Advanced Calculus Handout 1 March 14, 2011 Theorem 1 (Inverse Function Theorem) Suppose U ⊂ Rn is open, f : U → Rn is C1, x0 ∈ U and dfx0 is invertible. Then there exists a neighborhood V of x0 in U and a neighborhood W of f (x0) in Rn such that f has a C1 inverse g = f−1 : W → V. ( Thus f (g(y)) = y for all y ∈ W and g(f (x)) = x for all x ∈ V.) Moreover
h∂gi
∂yj(y)i
1≤i,j≤n
= dgy = (dfg(y))−1 =h∂fi
∂xj(g(y))i−1 1≤i,j≤n
for all y ∈ W and g is smooth whenever f is smooth.
Remark 1. The theorem says that a continuously differentiable function f between regions in Rn is locally invertible near points where its differential is invertible,
i.e. W = {y = (y1(x), . . . , yn(x)) = (f1(x), . . . , fn(x)) = f (x) | x ∈ V }, V = {x = (x1(y), . . . , xn(y)) = (g1(y), . . . , gn(y)) = g(y) | y ∈ W }, and each coordinate function is continuously differentiable.
Remark 2. Let 0 < a < 1, define f : R → R by f (x) =
ax + x2sin1
x if x 6= 0
0 if x = 0
. Compute f0(x) for all x ∈ R. Show that f0(0) > 0, yet f is not one-to-one in any neighborhood of 0 [by using (1) the fact that there exists a sequence of points {xn → 0} at which f0(xn) = 0, and f00(xn) 6= 0, (2) the observation that if f0(p) = 0, f00(p) 6= 0, then f cannot be one-to-one near p.].
This example shows that in the Inverse Function Theorem, the hypothesis that f is C1 cannot be weaken to the hypothesis that f is differentiable.
Remark 3. Define f : R2 → R2 by f (x, y) = (excos y, exsin y). Show that f is C1 and df(x,y) is invertible for all (x, y) ∈ R2and yet f is not one-to-one function globally. Why doesn’t this contradict the Inverse Function Theorem?
Example 1. Use Inverse Function Theorem to determine whether the system u(x, y, z) = x + xyz
v(x, y, z) = y + xy w(x, y, z) = z + 2x + 3z2 can be solved for x, y, z in terms of u, v, w near p = (0, 0, 0).
Solution: Set F (x, y, z) = (u, v, w). Then DF (p) =
ux uy uz
vx vy vz wx wy wz
(p) =
1 + yz xz xy
y 1 + x 0
2 0 1 + 6z
(p) =
1 0 0 0 1 0 2 0 1
and
1 0 0 0 1 0 2 0 1
= 1 6= 0.
By the Inverse Function Theorem, the inverse F−1(u, v, w) exists near p = (0, 0, 0), i.e. we can solve x, y, z in terms of u, v, w near p = (0, 0, 0).
Theorem 2 (Implicit Function Theorem)Let U ⊂ Rm+n ≡ Rm × Rn be an open set, f = (f1, . . . , fn) : U → Rn a C1 function, (a, b) ∈ U a point such that f (a, b) = 0 and the n × n matrix dyf |(a,b) = h∂fi
∂yj
(a, b)i
1≤i,j≤n
is invertible. Then there exists a neighborhood V of (a, b) in U a neighborhood W of a in Rm and a C1 function g : W → Rn such that
{(x, y) ∈ V ⊂ Rm× Rn| f (x, y) = 0} = {(x, g(x)) | x ∈ W } = the graph of g over W
i.e. Letting S = {(x, y) ∈ Rm × Rn | f (x, y) = 0} denote the solution set, then S ∩ V is of the dimension m, and it is equal to graph(g), the graph of a C1 function g, over W . Moreover
dgx = −(dyf )−1|(x,g(x))dxf |(x,g(x))
Advanced Calculus Handout 1 (Continued) March 14, 2011 and g is smooth whenever f is smooth.
Example 2. Let F : R4 → R2 be given by F (x, y, z, w) = (G(x, y, z, w), H(x, y, z, w))
= (y2 + w2− 2xz, y3+ w3+ x3 − z3), and let p = (1, −1, 1, 1).
(a) Show that we can solve F (x, y, z, w) = (0, 0) for (x, z) in terms of (y, w) near (−1, 1).
Solution: Since DF (p) =Gx Gy Gz Gw Hx Hy Hz Hw
(p) =−2 −2 −2 2
3 3 −3 3
and
Gx Gz Hx Hz
(p) =
−2 −2 3 −3
= 12 6= 0, we can write (x, z) in terms of (y, w) near (−1, 1) by Implicit Function Theorem.
