First appearance of Hadamard's lemma on smooth functions

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.

• Have you tried his Cours d'Analyse? That wouldn't be reproduced in the collected works. – Francois Ziegler Jun 26 '17 at 1:02
• If you can get your hands on this book, its page 206 appears to quote a reference by Hadamard himself. – Francois Ziegler Jun 26 '17 at 1:18
• @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. – KCd Jun 26 '17 at 17:50
• Ah, apologies for sending you on that wild goose chase. I was hoping [5] might confirm the Cours d'Analyse guess. – Francois Ziegler Jun 26 '17 at 17:56
• If the reference to [5] really is correct then it'd be great if someone else can find it in there. – KCd Jun 26 '17 at 18:06

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*}