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Definition: A quantity $z$ is called a functional of $f(x)$ in the interval $[a,b]$ if it depends on all the values of $f(x)$ in $[a, b]$.

So the first order of business is this:

Is a functional a function?

And which direction does it go?

Do you start with a variable and evaluate it at the functional level and then use that as input to the function or the other way around?

Evaluate the function at some value and then place that in the functional's output.

Or is it not a function at all?

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    $\begingroup$ Please ask one question at a time. $\endgroup$
    – Shaun
    Jun 29, 2018 at 21:27
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    $\begingroup$ @Shaun Those questions wouldn't make any sense independently and all have the same answer. Please, check whether your stamp comment "one question at a time makes sense". $\endgroup$
    – user569098
    Jun 29, 2018 at 21:28
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    $\begingroup$ @Sedumjoy It looks like you are reading from a source using either an antiquated, or imprecise definition of what functional is. Normally a functional is not said to depend on all the values of $f$, but on $f$ itself. The values form a set, while functionals, in its classical meaning, are functions, which imputs are functions. So, a particular set of functions on $[a,b]$ would be considered. Say, all real functions on $[a,b]$, or perhaps only all continuous functions, denoted $C[a,b]$. And then a functional would be a function $F:C[a,b]\to \mathbb{R}$. $\endgroup$
    – user569098
    Jun 29, 2018 at 21:30
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    $\begingroup$ Example 1: $\int_{a}^{b}$ is a functional on $C[a,b]$, by sending a continuous function $f\in C[a,b]$ to the real $\int_{a}^{b}f(x)dx$. Example 2: "Evaluation at $(a+b)/2$" is a functional on $C[a,b]$, by sending a function $f\in C[a,b]$ to the real $f((a+b)/2)$. $\endgroup$
    – user569098
    Jun 29, 2018 at 21:35

3 Answers 3

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Composition of functions is when you "feed" the result of one function into another function to produce yet a third function. For example, if $f(x) = x^2$ and $g(x) = e^x$ then the composition $g\circ f$ would be defined by $(g\circ f)(x) = g(f(x)) = g(x^2) = e^{x^2}$. As you can see, the result is a function of $x$.

A functional, on the other hand, is when you "feed" a function -- a whole function, not just the value of the function at a specific point -- into some kind of "machine" that assigns a single numerical value to it.

For example, here are some examples of functionals:

  • $F(f) = \int_0^6 f(x) dx$. For $f(x) = x^2$, we'd have that $F(f) = 72$.
  • $G(f) = \max\{f(x) | -5 \le x \le 3 \}$. For $f(x) = x^2$, we'd have that $G(f) = 25$.
  • $H(f) = \textrm{the number of critical points of }f(x) \textrm{ on }[-5, 3]$. For $f(x) = x^2$, we'd have $H(f) = 1$.

Notice that when you apply a functional to a function, the result is a single number. That's what is meant by the statement that the value of $F(f)$ depends, in some sense, on the "entirety" of $f(x)$ in a particular domain.

Notice also that in each of these examples the definition of the functional requires some choice of interval; different choices would lead to different results. Finally, a particular functional may only be defined for certain classes of functions; for example, neither of the examples $F$ and $G$ above are not defined for a discontinuous function with a vertical asymptote at $x=2$. So in defining a function, one usually needs to limit one's attention to some category of "nice" or "good" functions on which the functional will operate.

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  • $\begingroup$ i guess i needed an example.....but now I am seeing this almost seems like a form of "u substitution" why did they invent a new name...you basically just stick the whole function in ...and use the interval of the given "functional" to evaluate it ? did I get it correct ? $\endgroup$
    – Sedumjoy
    Jun 29, 2018 at 22:25
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A functional takes a function and gives you a number.

For example the functional $$ \int _a ^b f(x) dx $$ takes $f(x)=x^2$ and turns it into $\frac {b^3-a^3}{3}.$

Another functional is $z=f''(0)$ which takes $f(x)=3x^2+1$ and turns it into $6$

As you see a functional is not a composite function, but it is an operator whose domain is a vector space of functions and its range is the field of that vector space.

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It is true that the historical origins of the notion of "functional" were as a thing that would receive input of a function (in some vector space of functions) and produce an output that was a scalar.

This was abstracted more than 100 years ago to the following idea, which is now standard: for a vector space $V$ over a field $k$, a $k$-linear map $V\to k$ is called a ($k$-linear) functional on $V$. When $k$ has a topology and $V$ is a topological vector space over $k$, we may (or may not) require that a functional be continuous.

Yes, as in other answers and comments, very often interesting and useful functional are expressed in terms of integrals... which is the historical origin and indication of the usefulness of the abstracted idea.

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  • $\begingroup$ You are defining ($k$-) linear functional, but non-linear functionals (including the name) are also in modern use. $\endgroup$
    – user569098
    Jun 29, 2018 at 21:52

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