61CM Section: Week 10
for questions contact: Evangelie Zachos (ezachos [AT] . . . )
Today: discussion and examples for the inverse and implicit function theorems
1
Inverse Function Theorem
Theorem. Suppose U ⊂ Rn is open, f : U → Rn is C 1 and for a particular x0 ∈ U , Df (x0 )
is nonsingular (i.e. invertible as a matrix). Then there are open sets V, W around x0 , f (x0 ),
respectively, so that f |V is a 1:1 map onto W with a C 1 inverse g : W → V .
Note: the condition in the above problem, from which we can conclude that f is locally
invertible around x0 , is a sufficient, not a necessary condition. (Think of f (x) = x3 .)
Example. Here Df (x0 ) = 0 and, also, the function is not local invertible at x0 . The fact
that Df (x0 ) = 0 is not a proof that f is not locally invertible, but if f is not locally invertible
(as is clearly true in this case), the theorem tells us that we certainly must have Df (x0 ) = 0,
which is indeed the case. Hooray, math is not broken!
y
f (x) = 1
x
Example. Let us inspect here f (x) = log(x + 4) (we use of course the natural logarithm).
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Now if we inspect that point at x0 = 1, then Df (x0 ) 6= 0, so we can find a local inverse.
y
f (x) = log(x + 4)
x
Example. Here Df (x0 ) = 0 (and in fact f is not invertible around x0 ) but Df (x1 ) 6= 0, so
by the inverse function theorem, f is locally invertible around x1 .
y
x0
x1
f (x) = 5 − (x − 2)2
x
Two-dimensional examples are harder to see. Particular examples which are invertible are
rotations and dilations.
2
Implicit Function Theorem
Theorem. Suppose U ⊂ Rn is open, G : U → Rm ( where m < n) is C 1 , S := {[~x, ~y ] :
~x ∈ Rn−m , ~y ∈ Rm , G[~x, ~y ] = 0}, and [~a, ~b] a point so that the m × m matrix Dy G[~x, ~y ]
is nonsingular at [~a, ~b]. Then there are open sets V ⊂ Rn , W ⊂ Rn−m around [~a, ~b], ~a
respectively, and a C 1 map h : W → Rm so that S ∩ V = graph h.
2
2
Example. Let
h us take iour old friend the circle, G[x,√y] = x + y − 1 and let us consider the
point [a, b] = √12 , − √12 . Then Dy G[a, b] = 2b = − 2 6= 0. Therefore, by the theorem, we
can find V , W and X = range h as pictured below around x0 . V is the gray circle, W is the
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red set on the x-axis (as it is part of Rn−m = R1x and X is the green set on the y-axis (as it
is part of Rm = R1y .
y
x1
W
x2
X
x
x0
V
Similarly, we can perform the same procedure around x1 because Dy G(x1 ) 6= 0. However,
we run into a problem at Dy G(x2 ) = 0. It doesn’t look that different from the other points,
but if we insist on writing the y-variable as a function of the x-variable, we won’t be able to
do it.
The most important thing about the statement of this theorem is that it gives us a condition
(nonsingularity of Dy G[~a, ~b]) for which, say when we take m = 1, implicitly defined sets,
such as S 2 = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 − 1 = 0} are submanifolds. Well . . . almost. There
is one problem! What about the when the submanifold can not be expressed as a graph but
only as a permutation of a graph? For this matter, we will need to use the corollary from
the book:
Corollary. If k ∈ {1, . . . , n − 1}, U ⊂ Rn open, g1 , . . . , gn−k : U → R are C 1 , S := {~x ∈ U :
gj (~x) = 0 for each j}, M := {~y ∈ S : ∇g1 (~y ), . . . , ∇gn−k (~y ) are linearly independent}, then
M is a k-dimensional C 1 submanifold.
[You can relate this corollary to the earlier example by considering F = (g1 , . . . , gn−k ) where
F : U → Rn−k and n − k = m. Then, perform a permutation of coordinates in U ⊂ Rn ,
rewriting F to get G to which you can apply the implicit function theorem. We perform the
permutation in a way so that the fact that the linear independence of {∇gi } gives the linear
independence of the last n − k entries of each permuted ∇gi , that is the nonsingularity of
Dy G. ]
Example. The main reason to introduce this corollary is to talk more generally about
submanifolds, so let us return to the circle example from earlier, where we could not talk
about the point x2 = [a, b] = [1, 0]. However, we still need to give our setting: here g1 =
x2 − y 2 − 1, and ∇g1 (x, y) = [2x, 2y]. This vector is always nonzero at any point (x, y) ∈ S,
and so it is a linearly independent set of one vector. This means that at every point, the
circle S satisfies the definition of a 1-dimensional C 1 submanifold.
Example. Another good example to think about this theorem with is actually an nonexample, our old friend the cusp. Here, g1 = g(x, y) = x3 − y 2 . Then Dg1 (x, y) = [3x2 , −2y].
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Thus by the corollary, S\{~0} is a C 1 1-dimensional manifold. In fact, S containing the origin
is not a C 1 1-dimensional manifold, so that this corollary gives the optimal result, but that
point requires a bit more work.
y
x3 − y 2 = 0
x
Optional Exercise 1. Another good non-example to revisit is our old friend the x and y
axis in R2 . This set can be expressed implicitly as S = {(x, y) : g1 (x, y) = xy = 0}. Can you
find the M given by the corollary? Is this optimal?
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