Is the definite integral a special case of functionals?

[jack476]

So yesterday I learned about functionals, which my book claims are "machines that take a function and return a number", in contrast to functions, which take a number and return another number. I immediately thought of definite integration: it's an operation that takes a function, and returns a number.

I did a Google search for this and found the Wikipedia page on functional integration, which says that a functional integral is an integral where the domain is not a region of space but a space of functions. Again, could the definite integral be a special case of this where the "space of functions" is just all of the values between the bounds of integration treated as constant functions?

And finally, when you take the derivative of a function like 3x, which is of course 3, is that also a functional? This seems like a bit more of a stretch, because taking the derivative of a more complicated function like x^2 gives back 2x which is a function, not a number.

Just wondering.

 

[Einj]

Yes, the definite integral is a functional since, as you said, once you "feed" it with a function it gives you back a number. However, the derivative operator is not a functional since it does not give a number for every element of the space of functions.

 

[Stephen Tashi]

in contrast to functions, which take a number and return another number

Functionals are functions. They aren't "real valued functions of one real variable", but the definition of "function" is general enough to include functionals as being functions.

If you establish constant limits of integration on a definite integral, like $\int_0^1 f(x) dx$ or $\int_{-\infty}^{\infty} f(x) dx$ and consider $f$ the variable, then you have a functional. If you have a variable in the limits of integration, like $\int_0^x f(t) dt$ you don't get a functional.

You can define functionals that involve a function as a variable and other functions as non-variables. For example $\int_0^1 f(x) e^x dx$ is a functional if we consider $f$ the variable.

 

[micromass]

However, the derivative operator is not a functional since it does not give a number for every element of the space of functions.

Neither does the definite integral.

The derivative is certainly a functional, a functional does not mean that you need to assign a number to every function (since there would be very few functionals then). You can actually restrict the domain to a suitable space of functions.

 

[Svein]

A common functional in the L₂ space is integration against a square-summable kernel K: $h=Af$ given by $h(x)=\int_{-\infty}^{\infty}K(x,s)f(s)ds$.