(b) If (x, z) = Φ(y, w) is the solution in part (a), show that DΦ(−1, 1) is given by the matrix
−−2 −2 3 −3
−1
−2 2 3 3
=−1 0 0 1
Solution: The Implicit Function Theorem implies that, near p,the solution set {(x, y, z, w) | F (x, y, z, w) = (0, 0)} is the graph of (x, z) = Φ(y, w) near (−1, 1).
Hence, we have ∂F
∂y = (0, 0), and ∂F
∂w = (0, 0) near (−1, 1).
Therefore, 0 = Gx∂x
∂y + Gy+ Gz∂z
∂y, and 0 = Gx∂x
∂w + Gz∂z
∂w + Gw, which implies that −[Gy, Gw] = [Gx, Gz]
∂x
∂y
∂x
∂w
∂z
∂y
∂z
∂w
.
Similarly, we have −[Hy, Hw] = [Hx, Hz]
∂x
∂y
∂x
∂w
∂z
∂y
∂z
∂w
.
Thus, we have −Gy Gw Hy Hw
=Gx Gz Hx Hz
∂x
∂y
∂x
∂w
∂z
∂y
∂z
∂w
or DΦ =
∂x
∂y
∂x
∂w
∂z
∂y
∂z
∂w
= −Gx Gz Hx Hz
−1
Gy Gw Hy Hw
Hence, DΦ(−1, 1) is given by the matrix
−−2 −2 3 −3
−1
−2 2 3 3
=−1 0 0 1
Remark 4. The theorem says that if f : U ⊂ Rm+n → Rn is a map of the class C1(U ) , (a, b) is a point in U, and dyf |(a,b)=h∂fi
∂yj(a, b)i
1≤i,j≤n is invertible, then, locally, the solution set S = {(x, y) ∈ U | f (x, y) = f (a, b)} is a C1 “manifold” of dimension m.
Remark 5. Let f : U ⊂ Rm+n→ Rn is a map of the class C1 on an open subset U of R(m+n), and assume that the differential, dfp = h∂fi
∂xj i
1≤i≤n;1≤j≤(m+n), is of the constant rank k at
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Advanced Calculus Handout 1 (Continued) March 14, 2011 each p ∈ U. By relabeling, if necessary, we may assume that ˆf = (f1, . . . , fk) : U ⊂ Rm+n → Rk is a map with its differential d ˆfp =h∂fi
∂xj i
1≤i≤k;1≤j≤(m+n)
,of rank k. The Implicit Function Theorem says that locally the solution set S = {x ∈ U | ˆf (x) = 0} is a C1 “manifold” of dimension (m + n) − k.
Identify S with an open neighborhood W ⊂ R(m+n)−k of 0 ∈ R(m+n)−k and identify U with V × W, such that we write each x ∈ V × W ⊂ R(m+n) as x = (¯x, ˆx), where ¯x = (¯x1, . . . , ¯xk) ∈ V, and ˆ
x = (ˆxk+1, . . . , ˆxm+n) ∈ W. This implies ˆf (0, ˆx) = 0, and h∂ ˆfi
∂ ¯xj i
1≤i,j≤k is invertible on V × W. Using the Inverse Function Theorem, we may show that the range ˆf (V × W ) = f (U ) locally is a k−dimensional “manifold”.
Remark 6. Two mappings are said to be locally equivalent if under suitable choices of local coor- dinate systems in the domain and range spaces (with origin at 0) they can be written by the same formulas. The Implicit Function Theorem says that if f, g : U ⊂ Rm+n → Rn is a map of the class C1(U ) , p is a point in U, and rk[df (p)] = rk[dg(p)] = n, then f and g are equivalent, i.e. there exists diffeomorphisms h : Rm+n → Rm+n, k : Rn → Rn, such that f ◦ h = k ◦ g.
Definition 1. A map f : U ⊂ Rm → Rn is called a Lipschitz map on U if there exists a constant C ≥ 0 such that
kf (x) − f (y)k ≤ Ckx − yk for all x, y ∈ U.
If one can choose a (Lipschitz) constant C < 1 such that the above Lipschitz condition hold on U, then f is called a contraction map.
Example 3. Let U be a convex subset of Rm, and f : U ⊂ Rm → Rn be a map with bounded kdf k = sup{kdfp(x)kRn
kxkRm | p ∈ U, x 6= 0} (which implies that dfp is an n × m matrix, so if kdf k ≤ M then k∇fik ≤ M for i = 1, . . . , m.). Then f is Lipscitz on U.
Definition 2. Let U be a subset of Rm, and f be a map that maps U into U, i.e. f : U → U. A point p ∈ U is said to be a fixed point of f if f (p) = p.
