It is usual to introduce Fréchet and Gâteaux derivatives in Banach spaces. In this context, the familiar Taylor expansion with remainder is also at hand, as you can see on the picture below taken from this reference. in this Theorem from Now, I'd like to know if there exists an analogous theorem for locally convex spaces and where should I go to learn it if this is the case. Do we need Fréchet and Gâteaux derivatives as well? Any comments would be helpful. Thanks in advance!

EDIT: Let $E, F$ be locally convex spaces over some field $\mathbb{K}$ and $f: E \to F$. We say that $f$ is Gâteaux differentiable at $x \in E$ if there exists a continuous linear functional $Df[x]$ such that \begin{eqnarray} \lim_{\epsilon \to 0}\frac{f(x+\epsilon h)-f(x)}{\epsilon} = Df[x](h) \tag{1}\label{1} \end{eqnarray} for every $h \in E$.

The above definition makes sense and it is the definition of Gâteaux differentiability that I know. Now, here the author states a Taylor expansion result (pag. 25, Theorem 1.4.11) but his definition of Gâteaux differentiability is a little different (Definition 1.4.7 on page 24). However, I believe my definition and his definition are equivalent and, if so, this answers my question. Am I missing something here?

  • $\begingroup$ Why did you put the 'distributions' tag? $\endgroup$
    – YCor
    Apr 6, 2020 at 17:08
  • $\begingroup$ I thought that if such a result exits, it'd be possible to be best known for more especific spaces like $\mathcal{S}(\mathbb{R}^{n})$. $\endgroup$
    – IamWill
    Apr 6, 2020 at 17:13
  • $\begingroup$ I still don't see the link with distributions. 'schwartz-distribution' is a tag about distributions, not about Schwartz functions. $\endgroup$
    – YCor
    Apr 6, 2020 at 17:17
  • $\begingroup$ Ok. I'll replace it by a more adequate one. $\endgroup$
    – IamWill
    Apr 6, 2020 at 17:20
  • 1
    $\begingroup$ Will, just an observation: in the monograph linked by prof. Michor in his answer at pages 73-77 and 116, there are some interesting insights on how the functional calculus evolved. $\endgroup$ Apr 7, 2020 at 6:47

1 Answer 1


First of all the good news: There is an analogous theorem for locally convex spaces. It can be formulated in almost the same way using Gateaux derivatives (more on this notion in a moment). Obviously, the norm condition on the vanishing of the remainder makes no sense in the absence of a norm. However, you can formulate the vanishing of the remainder term of order $n$ via a homogeneity condition on the arguments of the remainder.

Before we come to the bad news, some comments on differential calculus beyond Banach spaces. Frechet derivatives make no sense beyond Banach spaces (again the norm condition breaks ones neck) and there are several (beyond Frechet spaces mutually inequivalent) ways of doing calculus in locally convex spaces. You can choose among dozens of approaches (all summarised in H.H. Keller: Calculus in locally convex spacesa), but most people stick with one of the following approaches: 1. The convenient calculus (popular since Kriegl and Michors book: The convenient setting of global analysis, from 1998 available online here 1 ). To my knowledge this does NOT feature a Taylor like theorem, though you find a section on Jet spaces (which is a more fancy way of talking about Taylor like expansions) 2. Bastiani calculus as used in most writings on infinite-dimensional Lie theory, see e.g. Section 1 of 2 for an overview and more references. In Bastiani calculus I am certain that there is a version of Taylors theorem. I give you a cartoon version since it is technical and long and I do not want to type endlessly:

Let $E,F$ be locally convex spaces, $k\in \mathbb{N}_0$ and $f\colon U \rightarrow F$ a Bastiani $C^k$-map on an open subset of $E$. Then the following holds:

(a) for each $x\in U$, there exists a unique polynomial $P_x^kf \colon E \rightarrow F$ of degree $\leq k$ such that $\delta_0^j(P_x^kf)=\delta_x^jf $ for each $j\leq k$.

(b) $\lim_{t\rightarrow 0} \frac{f(x+ty)-P_x^kf(ty)}{t^k}=0, \forall y \in E$ and the polynomial is unique with this property.

(c) There is a continuous remainder term $R_k(x,y,t)=\frac{1}{(k-1)!}\int_0^1 (1-r)^{k-1}(\delta_{x+rty}^kf(y)-\delta_x^kf(y))\mathrm{d}r$ satisfying equation (30.14) in the picture you uploaded

On to the bad news: I am certain that it exists because I have a nice pdf in front of me stating the theorem together with all the nitty gritty details and how to prove it. The unfortunate part is that I am not at liberty to share it since this is part of the book manuscript for the forthcoming* book by Glöckner and Neeb on infinite-dimensional Lie theory. This is unfortunately also the only source I know of, where one can find it in this form. You could try to write an email to Glöckner (see 2) and see whether he can either point you towards an older source or wants to share the pdf.

*: Unfortunately, the book was forthcoming from 2005 on. I have seen many revisions but it is anybodys guess when it will finally appear in public.

  • $\begingroup$ Thanks so much for the enlightening comment! I don't know if you noticed my edit, but I added a hyperlink on the mentioned work of Glockner and Neeb (I found it avaiable online). So this goes in the same direction as my edit suggests. $\endgroup$
    – IamWill
    Apr 6, 2020 at 18:31
  • $\begingroup$ As a final comment, I'd like to put Glockner and Neeb's Theorem in terms of my definition of Gâteaux differentiability. I think these definitions are equivalent. Do you agree? $\endgroup$
    – IamWill
    Apr 6, 2020 at 18:32
  • 2
    $\begingroup$ Indeed I missed the edit. The linked page is quite like one of the draft versions for the corresponding chapter in the book I was refering to, so this should have you covered. As to the definition of the Gateaux derivative: It should be the same. If you are reading the definition of Bastiani differentiable, the chapter you linked should have the relevant Lemma which shows that the derivative has the properties you claimed for the Gateaux derivative, so in essence that should be what you want. $\endgroup$ Apr 6, 2020 at 19:35

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.