# I What is a "functional"

1. Mar 12, 2017

### David112234

I have read the wiki page (https://en.wikipedia.org/wiki/Functional_(mathematics) but it is not helping.
I understand what a regular function is
input > do something to that input > output.
but not what functional is,
Wikipedia says "from a vector space into its underlying field of scalars"
That wording is confusing.
So the input is a vector but the output is a scalar? The only way I know to get scalar from vector is a dot product.
And a vector is just written as <a,b> , if you take that as an input isn't it just essentially a regular function of 2 variables? y=f(a,b)

2. Mar 12, 2017

### Staff: Mentor

The same with the restriction to those functions, whose output is in the field of scalars.
It is probably for traditional reasons. The "-al" ending simply shall indicate that scalars are the output.
This is one possibility, but you could do about anything with the coordinates of the vectors, e.g. applying the norm, which isn't a linear functional anymore. Another famous (linear) functional is the gradient or $\nabla-$operator, but this is again a dot product. The word functional is often used as a linear function only. I'm not aware that someone calls the norm a functional. Thus with the restriction to linear functionals, all are a dot product with a certain vector.
Yes.

3. Mar 12, 2017

### David112234

So a functional is just a special case of a function where the input is a vector and output is a scalar?
like Curvature?
Does being linear/ non linear has anything to do with it?

And if f(<a,b>) is the same as f(a,b) what is the significance of this?

Also, I read this topic https://www.physicsforums.com/threads/function-vs-functional.427168/ which seems to make is a lot more complicated
What do they mean by

"Fix a point a.
the assign to each function f its value f(a) at a. The map"

and
"The function f is the variable here. It varies in the space of all functions producing a different (in general) number for each function.
You should imagine the space of all functions - it is infinite dimensional, and draw a surface over this space."

How I interpreted that is, you chose one input, and then vary the "do something to that input" part, like select the point (3,0) and then try x^2, x^5 , 3x+1, and so on and compare what function will product the desired result.

4. Mar 12, 2017

### Staff: Mentor

As I tried to say with the example of the norm: without being restricted to linear functionals, there are plenty.
This depends on the definition you use. In the end it's just a word. Those terms usually describe the same principle, but are not always used in the exact same way. Similar could be discussed about operators. Linear? Non-linear? It depends on context, taste and consistency.
$f(<a,b>) \neq f(a,b)$. One has a one dimensional domain, the other one a two dimensional. Please remain clear and exact, if you start a discussion on the usage of a term, which isn't used the same anywhere by itself. A linear functional (on finite dimensional vector spaces) can be written by its matrix: $f(x_1,\ldots , x_n) = (f_1, \ldots , f_n) \cdot (x_1,\ldots x_n)^\tau$ and is therefore a dot product. The point to call it functional instead is for traditional reasons, and due to the fact, that it is often used on (infinite dimensional) function (vector) spaces. This way it can be distinguished from the functions it applies to.
See above. $F : C^\infty(\mathbb{R}) \rightarrow \mathbb{R}$ defined by $F(f)=f(0)$ is a linear(!) functional on $C^\infty(\mathbb{R})$. How would you write this as a dot product?
If you like. Unusual, but why not.

5. Mar 12, 2017

### David112234

"And a vector is just written as <a,b> , if you take that as an input isn't it just essentially a regular function of 2 variables? y=f(a,b)"
f(<a,b>) = f(a,b) is what I meant, you said "yes"
Does my original question mean something else than what I thought I was asking?
And I am not very familiar with matrix

and why is one a 1 dimensional domain, a and b are supposed to represent the x and why variables, wouldn't both domains be 2 dimensional then?
I do not understand what this means.

So a function where the input is a vector and output is a scalar is the more usual interpretation? are both these interpretations equivalent?

Last edited: Mar 12, 2017
6. Mar 12, 2017

### Staff: Mentor

Sorry, my fault, I only saw the $<a,b>$.
$C^\infty(\mathbb{R})$ is the vector space of all smooth (infinitely often differentiable) real valued functions. So $F$ assigns each vector from $C^\infty(\mathbb{R})$, which are functions $f$, a value $f(0)$. This is a linear functional, and the term functional for $F$ helps to distinguish it from its variables, the functions $f$.
Yes, and as one usually considers mappings, that preserve the structures, functionals are often required to be linear as well, as the domain in this case is a vector space. If the domain is simply a topological space, then the functionals are (usually) required to be continuous, and if the domain is a space of smooth functions, the functionals are (usually) required to be smooth as well. Often it is a combination of these. This is what I meant by context. It should be clear by the respective context (if not explicitly defined), what exactly is meant. Personally, I try to avoid the term outside functional analysis.
Not quite sure, what you mean by the other one.

