is the space of bounded linear mappings from X to Y where
here
(X, ∥⋅∥X )
and
(Y,∥⋅∥Y)
are normed linear spaces. Recall that this means that for each
L ∈ℒ
(X, Y)
∥L ∥ ≡ ∥sxu∥p≤1∥Lx∥ < ∞
As shown earlier, this makes ℒ
(X,Y )
into a normed linear space. In case X is finite dimensional, ℒ
(X, Y)
is the same as the collection of linear maps from X to Y . This was shown earlier. In what follows X,Y will
be Banach spaces. If you like, think of them as finite dimensional normed linear spaces, but if you like more
generality, just think: complete normed linear space and ℒ
(X,Y )
is the space of bounded linear
maps.
Definition 18.1.1A complete normed linear space is called a Banach space.
Theorem 18.1.2If Y is a Banach space, then ℒ(X,Y ) is also a Banach space.
Proof: Let {L_{n}} be a Cauchy sequence in ℒ(X,Y ) and let x ∈ X.
||Lnx − Lmx || ≤ ||x||||Ln − Lm ||.
Thus {L_{n}x} is a Cauchy sequence. Let
Lx = nl→im∞ Lnx.
Then, clearly, L is linear because if x_{1},x_{2} are in X, and a,b are scalars, then
. Therefore, the series converges to something in ℒ
(X, X)
by completeness of this normed linear space. Now why is it the inverse?
∞∑ n∑ ( ∑n n∑+1 ) ( )
Ak(I − A) = lim Ak(I − A ) = lim Ak − Ak = lim I − An+1 = I
k=0 n→ ∞ k=0 n→ ∞ k=0 k=1 n→ ∞
because
∥ ∥
∥An+1∥
≤
∥A∥
^{n+1}≤ r^{n+1}. Similarly,
∞
(I − A )∑ Ak = lim (I − An+1 ) = I
k=0 n→∞
and so this shows that this series is indeed the desired inverse.
Next suppose A ∈O so A^{−1}∈ℒ
(X, X)
. Then suppose
∥A − B∥
<
---r−1-
1+∥A ∥
,r < 1. Does it follow that
B is also invertible?
B = A − (A − B ) = A [I − A−1 (A − B )]
Then
∥∥A−1(A − B)∥∥
≤
∥∥A −1∥∥
∥A − B ∥
< r and so
[I − A− 1(A− B )]
^{−1} exists. Hence
B −1 = [I − A −1(A − B)]−1A− 1
Thus O is open as claimed. As to continuity, let A,B be as just described. Then using the Neuman
series,
∥ ∥
∥ℑA − ℑB ∥ = ∥∥A −1 − [I − A −1(A − B )]−1A− 1∥∥
∥ ∥ ∥ ∥
∥∥ −1 ∞∑ ( −1 )k −1∥∥ ∥∥∑∞ ( −1 )k − 1∥∥
= ∥∥A − A (A − B) A ∥∥ = ∥∥ A (A − B) A ∥∥
∞ k=0 k=1 ∞ ( )
∑ ∥∥ − 1∥∥k+1 k ∥∥ −1∥∥2 ∑ ∥∥ −1∥∥k ----r---- k
≤ A ∥A− B ∥ = ∥A − B ∥ A A 1 +∥A −1∥
k=1 ∥ ∥ 1 k=0
≤ ∥B − A∥∥A −1∥2-----.
1 − r
. Clearly we can continue in this way which shows ℑ is in C^{m}
(O)
for all
m = 1,2,
⋅⋅⋅
. ■
Here are the two fundamental results presented earlier which will make it easy to prove the implicit
function theorem. First is the fundamental mean value inequality.
Theorem 18.1.5Suppose U is an open subset of X and f : U → Y has the property that Df
(x)
existsfor all x in U and that, x + t
(y − x)
∈ U for all t ∈
[0,1]
. (The line segment joining the two points lies inU.) Suppose also that for all points on this line segment,
||Df (x+t (y− x))|| ≤ M.
Then
||f (y) − f (x)|| ≤ M |y − x |.
Next recall the following theorem about fixed points of a contraction map. It was Corollary
11.1.41.
Corollary 18.1.6Let B be a closed subset of the complete metric space
(X,d)
and let f : B → X be acontraction map
d(f (x),f (ˆx)) ≤ rd(x,ˆx), r < 1.
Also suppose there exists x_{0}∈ B such that the sequence of iterates
n
{f (x0)}
_{n=1}^{∞}remains in B. Then fhas a unique fixed point in B which is the limit of the sequence of iterates. This is a pointx ∈ B such that f
(x )
= x. In the case that B =B
(x0,δ)
, the sequence of iterates satisfies theinequality
d (fn(x ),x ) ≤ d(x0,f (x0))
0 0 1 − r
and so it will remain in B if
d(x0,f (x0))< δ.
1− r
The implicit function theorem deals with the question of solving, f
(x,y)
= 0 for x in terms of y and
how smooth the solution is. It is one of the most important theorems in mathematics. The proof I will give
holds with no change in the context of infinite dimensional complete normed vector spaces when suitable
modifications are made on what is meant by ℒ
(X, Y)
. There are also even more general versions of this
theorem than to normed vector spaces.
Recall that for X,Y normed vector spaces, the norm on X × Y is of the form
||(x,y)|| = max (||x ||,||y||).
Theorem 18.1.7(implicit function theorem)Let X,Y,Z be finite dimensional normed vector spaces andsuppose U is an open set in X × Y . Let f : U → Z be in C^{1}
(U )
and suppose
−1
f (x0,y0) = 0,D1f (x0,y0) ∈ ℒ (Z,X ). (18.2)
(18.2)
Then there exist positive constants, δ,η, such that for every y ∈ B
(y0,η)
there exists a uniquex
(y )
∈ B
(x0,δ)
such that
f (x (y),y) = 0. (18.3)
(18.3)
Furthermore, the mapping, y → x
(y)
is in C^{1}
(B (y0,η ))
.
Proof:Let T
(x,y)
≡ x − D_{1}f
(x0,y0)
^{−1}f
(x,y )
. Therefore,
D1T (x,y) = I − D1f (x0,y0)−1D1f (x,y ). (18.4)
(18.4)
by continuity of the derivative which implies continuity of D_{1}T, it follows there exists δ > 0 such that if
∥x− x0∥
< δ and
∥y − y0∥
< δ, then
1
||D1T (x,y)|| < -, D1f (x,y)−1 exists (18.5)
2
(18.5)
The second claim follows from Lemma 18.1.4. By the mean value inequality, Theorem 18.1.5, whenever
x,x^{′}∈ B
(x0,δ)
and y ∈ B
(y0,δ)
,
||T (x,y)− T (x ′,y)|| ≤ 1||x − x′||. (18.6)
2
(18.6)
Also, it can be assumed δ is small enough that for some M and all such
and now, the continuity of the partial derivatives D_{1}f,D_{2}f, continuity of the map A → A^{−1}, along with the
continuity of y → x
(y)
shows that y → x
(y)
is C^{1} with derivative equal to
− D1f (x (y ),y )− 1D2f (x (y ),y ) ■
The next theorem is a very important special case of the implicit function theorem known as the inverse
function theorem. Actually one can also obtain the implicit function theorem from the inverse function
theorem. It is done this way in [22], [24] and in [2].
Theorem 18.1.8(inverse function theorem)Let x_{0}∈ U, an open set in X , and let f : U → Y whereX,Y are finite dimensional normed vector spaces. Suppose