I understand 'functionals' as functions of functions, for example:

$$ S[y(x)]= \int_{t_1}^{t_2} \sqrt{1+(y')^2} dx$$

Which is the famous arc length integral

Now, in a similar way, a limit we can write as:

$$L(a, [y(x)] ) = \lim_{x \to a} y(x) \tag{1}$$

In this way, we can think of a limit as a function of a 'function' and a 'number'. So, would it be correct to call the above object a functionals? Why/Why not?

Examples of (1):

$$L(0,\frac{\sin x}{x}) = \lim_{x \to 0} \frac{\sin x}{x} = 1$$

$$L(0,e^x) = 1$$


This doubt mainly emerged while I was answering through this post

  • 2
    $\begingroup$ You may feel to do so: $L$ is a function whose argument is a function. But who cares? ;) $\endgroup$ Apr 24, 2021 at 19:09
  • $\begingroup$ I've added some more context @Micheal Hoppe $\endgroup$ Apr 24, 2021 at 19:14
  • 1
    $\begingroup$ Evaluation is a functional, so yes, evaluating a limit at a point is also a functional. the usage does sound weird, though, perhaps because it's hard to imagine a natural situation where one wants a limit to mean something different than a value. Instead, if $f$ were discontinuous at $a$ one would redefine $f$ so that $f(a) = \lim(f, a)$. $\endgroup$ Apr 24, 2021 at 19:25
  • $\begingroup$ Sure you can do this, but it adds very little value in thinking about limits. In the arclength example, the functional viewpoint adds a lot of value in the sense that it leads to the Euler Lagrange equations. $\endgroup$
    – shalop
    Apr 28, 2021 at 23:25

3 Answers 3


A functional is just a function whose domain and target space are of a specific type. Typically the domain is assumed to be a space like $L^p, C^k$, or some other Hilbert/Banach/Frechet space (or maybe you may not even want all this structure, and just assume the domain is some real/complex vector space) etc, while the target space is assumed to be a field $\Bbb{R}$ or $\Bbb{C}$. So, tbh this whole "function vs functional" business is in my opinion a pointless terminology chasing game with no mathematical content.

Anyway, here's what you can do. Let \begin{align} S:=\{f:\Bbb{R}\to\Bbb{C}\,|\,\text{for every $a\in \Bbb{R}$, $\lim\limits_{x\to a}f(x)$ exists in $\Bbb{C}$}\} \end{align}

Now, we can define the "limiting" operation as $L:\Bbb{R}\times S\to \Bbb{C}$ as \begin{align} L(a,f):=\lim_{x\to a}f(x). \end{align} So, sure, the domain of $L$ is a vector space and it maps into a field, so you can call this a functional (though it is not a linear functional, only for fixed $a$, is $L(a,\cdot)$ linear). If you want to talk about limits at $\infty$, things can get more tricky since the domain will no longer be a vector space, so I guess you can't call them a functional in the typical sense of the word.

Note also, that sometimes the term "functional" means "continuous linear map from a (topological/normed) vector space into the field ($\Bbb{R}$ or $\Bbb{C}$)". So, if you want to talk about continuity, you of course have to equip $S$ with a topology, you can of course do this, but as of right now I don't see a point to this. Terminology can be confusing and is not always standardized; one must always refer to contextual clues for which meaning is intended.

As you can see, it is a little unweidly to define $L$ in full generality; this is not to say it isn't useful to think of it in this way. There are some problems where viewing a "basic vanilla" object from a more abstract perspective leads to quicker solutions (sometimes it's just a simple rephrasing which allows us to invoke more power theorems, eg from functional analysis)... though I can't think of any super convincing examples off the top of my head.

  • 1
    $\begingroup$ Nice answer! +1 As an example where one paraphrases the problem to be able to apply deeper theorems in that field: take the fundamental theorem of calculus in the case of Lesesgue integral. To prove that an absolute continuous function has derivative a.e. that is integrable and satisfies the identity $f(a) - f(b)= \int_a^b f'(x)dx$ Lebesgue had to go into the setting of measure theory, define the notion of derivative of a measure and then go back to functions. $\endgroup$
    – Son Gohan
    Apr 24, 2021 at 19:45
  • 1
    $\begingroup$ @SonGohan thank you that is a nice example of how a different perspective takes us deeper into the field. When writing this I had something else in mind (though I couldn't recall precisely nor could I put into words). I was thinking along the lines of for example proving the existence and uniqueness of ODEs. A very common method of proof is to rephrase the problem into the language of a fixed-point problem, from which the machinery of Banach's fixed point theorem gives a slick proof (similar story for proving the inverse function theorem in Banach spaces). $\endgroup$
    – peek-a-boo
    Apr 24, 2021 at 19:51
  • 1
    $\begingroup$ Wonderful example as well! Banach-Caccioppoli fixed point theorem goes exactly in the functional analysis direction! So many examples where one rephrases the problem in another field, then the problem is solved in that environment, and then it's all about to go back to the main issue: Perelman with a topological problem, solved with theory of PDEs and Ricci flows. Fermat's last theorem in algebraic geometry. And we could keep going for hours... $\endgroup$
    – Son Gohan
    Apr 24, 2021 at 20:00
  • 1
    $\begingroup$ @Acccumulation I agree that giving a different name to indicate a certain meaning can be helpful (eg measures, "multivalued functions", vector fields, tensor fields, differential forms, operators are strictly speaking functions in the modern sense. THey have a domain and a target space, but of course giving such names tells us something extra in few words), but what I don't agree with is the insistence (of some people) who make statements like "functionals are NOT functions" without providing any other context, hence I (think rightfully so) I called that a pointless definition chasing game. $\endgroup$
    – peek-a-boo
    Apr 25, 2021 at 3:46
  • 1
    $\begingroup$ I also think asking the question of "is blablabla a functional" serves no purpose particularly for the reason I (indirectly) describe in my answer. The answer to this question depends very very heavily on context: do we require continuity in the definition, do we require linearity? Do we require a special form (such as the action in Lagrangian mechanics) as to be specified by an integral? Because of so many inconsistencies/lack of uniformity or however better one can phrase it, I don't think its meaningful to provide a yes/no answer to that question. $\endgroup$
    – peek-a-boo
    Apr 25, 2021 at 3:50