Theorem 3. (Fixed point theorem for contractions) Let f : Rm → Rm be a contraction map.
Then f has a fixed point.
Note that Rm is complete which implies that any Cauchy sequence (is bounded and has a limiting point by Bolzano-Weierstrass theorem, and Cauchy condition implies that it) converges.
Definition 3. (Uniform Boundedness) Let K be a compact subset of Rp, and Cpq(K) = BCpq(K) = {f : K ⊂ Rp → Rq} denotes the set of all continuous (and bounded) functions from K into Rq. We say that a set F ⊂ Cpq(K) is bounded (or uniformly bounded) on K if there exists a constant M such that kf kK = sup{f (x) | x ∈ K} ≤ M, for all f ∈F .
Definition 4. (Uniform Equicontinuity) A setF of functions on K to Rq is said to be uniformly equicontinuous on K if, for each real number > 0 there is a number δ() > 0 such that if x, y belong to K and kx − yk < δ() and f is a function inF , then kf(x) − f(y)k <
Definition 5. (Uniform Convergence) A sequence (fn) of functions on U ⊂ Rp to Rq converges uniformly on a subset D ⊂ U to a function f if for each > 0 there is a natural number L() such
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Advanced Calculus Handout 1 (Continued) March 14, 2011 that for all n ≥ L() and x ∈ D, then
kfn(x) − f (x)k < .
In this case, we say that the sequence is uniformly convergent on D. It is immediate from the defi- nition that a sequence fn ∈ Bpq(D) = {the set of bounded functions from D ⊂ Rp to Rq} converges uniformly to f ∈ Bpq(D) on D if and only if lim
n→∞kfn− f kD = lim
n→∞sup{kfn(x) − f (x)k | x ∈ D} = 0.
Examples 4. (a) For each n ∈ N, let fn: R → R be defined by fn(x) = x
n for each x ∈ R.
(b) For each n ∈ N, let fn: I = [0, 1] ⊂ R → R be defined by fn(x) = xn for each x ∈ I.
(c) For each n ∈ N, let fn : R → R be defined by fn(x) = x2+ nx
n for each x ∈ R.
(d) For each n ∈ N, let fn: R → R be defined by fn(x) = 1
n sin(nx + n) for each x ∈ R.
One can easily see that the limiting functions are f ≡ 0, f (x) =
(0, 0 ≤ x < 1
1, x = 1 , f (x) = x, and f ≡ 0, for (a), (b), (c), and (d), respectively, and only the convergence in (d) is uniform.
Theorem 4. (Arzela-Ascoli Theorem). Let K be a compact subset of Rp and let F be a collection of functions which are continuous on K and have values in Rq. The following properties are equivalent:
(a) The family F is bounded and uniformly equicontinuous on K.
(b) Every sequence fromF has a subsequence which is uniformly convergent on K.
Example 5. Consider the following sequences of functions which show that Arzela-Ascoli Theo- rem may fail if the various hypothesis are dropped.
(a) fn(x) = x + n for x ∈ [0, 1]
(b) fn(x) = xn for x ∈ [0, 1]
(c) fn(x) = 1
1 + (x − n)2 for x ∈ [0, ∞)
Example 6. Let fn : [0, 1] → R be continuous and be such that |fn(x)| ≤ 100 for every n and for all x ∈ [0, 1] and the derivatives fn0(x) exist and are uniformly bounded on (0, 1).
(a) Show that there is a constant M such that | fn(x) − fn(y) | ≤ M | x − y | for any x, y ∈ [0, 1] and any n ∈ N.
Solution: Let M be a constant such that |fn0(x)| ≤ M for all x ∈ (0, 1). By the mean value theorem, we get | fn(x) − fn(y) | ≤ M | x − y | for any x, y ∈ [0, 1] and any n ∈ N.
(b) Prove that fn has a uniformly convergent subsequence.
Solution: We apply the Arzela-Ascoli Theorem by verifying that {fn} is equicontinuous and bounded. Given , we can choose δ = /M, independent of x, y, and n. Thus {fn} is equicon- tinuous. It is bounded because kfnk = sup
x∈[0,1]
|fn(x)| ≤ 100.
Example7. Let the functions fn: [a, b] → R be uniformly bounded continuous functions. Set Fn(x) =
Z x a
fn(t) dt, for x ∈ [a, b]. Prove that Fn has a uniformly convergent subsequence.
Solution: Since kFnk ≤ kfnk(b − a), Fn is uniformly bounded. Also, since |Fn0(x)| ≤ kfnk, Fn is equicontinuous by the preceding result. Therefore, Fn has a uniformly convergent subsequence by Arzela-Ascoli Theorem.
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