Supplement for Manifolds and Differential Geometry Jeffrey M. Lee

38 0. Background Material
0.7.1. Chain Rule, Product rule and Taylor’s Theorem.
Theorem 0.56 (Chain Rule). Let U 1 and U 2 be open subsets of Banach
spaces E 1 and E 2 respectively. Suppose we have continuous maps composing
as
U 1
f
→ U2
g
→ E3
where E 3 is a third Banach space. If f is differentiable at p and g is differentiable
at f(p) then the composition is differentiable at p and D(g ◦ f) =
Dg(f(p)) ◦ Dg(p). In other words, if v ∈ E 1 then
D(g ◦ f)| p · v = Dg| f(p) · (Df| p · v).
Furthermore, if f ∈ C r (U 1 ) and g ∈ C r (U 2 ) then g ◦ f ∈ C r (U 1 ).
Proof. Let us use the notation O 1 (v), O 2 (v) etc. to mean functions such
that O i (v) → 0 as ‖v‖ → 0. Let y = f(p). Since f is differentiable at p we
have
f(p + h) = y + Df| p · h + ‖h‖ O 1 (h) := y + ∆y
and since g is differentiable at y we have g(y + ∆y) = Dg| y · (∆y) +
‖∆y‖ O 2 (∆y). Now ∆y → 0 as h → 0 and in turn O 2 (∆y) → 0 hence
g ◦ f(p + h) = g(y + ∆y)
= Dg| y · (∆y) + ‖∆y‖ O 2 (∆y)
= Dg| y · (Df| p · h + ‖h‖ O 1 (h)) + ‖h‖ O 3 (h)
= Dg| y · Df| p · h + ‖h‖ Dg| y · O 1 (h) + ‖h‖ O 3 (h)
= Dg| y · Df| p · h + ‖h‖ O 4 (h)
which implies that g ◦ f is differentiable at p with the derivative given by
the promised formula.
Now we wish to show that f, g ∈ C r r ≥ 1 implies that g ◦ f ∈ C r also.
The bilinear map defined by composition, comp : L(E 1 , E 2 ) × L(E 2 , E 3 ) →
L(E 1 , E 3 ), is bounded. Define a map on U 1 by
m f,g : p ↦→ (Dg(f(p), Df(p)).
Consider the composition comp ◦m f,g . Since f and g are at least C 1 this
composite map is clearly continuous. Now we may proceed inductively.
Consider the r th statement:
compositions of C r maps are C r
Suppose f and g are C r+1 then Df is C r and Dg ◦ f is C r by the
inductive hypothesis so that m f,g is C r . A bounded bilinear functional is
C ∞ . Thus comp is C ∞ and by examining comp ◦m f,g we see that the result
follows.
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