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The Wikipedia page for the Gateaux Derivative contains the following statement:

A version of the fundamental theorem of calculus holds for the Gateaux derivative of $F$, provided $F$ is assumed to be sufficiently continuously differentiable. Specifically:

Suppose that $F:X\to Y$ is $C^1$ in the sense that the Gateaux derivative is a continuous function $dF:U\times X \to Y$. Then for any $u \in U$ and $h\in X$,

$$ F(u+h)-F(u) = \int_0^1 dF(u+th;h)dt \label{a}\tag{1} $$

I stumbled on this result and proved it myself (in a less general context) around a year ago, but no one I asked about it recognized it. I am now writing up my work (of which this forms a small but key part) for publication. Now that I have found a statement of the result in Wikipedia, I would like to be able to cite it. But the Wikipedia page doesn't provide a citation or link or name for the theorem. I checked the paper by Gateaux, and didn't see it there.

Does anyone known where I can find it?

Thanks for any help...

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    $\begingroup$ Leon, while this Q&A deals a entirely different matter related to functional derivatives, I am almost 100% sure that the reference cited, in particular the monographs by Mikhail Vainberg (and in particular reference [3] cited therein), include a complete theory of the functional integration pertaining functional derivatives. You could have a look there just to start with something. $\endgroup$ Dec 3, 2020 at 15:43
  • $\begingroup$ Thank you so much! Not an easy monograph to get hold of, alas. $\endgroup$
    – Leon Avery
    Dec 4, 2020 at 15:09
  • $\begingroup$ Found a used copy by an Amazon seller... $\endgroup$
    – Leon Avery
    Dec 4, 2020 at 15:30
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    $\begingroup$ Leon, I am glad that you found a used copy though I expected there were a copy in the university library. However, I had a quick look to the monograph, and the result you need is theorem 2.7 in §2.7 pp. 34-35. Note that this theorem does not even require the existence of a Gâteaux derivative: it is only needed the continuity of a functional derivative (apart from its obvious existence)). $\endgroup$ Dec 4, 2020 at 16:44
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    $\begingroup$ Thanks for looking that up--it'll save me time once I get the book. Unfortunately, I no longer have easy access to a university library. I graduated from U Waterloo last February. In nonpandemic times, alumnus status would give me borrowing privileges, but as the world is now, no dice. $\endgroup$
    – Leon Avery
    Dec 4, 2020 at 21:49

1 Answer 1

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This answer arises from my comments above, in order to fulfill a request of the Asker. The result above is a consequence of a more general one, the following theorem (found in reference [1] §2.7, theorem 2.7, pp. 34-35; below I use equation numbering as in this reference) pertaining integration of continuous abstract functions of a real variable i.e. functions $\Phi: \Bbb R\to Y$, where $Y$ is a real Banach space:

Theorem. If the abstract function $\Phi(t)$ has continuous derivative on the open interval $]a,b[$ (i.e. $\Phi\in C^1(]a,b[)$), then for any number $\alpha, \beta\in]a,b[$ the formula $$ \int\limits_{\alpha}^{\beta}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s= \Phi(\beta)-\Phi(\alpha). \label{1}\tag{2.6} $$ is valid.
Sketch of proof: the first step consist in proving the formula
$$ \frac{\mathrm{d}}{\mathrm{d}t}\int\limits_{a}^{t}\Psi(s)\mathrm{d}s= \Psi(t)\quad\forall t\in]a,b[,\label{2}\tag{2.7} $$ for any continuous abstract function $\Psi:[a,b]\to Y$, by using an inequality for difference quotients ([1], §2.3, theorem 2.2, pp. 28-29). Then, by putting $$ \Psi(t) = \int\limits_{a}^{t}\Phi(s)\mathrm{d}s $$ we get $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\int\limits_{\alpha}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s -\Phi(t)\right] = 0. $$ and, by applying the mean value theorem for abstract functions, the relation $$ \int\limits_{\alpha}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s = \Phi(t) +C $$ holds for a given real constant $C$ which, evaluated by choosing $t=\alpha$, gives equation \eqref{1}. $\blacksquare$

Now equation \eqref{a} is a direct consequence of \eqref{1} obtained by putting $$ \Phi(t) = F (u+th)\quad \forall u\in U,\,h\in X. $$

Notes

  • Equation \eqref{1} is more general than \eqref{a} since its proof requires only the existence and continuity (respect to only to the real variable $t$) of the functional derivative (or better the functional differential) $$ \frac{\mathrm{d}}{\mathrm{d}t} F(u+th)= \lim_{t\to 0}\frac{F(u+th)-F(u)}{t}, $$ which could also be nonlinear respect to $h\in X$ (and thus it could not be a Gâteaux nor a Frechet derivative, see this Q&A on such matters).
  • Vainberg, despite citing several Authors like Norbert Wiener, Lawrence Graves, Michael Kerner, Israil’ Moiseevich Gel’fand and Mark Konstantinovich Gavurin ([1] §2 p. 24), does not give a precise reference for the result, and this perhaps means that it is due to himself, in the work of synthesis inevitably done during the writing of a monograph.

Bibliography

[1] Vaĭnberg, Mikhail Mordukhovich, Variational methods for the study of nonlinear operators. With a chapter on Newton’s method by L.V. Kantorovich and G.P. Akilov, translated and supplemented by Amiel Feinstein, Holden-Day Series in Mathematical Physics. San Francisco-London- Amsterdam: Holden-Day, Inc. pp. x+323 (1964), MR0176364, ZBL0122.35501.

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  • $\begingroup$ Excellent. Thanks. $\endgroup$
    – Leon Avery
    Dec 5, 2020 at 20:19
  • $\begingroup$ You are welcome, @LeonAvery: I am glad to have been of some help. $\endgroup$ Dec 5, 2020 at 20:20

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