1. Proof of Inverse Function Theorem
Theorem 1.1. Let E be an open subset of Rn and f : E → Rn be a C1-map. Suppose a ∈ E and Df (a) is invertible. Then there exist an open neighborhood U of a and an open neighborhood V of f (a) so that f : U → V is a bijection whose inverse g : V → U is also C1 with Dg(y) = (Df (g(y)))−1 for all y ∈ V.
Proof. Let A = Df (a). Then A : Rn→ Rn is a linear transform. Since f is C1, Df : E → L(Rn) is continuous. Choose δ > 0 so that
(1.1) kDf (x) − Df (a)k < 1
2kA−1kfor any x with kx − ak < δ.
Let U = B(a, δ) and V = f (U ). Let us show that f : U → V is a bijection and that V is open. We only need to prove the injectivity of f ; surjectivity of f follows from V = f (U ).
To show that V is open, we prove that every point of V is an interior point of V. To prove this, let y0 be any point of V ; we need to find δ00 > 0 so that the open ball B(y0, δ00) ⊆ V.
Since V = f (U ), y0 = f (x0) for some x0 ∈ U. Since U is open, we choose δ0 > 0 such that B(x0, δ0) ⊆ U. We choose δ00 = δ0/(4kA−1k) and claim that B(y0, δ00) ⊆ V. In other words, we need to show that
for any y ∈ B(y0, δ00), y ∈ V .
In fact, we can prove further that B(y0, δ00) ⊆ V. For each y ∈ B(y0, δ00), i.e. ky − y0k ≤ δ00, we define a function
ϕy : U → Rn by ϕy(x) = x + A−1(y − f (x)) for x ∈ U.
Then ϕy : U → Rn is a C1-function and
Dϕy(x) = I − A−1Df (x) = A−1(A − Df (x)) = A−1(Df (a) − Df (x)).
For any x ∈ U, by (1.1), we obtain that
kDϕy(x)k ≤ kA−1kkDf (a) − Df (x)k < kA−1k · 1
2kA−1k = 1 2. This shows that ϕy : U → Rn is a contraction: (proved by mean value inequality)
kϕy(x1) − ϕy(x2)k ≤ 1
2kx1− x2k.
Observe that for any x1, x2 ∈ U, one has
ϕy(x1) − ϕy(x2) = x1− x2− A−1(f (x1) − f (x2)).
By triangle inequality,
kx1− x2k ≤ k(x1− x2) − A−1(f (x1) − f (x2))k + kA−1(f (x1) − f (x2))k
= kϕy(x1) − ϕy(x2)k + kA−1(f (x1) − f (x2))k
≤ 1
2kx1− x2k + kA−1kkf (x1) − f (x2)k.
Therefore we find that
(1.2) kf (x1) − f (x2)k ≥ 1
2kA−1kkx1− x2k
for any x1, x2 ∈ U. This implies that f is an injection and hence f : U → V is a bijection.
1
2
Let B = B(x0, r) where r = δ0/2. Let us prove that ϕy(B) ⊆ B. Let y0 ∈ ϕy(B). Then y0= ϕy(x0) for some x0 ∈ B. To show that y0 ∈ B, we need to show that ky0− x0k ≤ r :
ky0− x0k = kϕy(x0) − x0k ≤ kϕy(x0) − ϕy(x0)k + kϕy(x0) − x0k
≤ 1
2kx0− x0k + kϕy(x0) − x0k.
Since x0 ∈ B, kx0− x0k ≤ r. Since ky − y0k ≤ δ00,
kϕy(x0) − x0k = kA−1(y − f (x0))k ≤ kA−1kky − y0k
≤ kA−1kδ00= kA−1k · δ0 4kA−1k
= δ0 4 = r
2.
We conclude that ky0− x0k ≤ r/2 + r/2 = r. Therefore y0 ∈ B. This shows that ϕy(B) ⊆ B and hence
ϕy : B → B.
Since B is a closed subset of Rn and Rn is complete, B is also complete. By contraction mapping principle, there exists x ∈ B such that ϕy(x) = x. Notice that ϕy(x) = x is equivalent to
x + A−1(y − f (x)) = x.
This implies that y = f (x) for x ∈ B. We conclude that
y = f (x) ∈ f (B) ⊆ f (U ) = V for any y ∈ B(y0, δ00).
This shows that B(y0, δ00) ⊂ V and hence
B(y0, δ00) ⊆ B(y0, δ00) ⊂ V.
We prove that y0 is an interior point of V. Therefore V is open.
Let g : V → U be the inverse of f : U → V. Let us show that g is differentiable at every point of V. Let T = (Df (x0))−1 and y0 ∈ V be a point. Choose δ00 = δ0/4kA−1k as above so that B(y0, δ00) is contained in V. For kkk < δ00,
g(y0+ k) − g(y0) − T (k) = h − T (k) = T (Df (x0)h − k).
where h = g(y0+ k) − g(y0). Then y0+ k = f (x0+ h), and hence k = f (x0+ h) − f (x0).
The above equation can be rewritten as
g(y0+ k) − g(y0) − T (k) = −T (f (x0+ h) − f (x0) − Df (x0)(h)).
Let us estimate khk. By inequality (1.2), we find
khk = k(x0+ h) − x0k ≤ 2kA−1kkf (x0+ h) − f (x0)k = 2kA−1kkkk.
Hence we see that
kg(y0+ k) − g(y0) − T (k)k
kkk ≤ kT k
kkkkf (x0+ h) − f (x0) − Df (x0)(h)k
≤ 2kT kkA−1kkf (x0+ h) − f (x0) − Df (x0)(h)k
khk .
This implies the differentiability of g and Dg(y0) = (Df (x0))−1 for any x0 ∈ U. The continuity of Dg is guaranteed by the fact that the map GLn(R) → GLn(R) sending B to B−1 is continuous for any B ∈ GLn(R) where GLn(R) = {A ∈ Mn(R) : det A 6= 0}.
