/\partial x^\mu \,$, known as the Lagrangian density $\mathscr{L}$:

$$ L[\Psi(t), \dot{\Psi}(t)]= \int d^3x \, \mathscr{L}\bigr(\Psi({\bf x},t), \nabla \Psi({\bf x},t), \dot{\Psi}({\bf x},t)\bigl) $$

So he says that $\mathscr{L} \, $ is a function. But Gelfand and Formin at page one of their book Calculus of variations say:

By a functional we mean a correspondence which assigns a definite (real) number to each function (or curve) belonging to some class.

So from that I'd say it is a functional. The notes of quantum field theory of my professor stay on this side, explicitly calling the lagrangian density a functional.

I'm very confused at the moment. I always used this latter way of defining functionals (the Gelfand way) so Weinberg saying that $\mathscr{L}$ is a function confuses me.

Can someone makes some clarity about this?

Answer

The Lagrangian density is a function.

Consider the following examples: $$ A[f]=\int_0^1\mathrm dx\ f(x) $$ and $$ B(f(x))=f(x) $$

It is clear that $A$ is a functional, because for example $$ A[\sin]=1-\cos1=0.45\in\mathbb R $$ is a number, while $B$ is a function, because $$ B(\sin)=\sin $$ is not a number, but a function.

In your notation, $L$ is a functional, because given a certain field configuration, you get a number. But $\mathscr L$ is a function, because given a certain field configuration, you get another function, not a number.

In some cases, such as QED in the Coulomb gauge, you may want to include non-local terms in the Lagrangian density, which makes it into a function of some of its arguments, and a functional of the others. This is an exception to the rule above.