Math 174a: Lecture Notes on the Inverse Function
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چکیده
by the chain rule, so DG(0) = Id. Thus, G satisfies the assumptions of the theorem with p0, q0 resp. DF (p0) replaced by 0, 0, resp. Id. If we show that G is invertible as stated, then F(z) = p0 +G ((DF )(p0) (z − q0)) for z near q0. Indeed, substituting in z = F (p0 + y) gives F (F (p0 + y)) = p0 +G (G(y)) = p0 + y, and F (F (q0 + z)) = q0 + z is also easy to check. So from now on we assume that F (0) = 0, DF (0) = Id. Let δ0 > 0 be such that B(0, δ0) = {u : ‖u‖ ≤ δ0} ⊂ Ω. Then F (u) = u+R(u) with lim‖u‖→0 ‖Ru‖ ‖u‖ = 0. We are assuming also thatDR(u) = DF (u)− Id is continuous and DR(0) = 0. Combining these two we see that given > 0 there is δ = δ( ) > 0 such that δ ≤ δ0 and
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