Bourbaki

Lemma 6

Tag 030U, permanent. Cite it as [Bourbaki, Tag 030U].

Théories spectrales, chapter IV, § 3. Distributions et distributions tempérées

Let kk be a natural number. Let f∈Ck(U)f\in \mathscr{C}^k(U) and let λ\lambda be the distribution associated with ff. For every β∈Nn\beta \in \mathbf{N}^n such that ∣β∣⩽k|\beta |\leqslant k, the distribution ∂βλ\partial^{\beta}\lambda is the distribution associated with the function ∂βf\partial^{\beta}f.

By induction on kk, it is enough to prove this property when β\beta satisfies ∣β∣=1|\beta |= 1, and one may even suppose that β=(0,...,0,1)\beta = (0, . . . ,0,1). Since distributions define a sheaf (prop. 5 of IV, p. 204), it is enough to verify the assertion when there exist an open set V⊂Rn−1V\subset \mathbf{R}^{n-1} and an open interval I⊂RI\subset \mathbf{R} such that U=V×IU = V\times I.

For every test function φ∈D\varphi \in \mathscr{D}(U), one has by definition

⟨∂βλ,φ⟩=−∫Uf(x)∂βφ(x)dx=−∫V(∫If(y,t)∂βφ(y,t)dt)dy \langle \partial^{\beta}\lambda , \varphi \rangle =-\int_Uf(x)\partial^{\beta}\varphi (x)dx=-\int_V(\int_If(y, t)\partial^{\beta}\varphi (y, t)dt)dy

from the Lebesgue–Fubini theorem (INT, V, p. 96, § 8, no4^o4, th. 1). By integration by parts (FVR, II, p. 10), one has

−∫If(y,t)∂βφ(y,t)dt=∫I∂βf(y,t)φ(y,t)dt -\int_If(y, t)\partial^{\beta}\varphi (y, t)dt=\int_I\partial^{\beta}f(y, t)\varphi (y, t)dt

since t↦φ(y,t)t\mapsto \varphi (y, t) has compact support in I. One therefore obtains

⟨∂βλ,φ⟩=∫V(∫I∂βf(y,t)φ(y,t)dt)dy=⟨∂βf,φ⟩ \langle \partial^{\beta}\lambda , \varphi \rangle =\int_V(\int_I\partial^{\beta}f(y, t)\varphi (y, t)dt)dy=\langle \partial^{\beta}f, \varphi \rangle

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5-mini, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.