Ok so I just started Calc I this summer and since I already feel pretty comfortable with it from high school, I'm trying to gain a more rigorous perspective on it. I already know that limits behave linearly in the sense that $$ \lim_{x \to a}[f(x)+g(x)]=\lim_{x \to a}f(x)+\lim_{x \to a}g(x) $$ and $$ \lim_{x \to a}[af(x)]=a \left(\lim_{x \to a}f(x)\right) $$ but I have never seen them formally described as a linear functional (or linear operator if the output is a function as in the case of the derivative) in the sense that they take an element of a suitable function space (for simplicity, take the continuous functions which form an infinite dimensional normed vector space, lets say $E$) such that $L:E \to \mathbb{R}$ where $L$ is defined by $$ L=\lim_{x \to a} $$ My gut instinct on this is that it may have never been useful to formalize the notion of a limit as a linear functional or that the definition of the derivative operator $D:C^{k} \to C^{k-1}$ as $$ Df=\lim_{h \to 0} \frac{f(x+h)+f(x)}{h} $$ makes this so obvious that no one talks about it explicitly. Another way to phrase my question would be:

"Limits belong to which class of mathematical objects?"

I tried asking my teacher but she didn't even understand what I was asking (she is a TA type who is well intentioned but clearly not comfortable enough with the material to teach) so any additional insights would be of great help here.

Martin Argerami
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    This is just an aside, but I do find the last paragraph to be pretty unfair to your teacher. – pjs36 Jun 05 '16 at 18:22
  • You wouldn't expect someone teaching a calculus class at the university level to understand how concepts fit into the bigger picture? Maybe I do have unfair expectations and if I were still in high school that would certainly be true. I just feel that someone teaching a calculus class should have an understanding of "linearity" (and not equate this notion with an affine function) and should have enough background to know that the derivative can be viewed as a linear operator between $C^{k}$ and $C^{k-1}$. Like I said, I could be wrong but I feel pretty strongly on this. – AnalysisStudent Jun 05 '16 at 19:42
  • In her defense, equating the notion of linearity with an affine function (i.e. "linear just means a line") is mainly the fault of James Stewart and his joke of a textbook. – AnalysisStudent Jun 05 '16 at 19:47
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    I think my real point was just that it doesn't *really* seem relevant to the question, and thus comes across as more of a slight than a worthwhile detail. – pjs36 Jun 05 '16 at 20:35
  • Ah I certainly see how it might come across that way. It was intended mainly to avoid the "did you ask your teacher?" or "what does your teacher have to say about it?" type responses. I definitely wasn't expecting the helpful replies I got so I guess it did indeed become an irrelevant detail. – AnalysisStudent Jun 05 '16 at 21:39
  • @AnalysisStudent Which book you recommend for calculus? – user46372819 Dec 20 '17 at 21:04

1 Answers1


Limits can definitely be seen as functionals. The problem is that to consider them as a functional, you need the limit to be defined on all elements of a vector space.

If you are dealing with continuous functions on a compact set (or an interval, to make things simpler), then the limit is just evaluation at a point. As soon as you are dealing with non-continuous functions (a very common occurrence in functional analysis, as in $L^p$ spaces for instance), limits as-is make no sense.

Still, one can use deep ideas in functional analysis to extend a limit from some objects where it exists to a more general setting. For instance, you could consider $\ell^\infty(\mathbb N)$, the set of bounded sequences. Of course, not every sequence has a limit; but there is a way to define a linear functional (many, actually) that agree with the limit where it exists. Some keywords for these ideas (I'll let you read about them if you have the background) are

