Lecture 12
Definition 1.
Let
U
be open in
X
. When
F
:
U
→
Y
is differentiable at all
points
a
∈
U
, its
derivative
is the map
F
0
:
U
→
L
(
X, Y
) :
a
7→
F
0
a
and we call
F
continuously differentiable
when
F
0
is continuous.
Theorem 1.
Suppose
U
contained in
X
1
×
X
2
; then
F
:
U
→
Y
is
continu
ously differentiable
if and only if
∂
1
F
and
∂
2
F
are defined and continuous
at each point of
U
.
Proof.
”Scribal Exercise”
For any
a
= (
a
1
, a
2
)
∈
U
, let

a

= max
{
a
1

,

a
2
}
.
(
⇒
)
First, suppose that
F
is continuously differentiable. This means that
F
0
a
is
defined for all
a
∈
U
and that
F
0
is continuous at
a
. By the last theorem of
“Lecture 9”
∂
1
F
a
and
∂
2
F
a
exist at every
a
∈
U
and
F
0
a
(
h
1
, h
2
) =
∂
1
F
a
(
h
1
) +
∂
2
F
a
(
h
2
)
.
Specifically,
∂
1
F
a
(
h
1
) =
F
0
a
(
h
1
,
0)
∂
2
F
a
(
h
2
) =
F
0
a
(0
, h
2
)
.
Now consider the maps defined below:
α
1
:
X
1
→
X
1
×
X
2
:
h
1
7→
(
h
1
,
0)
α
2
:
X
2
→
X
1
×
X
2
:
h
2
7→
(0
, h
2
)
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
α
1
and
α
2
are linear and
k
α
1
(
h
1
)
k
=
k
(
h
1
,
0)
k
=
k
h
1
k
k
α
2
(
h
2
)
k
=
k
(0
, h
2
)
k
=
k
h
2
k
.
It follows that
k
α
1
k
= 1 =
k
α
2
k
. We can see that
∂
1
F
:
U
→
L
(
X
1
, Y
) :
a
7→
F
0
a
◦
α
1
∂
2
F
:
U
→
L
(
X
2
, Y
) :
a
7→
F
0
a
◦
α
2
.
Since
F
0
is continuous at each
a
let
ε >
0 and choose
δ
such that if
b
∈
U
with

a

b

< δ
then
k
F
0
a

F
0
b
k
< ε,
thus for
i
= 1 or 2
k
∂
i
F
a

∂
i
F
b
k
=
k
F
0
a
◦
α
i

F
0
b
◦
α
i
k
=
k
(
F
0
a

F
0
b
)
◦
α
i
k
≤ k
F
0
a

F
0
b
k k
α
i
k
=
k
F
0
a

F
0
b
k
< ε.
So
∂
1
F
and
∂
2
F
are continuous at all
a
∈
U
.
(
⇐
)
Suppose
∂
1
F
and
∂
2
F
are defined and continuous for all
a
∈
U
. From the
last Theorem of “Lecture 11” we have that
F
is differentiable at each
a
∈
U
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 Staff
 Algebra, Derivative, Open set, Compact space, Banach space

Click to edit the document details