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 . In what
follows X,Y will be Banach spaces, complete normed linear spaces. Thus these
are complete normed linear space and ℒ
(X, Y)
is the space of bounded linear maps. I
will also cease trying to write the vectors in bold face partly to emphasize that these are
not in ℝ^{n}.
Definition 20.10.1Let
(X,∥⋅∥X)
and
(Y,∥⋅∥Y)
be two normed linear spaces. Thenℒ
(X,Y )
denotes the set of linear maps from X to Y which also satisfy the followingcondition. For L ∈ℒ
(X,Y )
,
∥xli∥m≤1∥Lx ∥Y ≡ ∥L ∥ < ∞
X
Recall that this operator norm is less than infinity is always the case where X is finite
dimensional. However, if you wish to consider infinite dimensional situations, you assume
the operator norm is finite as a qualification for being in ℒ
(X, Y)
. Then here is an
important theorem.
Theorem 20.10.2If 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.
||L x− L x|| ≤ ||x||||L − L ||.
n m n m
Thus {L_{n}x} is a Cauchy sequence. Let
Lx = ln→im∞ Lnx.
Then, clearly, L is linear because if x_{1},x_{2} are in X, and a,b are scalars, then
L(ax1 + bx2) = nli→m∞ Ln (ax1 + bx2)
= lim (aL x + bL x )
n→∞ n 1 n 2
= aLx1 + bLx2.
Also L is bounded. To see this, note that {||L_{n}||} is a Cauchy sequence of real numbers
because
|||Ln ||− ||Lm |||
≤||L_{n}−L_{m}||. Hence there exists K > sup{||L_{n}|| : n ∈ ℕ}. Thus,
if x ∈ X,
∥Lx∥ = lim ||Lnx|| ≤ K ∥x∥.■
n→ ∞
The following theorem is really nice. The series in this theorem is called the Neuman
series.
Lemma 20.10.3Let
(X, ∥⋅∥)
is a Banach space, and if A ∈ℒ
(X, X )
and
∥A∥
= r < 1,then
∞
(I − A )− 1 = ∑ Ak ∈ ℒ(X, X)
k=0
where the series converges in the Banach space ℒ
(X, X)
. If O consists of the invertiblemaps in ℒ
(X,X )
, then O is open and if ℑ is the mapping which takes A to A^{−1}, then ℑis continuous.
Proof:First of all, why does the series make sense?
(A + B)−1 = (A(I +A −1B ))−1 = (I + A− 1B )−1A −1 ∈ ℒ (Y, X).
Thus O is an open set.
Thus
−1 ( −1 )−1 − 1 ∑∞ n( −1 )n −1
(A + B) = I + A B A = (− 1) A B A
n=0
[ −1 ] −1
= I − A B + o(B ) A
which shows that O is open and, also,
∑∞
ℑ (A + B )− ℑ(A) = (− 1)n (A−1B )nA−1 − A−1
n=0
= − A−1BA −1 + o(B)
= − ℑ(A)(B )ℑ(A)+ o (B )
which demonstrates 20.10.19. It follows from this that we can continue taking derivatives
of ℑ. For
||B1 ||
small,
− [D ℑ(A + B1)(B)− D ℑ(A) (B )] =
ℑ(A + B )(B)ℑ (A+ B )− ℑ(A)(B )ℑ(A)
1 1
= ℑ (A + B1 )(B )ℑ(A + B1)− ℑ (A )(B )ℑ(A + B1)+
ℑ (A )(B)ℑ (A + B ) − ℑ(A)(B)ℑ (A)
1
= [ℑ (A)(B1)ℑ (A )+ o(B1)](B )ℑ(A + B1)+
ℑ (A )(B)[ℑ(A)(B1) ℑ(A)+ o(B1)]
[ − 1 −1 −1 ]
= [ℑ(A)(B1)ℑ (A)+ o(B1)](B) A − A B1A + o (B1 ) +
ℑ (A )(B)[ℑ(A)(B1)ℑ (A )+ o(B1)]
= ℑ (A )(B1 )ℑ(A)(B )ℑ(A)+ ℑ (A)(B)ℑ (A )(B1)ℑ(A )+ o(B1)
and so
D2ℑ(A )(B )(B ) = ℑ (A )(B )ℑ(A )(B )ℑ(A )+ ℑ(A)(B) ℑ(A)(B )ℑ (A )
1 1 1
which shows ℑ is C^{2}
(O )
. 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 20.10.5Suppose U is an open subset of X and f : U → Y has the propertythat Df
(x)
exists for all x in U and that, x + t
(y − x)
∈ U for all t ∈
[0,1]
. (The linesegment joining the two points lies in U.) Suppose also that for all points on this linesegment,
||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 6.7.3.
Corollary 20.10.6Let B be a closed subset of the complete metric space
(X,d)
and letf : B → X be a contraction 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 f has a unique fixed point in B which is the limit of the sequence ofiterates. This is a point x ∈ B such that f
(x )
= x. In the case that B =B
(x0,δ)
, thesequence of iterates satisfies the inequality
d (x ,f (x ))
d(fn(x0),x0) ≤---0----0--
1 − r
and so it will remain in B if
d(x0,f (x0))
1 − r < δ.
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 20.10.7(implicit function theorem)Let X,Y,Z be Banach spaces andsuppose U is an open set in X × Y . Let f : U → Z be in C^{1}