  • Banach limits
  • Free ultrafilters
  • Stone-Cech compactification.
Martin Argerami
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  • I really appreciate the answer and I will definitely check out those concepts. Just to be certain, it is correct to say that on the space of continuous functions defined over a compact interval $C[a,b]$, you CAN think of the limit as a linear functional and it is equivalent to the evaluation-at-a-point functional? – AnalysisStudent Jun 04 '16 at 21:00
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    Yes indeed. Something that I forgot to mention in the answer is that, as a linear functional, the limit is fairly peculiar because it is also multiplicative (this is usually called a *character* in some contexts; it is also an example of an *irreducible representation* in others). For an example of a non-multiplicative linear functional that you know well, consider in $C[a,b]$ the map $$f\longmapsto \int_a^b f.$$ – Martin Argerami Jun 04 '16 at 21:04
  • Also, I know that for most function spaces you have to think of limits of sequences of functions to have any meaning (for example the space of continuous functions is not a Banach space since a sequence of "tent functions" $f_{n}$ whose support converges to zero as $n \to \infty$ is not continuous in the limit) and it seems obvious to me that these are linear operators but is taking limits of sequences of functions the "function space analog" of the limits we take in your typical calculus class? – AnalysisStudent Jun 04 '16 at 21:06
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    Not really. When you think a bit about it, the notion of limit doesn't really require that your objects are numbers; all you use is that you have a notion of distance, and your sequence is "closer and closer" to some point . Bottom line is that "limit" makes sense in a *metric space*. And even further, then you realize that you don't really need a distance, just a notion of "close", and then you move on to *topological spaces*. In any case, limits of functions are usually taken in metric spaces (more concretely, *normed spaces*) – Martin Argerami Jun 04 '16 at 21:13
  • " there is a way to define a linear functional (many, actually) that agree with the limit where it exists." ok, so define it please – reuns Jun 04 '16 at 21:23
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    @user1952009: Really? Are you suggesting a lecture on weak topologies on a question by an undergraduate, that is tagged *calculus*? – Martin Argerami Jun 04 '16 at 21:25
  • Basically, what I meant is that I've just always thought of the standard $\mathbb{R}^{n}$ and functions $f:\mathbb{R}^{n} \to \mathbb{R}^{m}$ as forming a base for the rest of analysis in the sense that you "zoom out" (in an abstract sense) while retaining all of the structures and then sort of "bumping" everything down a level and "throwing out" the numbers (or vectors) and replacing the elements of the original space with the functions between the elements of the original space. Since the structures of the original space are preserved, there should be an analog of all the operations – AnalysisStudent Jun 04 '16 at 21:26
  • possible operations on the previous space. – AnalysisStudent Jun 04 '16 at 21:26
  • I was just asking if you can define it, or if you'll need the axiom of choice for saying it exists. Because for extending the limit to non convergent sequences, personally I only know $T(a_n) = \lim_{N \to \infty} \frac{\sum_{n=1}^N a_n c_n}{\sum_{n=1}^N c_n}$ where $c_n$ is a $ \ge 0$ and non summable sequence, and it won't extend the limit to ***any*** bounded sequence – reuns Jun 04 '16 at 21:31
  • @user1952009: it does require the axiom of choice. There is little to do in infinite-dimensional spaces without it. The most straightforward way is probably to define the limit as a functional on the set of convergent sequences of $\ell^\infty(\mathbb N)$, and then extend it by Hahn-Banach. There are other (equivalent) ways of doing the same, via free-ultrafilters and via the Stone-Cech compactification. They all require the axiom of choice. – Martin Argerami Jun 04 '16 at 21:35
  • For a concrete example, $\mathbb{R}^{n}$ is a complete, normed vector space with the notion of orthogonality defined on it so in some sense, it is a "prototype" for $L^{2}$ in the sense that except for dimensionality and the actual elements of the space, the structures are the same. Just as there is the standard euclidean distance on $\mathbb{R}^{n}$, there is the same notion of "size" in $L^{2}$ given by $(\int \vert f(x) \vert ^{2} dx )^{\frac{1}{2}}$. So since $L^{2}$ is complete (though it contains discontinuous functions), there should be an analog for the taking of limits. – AnalysisStudent Jun 04 '16 at 21:36
  • Yes I know, but then ***there is no way to define such a linear functional***, since the axiom of choice (and his cousins Zorn's lemma or even Hahn Banach in functional analysis) is precisely for un-definable things – reuns Jun 04 '16 at 21:37
  • "Un-definable"? I have no idea what you mean by that. – Martin Argerami Jun 04 '16 at 21:39
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    https://en.wikipedia.org/wiki/Definable_real_number , https://en.wikipedia.org/wiki/Definable_set , https://en.wikipedia.org/wiki/Definable , it is not a very well-defined ( lol ) concept, but it is still important. So what I just wanted to say is that there is a difference between "defining" and "proving something exists under the axiom of choice" – reuns Jun 04 '16 at 21:43
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    Ok, good luck doing calculus then, since you only want to work on those "definable" real numbers. And I guess you only want to consider functions that are defined by explicit formulas? – Martin Argerami Jun 04 '16 at 21:46
  • @user1952009: It's also an often misunderstood concept; I would not trust ***any*** lay description of such things, including wikipedia. For example, did you know that you can explicitly write down a formula such that you cannot disprove the proposition it defines a well-ordering on the reals? I think you can even manage the same thing with global choice! –  Jun 05 '16 at 04:42
  • @AnalysisStudent: I didn't know you knew about $L^2$. With the distance $$d (f,g)=\left (\int|f-g|^2\right)^{1/2},$$ this is an example of a metric space. And of course one considers limits of functions there, but the notion has little to do with the limit of a function at a point: you take limits of sequences of functions and the limit is another function, so you are not dealing with a linear functional. – Martin Argerami Jun 05 '16 at 06:39
  • Why is it that "to consider them as a functional, you need the limit to be defined on all elements of a vector space"? As far as I know, there is plenty of theory where one deals with functionals that are not defined everywhere, for instance in quantum mechanics. – Federico Poloni Jun 05 '16 at 07:21
  • @FedericoPoloni: I don't know why I only saw your question five years later, but I'll answer for anyone who is interested. I never said "closed" when I mentioned vector spaces, although things are more reasonable on a closed space. But you do need a vector space to even be able to talk about linearity. – Martin Argerami Apr 24 '21 at 22:23