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 ∥ ≡ sup ∥Lx∥ < ∞
∥x∥≤1
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. In either case, this
symbol is given in the following definition.
Definition 6.1.1Let
(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 )
,
lim ∥Lx∥Y ≡ ∥L∥ < ∞
∥x∥X≤1
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)
.
Definition 6.1.2A complete normed linear space is called a Banach space.
Theorem 6.1.3If 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 = nli→m∞ Lnx.
Then, clearly, L is linear because if x_{1},x_{2} are in X, and a,b are scalars, then
L (ax + bx ) = lim L (ax + bx)
1 2 n→∞ n 1 2
= lnim→∞ (aLnx1 + bLnx2)
= aLx + bLx .
1 2
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 || = nl→im∞ ||Lnx|| ≤ K ||x||.■
The following theorem is really nice. The series in this theorem is called the Neuman
series.
Lemma 6.1.4Let
(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 invertible mapsin ℒ
(X,X )
, then O is open and ifℑis the mapping which takes A to A^{−1}, then ℑ iscontinuous.
Proof:First of all, why does the series make sense?
. 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,
∞∑ k ( n+1)
(I − A) A = ln→im∞ I − A = I
k=0
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+∥A−1∥
,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,
∥∥ −1 [ − 1 ]− 1 −1∥∥
∥ℑA − ℑB ∥ = ∥A − I − A (A− B ) A ∥
∥ ∥ ∥ ∥
∥∥ −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
[ −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
D2 ℑ(A)(B1) (B ) = ℑ(A )(B1 )ℑ(A)(B )ℑ(A)+ ℑ (A )(B)ℑ (A )(B1)ℑ(A )
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 6.1.6Suppose 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 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 2.8.3.
Corollary 6.1.7Let B be a closed subset of the complete metric space
(X, d)
and letf : B → X be a contraction map
d(f (x),f (xˆ)) ≤ rd (x,xˆ) , r < 1.
Also suppose there exists x_{0}∈ B such that the sequence of iterates
{fn(x0)}
_{n=1}^{∞}remainsin B. Then f has a unique fixed point in B which is the limit of the sequence of iterates. Thisis a point x ∈ B such that f
(x)
= x. In the case that B =B
(x0,δ)
, the sequence of iteratessatisfies the inequality
n d(x0,f (x0))
d (f (x0),x0) ≤ ---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 6.1.8(implicit function theorem)Let X,Y,Z be finite dimensionalnormed vector spaces and suppose U is an open set in X ×Y . Let f : U → Z be in C^{1}
(U )
andsuppose
f (x0,y0) = 0,D1f (x0,y0)−1 ∈ ℒ (Z,X ). (6.2)
(6.2)
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. (6.3)
(6.3)
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) . (6.4)
(6.4)
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
||D1T (x,y)|| < 1, D1f (x,y)− 1 exists (6.5)
2
(6.5)
The second claim follows from Lemma 6.1.5. By the mean value inequality, Theorem 6.1.6,
whenever x,x^{′}∈ B
(x0,δ)
and y ∈ B
(y0,δ)
,
1
||T (x,y )− T(x′,y)|| ≤- ||x − x′||. (6.6)
2
(6.6)
Also, it can be assumed δ is small enough that for some M and all such