Why not? There are lots of examples in which old concepts can be seen as functionals or operators from some abstract space to another abstract space or the real numbers. And, indeed, now this seems obvious to us but this is the case just because we are used to functional analysis point of view, where it is natural to consider functions just as points of that space; this was a great revolution in analysis though.

The only one needs to check is that the definition is well defined. As one defines the derivative as an operator that to every function (with some regularity) associates another function (its derivative), one can define the operator in the following way:

$$L : \mathbb{\overline{R}} \times V \rightarrow \mathbb{\overline{R}}$$ defined in the following way (note we allow it to take values in the extended real line, since we want to allow the cases of $+\infty $ and $ - \infty$):

$$ L (y, f) := \lim_{x \rightarrow y}f(x)$$

Now, to have a well-defined operator $L$ we need the limit to be unique, and this is not an issue in the real line since the real line is a Hausdorff space. Nevertheless, we need to be sure the limit exists, otherwise what would be the real number that the $L$ operator gives us in the case we consider the point $(\infty, sin(x))$? It would be:

$$ L (\infty, sin(x)) := \lim_{x \rightarrow \infty}\sin(x)$$ which does not exist!

Solved this problem, there is no limitation in viewing the limit as an operator. The only thing is that, indeed, to solve this existence issue one needs to impose strong restrictions on the vector space $V$ in which we take functions or to the space where we pick our points (maybe a compact subset of $\mathbb{\overline{R}}$).


It's not been pointed out yet, but your syntax is wrong. In "$L(0,e^x)$" the variable "$x$" is undefined, so the whole thing is meaningless if you want $L$ to be a function. The point is that not all syntax is literally a function. The limit notation includes a limiting variable, which is not expressible in terms of functions in the usual sense. It is the same with summation; in "$\sum_{k=1}^n f(k)$" the "$f(k)$" is an expression with one free variable $k$, not a function, and neither is "$\sum_{k=1}^n$" a function since it binds the variable $k$. It is also the same with quantifiers; in "$∀x ∃y ( Q(x,y) )$" the "$∀x$" is certainly not a function.

So if you want to treat the limiting operation as an abstract mathematical object, you need to reify the relevant expressions and syntactic structure. peek-a-boo has shown you one way to do that for limits, but I will show you how to do that for summation to make the underlying concept clearer. Reification simply means to capture the 'essence' of a concept by an object.

We typically define summation for commutative rings. Given any commutative ring $(R,0,1,+,·)$ and any function $f : D→R$ with $D⊆ℤ$, we can define $S(f,m,n) = \sum_{k=m}^n f(k)$ for every $m,n∈D$. Here $f$ reifies the expression "$f(k)$" with free variable $k$, in the sense that applying $f$ to the value of an expression "$t$" captures the essence of the expression "$f(t)$" (i.e. the value of "$f(k)$" after substituting the free variable by the term "$t$"). And $S$ reifies the concept of $\sum_{k=m}^n E$ built from expressions E,m,n where $E$ has one free variable $k$.

Importantly, notice that the free variable does matter in the syntactic constructions; "$\sum_{k=m}^n f(i)$" would not mean $\sum_{k=m}^n f(k)$. But this issue of matching free variables does not appear in the reified parts. This commonly occurs in reifying most mathematical notation, including quantifiers, summations/products, limits and so on.

Also, if you do want to formalize reasoning about limits in a clean algebraic manner and also be able to deal with undefined limits algebraically, then the best approach is to extend the possible limit values to include some sentinel value, say $null$, and then define $L(f,x)$ to be the limit of $f$ at $x$ if it exists but $null$ otherwise, for any function $f : D→ℂ$ with $x∈D⊆ℂ$.

  • $\begingroup$ I'll leave commenting on this answer till I learn more rigorous mathematics xD $\endgroup$ Apr 25, 2021 at 10:21
  • $\begingroup$ @Buraian: Sure, no problem! =) $\endgroup$
    – user21820
    Apr 25, 2021 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.