7. Mar 12, 2017

### Stephen Tashi

Yes, if you keep in mind that some sets of real valued functions can be regarded as vector spaces.

Some sets of functions obey the definition of a vector space - after all, you can add two functions and get a function, you can multiply a function times a scalar and get a function, etc. One intuitive view is that a real valued function is a vector with an uncountably infinite number of coordinates.

It's common to encounter functionals defined on vector spaces of functions. For example, a functional $F(f)$ can be defined by taking a fixed function $g(x)$ and defining $F(f) = \int_{-\infty}^{\infty} f(x) g(x) dx$. The fixed function $g(x)$ is often called the "kernel of integration". In the case where $g(x) = x$ and $f$ is a probability distribution, the functional $F$ gives the expected value of the distribution $f$.

8. Mar 13, 2017

### David112234

So from doing further research this is my current understanding, a functional comes from the calculus of variations.
You have the formula for the length of a curve
∫√(dx2+dy2) = L
∫√((dy/dx)2+1) dx = L

So the curve length of a some function = some length.
and what ever that function is, is the input and you find the function that will give you the smallest length?
and by the definition previously stated that input is written as a vector function r(t) ?

so F(ƒ)
F is the curve length formula
ƒ is the function that gives the curve you are finding the arc length of

Is this all correct?

9. Mar 13, 2017

### Stephen Tashi

Do you mean that one example of a functional is studied in the calculus of variations? Yes, that's true.

Yes, provided those are definite integrals.

Are you trying to describe a functional or are you describing the goal of the calculus of variations? Yes, one goal of the calculus of variations is to minimize various functionals.

What do you mean by a "vector function"?

That's one example of a functional.

10. Mar 14, 2017

### David112234

I suppose both since calculus of variations is how I was introduced to a functional. If ∫√((dy/dx)2+1) dx = L is a functional and ∇operator or curvature function are both functionals, I do not see the similarity between them, the input for ∇ and curvature (K) are vector functions. In order for ∫√((dy/dx)2+1) dx = L to satisfy the functional definition that the input has to be a vector, does y have to written as a vector function?
And the ∇operon does not output a scalar, so how is it also a functional?

And by vector function I mean something like this r(t) = < sin(t) , t^2 > where as t varies the tip of the vector traces out a curve.

Also in the calculus of a variation L = ∫√((dy/dx)2+1) dx is written as L = ∫ F( x, y, y' ) dx
Is this the general form of all functionals?

11. Mar 14, 2017

### Stephen Tashi

No. Some sets of real valued functions of a real variable can be considered vector spaces. For example, consider the set of all functions defined by polynomials of finite degree (which includes the constant polynomial $p(x) = 1$ of degree 0). A polynomial in this set can be considered to be a vector, since the set satisfies the properties of a vector space with the appropriate definitions for vector addition and scalar multiplication. We say that to add two vectors $p_1(x)$ and $p_2(x)$ we form the function $p_1(x) + p_2(x)$ which is equal to another polynomial function $p_3(x)$. To multiply a "vector" $p_1(x)$ in this space by a scalar $c$ we form the function $c\ p_1(x)$ , which is also a polynomial and hence also a "vector" in this vector space.

An example of a basis for this vector space is the set of "vectors" $\{1, x, x^2, x^3,...\}$. In that basis, the polynomial $f(x) = 3x^2 - 2x + 5$ has coordinates (5, -2, 3, 0,0,0,...).

The gradient operator isn't a functional.

No. The only "general form" for a functional is that it must a function who domain is a set of functions and whose co-domain is set of real numbers.

Also, you statement that "L = ∫√((dy/dx)2+1) dx is written as L = ∫ F( x, y, y' ) dx" is incorrect. The calculus variations deals with various different types of problems. One type of problem can involve finding a curve of minimum length. A different kind of problem can be finding a curve that a particle travels in the minimum amount of time. Another type of problem involves finding the curve of "least action". Different problems have different "L"'s.