Hadamard's lemma, in one dimension, says for any smooth function $f \colon \mathbf R \rightarrow \mathbf R$ there is a first-order expansion of $f$ at $0$: $f(x) = f(0) + xg(x)$ where $g \colon \mathbf R \rightarrow \mathbf R$ is smooth. Explicitly, $g(x) = \int_0^1 f'(tx)\,dt$. (The analogue in higher dimensions is on the Wikipedia page for Hadamard's lemma.) My question is: where/when did Hadamard first present this result, or is it not due to him at all (Stigler's law)?

I looked through the four volumes of Hadamard's Oeuvres and the only thing I could find resembling this result was in his 1892 PhD thesis on analytic functions. Sections 32-35 contain some integrals of the form $\int_0^1 V(t)f(tx)\,dt$ for different choices of $V(t)$. Perhaps these motivated someone else (Whitney?) later on when the general foundations of manifolds were laid down in the 20th century.

  • $\begingroup$ Have you tried his Cours d'Analyse? That wouldn't be reproduced in the collected works. $\endgroup$ Jun 26, 2017 at 1:02
  • $\begingroup$ If you can get your hands on this book, its page 206 appears to quote a reference by Hadamard himself. $\endgroup$ Jun 26, 2017 at 1:18
  • $\begingroup$ @FrancoisZiegler I got my hands on the book in your second comment. There are two references by Hadamard in the bibliography. Item [5], which is in the footnote on page 206 that you point out, is his book on wave propagation, which is available from the Internet Archive at archive.org/details/leonssurlaprop00hada. So that reference is to a 370+ page book with no page indicated. Why do people do such things? Well, I clicked on every page of the book but nothing stuck out as this result. If you or someone else has the time to try, tell me if you find the result in some form there. $\endgroup$
    – KCd
    Jun 26, 2017 at 17:50
  • $\begingroup$ Ah, apologies for sending you on that wild goose chase. I was hoping [5] might confirm the Cours d'Analyse guess. $\endgroup$ Jun 26, 2017 at 17:56
  • $\begingroup$ If the reference to [5] really is correct then it'd be great if someone else can find it in there. $\endgroup$
    – KCd
    Jun 26, 2017 at 18:06

1 Answer 1


The reference to the book on wave propagation (Leçons sur la propagation des ondes et les équations de l'hydrodynamique. Cours du collège de France, Hermann, 1903) available at https://archive.org/details/leonssurlaprop00hada/page/n12 indeed contains this lemma in the Note I, on p. 352, under the form:

if $ p \in \mathbb{N}^* $ and $ F \in \mathcal{C}^p (\mathbb{R}^d) $, then there exist $ \varphi_1 , \dotsc , \varphi_d \in \mathcal{C}^{p-1} (\mathbb{R}^d \times \mathbb{R}^d ) $ such that for every $(x_1, \dotsc , x_d )$, $( y_1 , \dotsc , y_d ) \in \mathbb{R}^d $, we have $$ F( x_1 , \dotsc , x_d ) - F( y_1 , \dotsc , y_d ) = \sum_{j=1}^d (x_j - y_j ) \varphi_j ( x_1 , \dotsc , x_d , y_1 , \dotsc , y_d ) . $$ For the proof, J. Hadamard claims that for $ d =1 $, it is "easy to verify" that $ (x,y) \mapsto \frac{F(x)-F(y)}{x-y} $ is of class $ \mathcal{C}^{p-1} (\mathbb{R}) $ when $ F \in \mathcal{C}^p (\mathbb{R}) $, and deduce the general case by writing \begin{align*} F( x_1 , \dotsc , x_d ) - F( y_1 , \dotsc , y_d ) & = [ F( x_1 , x_2 , \dotsc , x_d ) - F( y_1 , x_2 , \dotsc , x_d ) ] \\ & \quad + [ F( y_1, x_2 , \dotsc , x_d ) - F( y_1 , y_2 , x_3 , \dotsc , x_d ) ] \\ & \quad + \dotsc + [ F( y_1, y_2 , \dotsc , y_{d-1} , x_d ) - F( y_1 , y_2 , \dotsc , y_{d-1} , y_d ) ] . \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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