Bourbaki

Remark 2

Tag 083M, permanent. Cite it as [Bourbaki, Tag 083M].

Functions of a Real Variable, chapter I, § 1. FIRST DERIVATIVE

The preceding definitions and propositions (except for those concerning right and left derivatives) extend easily to the case where one replaces R\mathbf{R} by an arbitrary commutative non-discrete topological field KK, and the topological vector spaces (resp. topological algebras) over R\mathbf{R} by topological vector spaces (resp. topological algebras) over KK. In def. 1 and props. 1, 2 and 3 it is enough to replace I by a neighbourhood of x0x_0 in KK; in prop. 4 one must assume further that the map y↦y′y \mapsto y' is defined and continuous on a neighbourhood of f(x0)f(x_0) in E. Prop. 5 generalizes in the following manner: let K′K' be a non-discrete subfield of the topological field KK, let E be a topological vector space over KK; let ff be a function defined on a neighbourhood V⊂K′V \subset K' of x0∈K′x_0 \in K', with values in KK (considered as a topological vector space over K′K'), differentiable at x0x_0, and let g be a function defined on a neighbourhood of f(x0)∈Kf(x_0) \in K, with values in E, and differentiable at the point f(x0)f(x_0); then the map g∘fg \circ f is differentiable at x0x_0 and has derivative g′(f(x0))f′(x0)g'(f(x_0))f'(x_0) there (E being then considered as a topological vector space over K′K').

With the same notation, let ff be a function defined on a neighbourhood V of a∈Ka \in K, with values in E, and differentiable at the point aa; if a∈K′a \in K', then the restriction of ff to V∩K′V \cap K' is differentiable at aa, and has derivative f′(a)f'(a) there. These considerations apply above all, in practice, to the case where K=CK = \mathbf{C} and K′=RK' = \mathbf{R}.

Finally, prop. 6 extends to the case where one replaces I by a neighbourhood of x0∈Kx_0 \in K, and ff by a homeomorphism of I onto a neighbourhood J=f(I)J = f(I) of y0=f(x0)y_0 = f(x_0) in KK.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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