ECE 580 / Math 587
SPRING 2011
Correspondence # 19
May 3, 2011
Solution Set 7
48.
Let us first prove the following Lemma, which was given as a hint:
Lemma.
Let
x
∈
C
[0
,
1]
have a unique maximum at
t
0
∈
(0
,
1)
, which is also the unique
maximum of

x
(
t
)

, and let
h
be an arbitrary element of
C
[0
,
1]
. Then,
lim
α
→
0
1
α
{∥
x
+
αh
∥ − 
x
(
t
0
) +
αh
(
t
0
)
}
=
0
.
Toward this end assume, without any loss of generality, that
x
(
t
0
)
>
0, and let
ϵ >
0 be a
suﬃciently small scalar (in particular,
ϵ < t
0
,
t
0
+
ϵ <
1). Then, for suﬃciently small
α
,
∥
x
+
αh
∥
=
max
0
≤
t
≤
1

x
(
t
) +
αh
(
t
)

=
max
t
0
−
ϵ
≤
t
≤
t
0
+
ϵ

x
(
t
) +
αh
(
t
)

which follows because of the continuity of
x
and
h
. Furthermore, since
x
(
t
0
)
>
0, for suﬃ
ciently small
ϵ
and
α
,
x
(
t
) +
αh
(
t
)
>
0
∀
t
∈
[
t
0
−
ϵ, t
0
+
ϵ
]
.
Case 1.
α >
0
Then,
max
t
0
−
ϵ
≤
t
≤
t
0
+
ϵ

x
(
t
) +
αh
(
t
)

=
max
t
∈N
ϵ
(
t
0
)
[
x
(
t
) +
αh
(
t
)]
≤
max
t
∈N
ϵ
(
t
0
)
x
(
t
) +
α
max
t
∈N
ϵ
(
t
0
)
h
(
t
)
=
x
(
t
0
) +
α
max
t
∈N
ϵ
(
t
0
)
h
(
t
)
.
Similarly, replacing the maximum of
h
with minimum at the 2nd line:
max
t
∈N
ϵ
(
t
0
)

x
(
t
) +
αh
(
t
)

≥
x
(
t
0
) +
α
min
t
∈N
ϵ
(
t
0
)
h
(
t
)
.
Hence,
x
(
t
0
) +
α
min
t
∈N
ϵ
(
t
0
)
h
(
t
)
≤
max
t
∈N
ϵ
(
t
0
)
[
x
(
t
) +
αh
(
t
)
 ≤
x
(
t
0
) +
max
t
∈N
ϵ
(
t
0
)
h
(
t
)
·
α.
This now leads to
lim
α
→
0
α>
0
∥
x
+
αh
∥ − 
x
(
t
0
) +
αh
(
t
0
)

α
=
lim
α
→
0
α>
0
∥
x
+
αh
∥ −
x
(
t
0
)
−
αh
(
t
0
)
α
≤
̸
α
̸
α
[
max
t
∈N
ϵ
(
t
0
)
h
(
t
)
−
h
(
t
0
)
]
=
max
t
∈N
ϵ
(
t
0
)
h
(
t
)
−
h
(
t
0
)
.
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Likewise, the limit above can be bounded below by
≥
min
t
∈N
ϵ
(
t
0
)
h
(
t
)
−
h
(
t
0
)
.
But since
ϵ >
0 was arbitrary, we can let it go to zero, which makes both bounds go to zero,
since
h
is continuous. This then verifies the desired result as
α
→
0 on the positive side.
Case 2.
α <
0.
Let
β
=
−
α
,
¯
h
=
−
h
in [0
,
1]. Then
lim
α
→
0
α>
0
∥
x
+
αh
∥ − 
x
(
t
0
) +
αh
(
t
0
)

α
=
−
lim
β
→
0
∥
x
+
β
¯
h
∥ − 
x
(
t
0
) +
β
¯
h
(
t
0
)

β
→
0 by the argument of Case 1.
Remark.
If
t
0
is a boundary point, say
t
0
= 0, then replace
N
ϵ
(
t
0
) by
{
t
: 0
≤
t
≤
ϵ
}
; and
observe that similar arguments as above apply to prove an extension of the Lemma to this
more general case.
⋄
Now, coming back to the original problem, where we are given the function
f
(
x
) = max
0
≤
t
≤
1

x
(
t
)

=
∥
x
∥
,
x
∈
C
[0
,
1]
,
let us take
D
⊂
C
[0
,
1] to be the class of continuous functions
x
with the property that

x
(
t
)

has a unique maximum on [0
,
1]. Given
x
∈
D
, let
t
0
be the point where

x
(
t
)

achieves its
unique maximum on [0
,
1]. We now claim that,
δf
(
x
;
h
) = sgn[
x
(
t
0
)]
h
(
t
0
)
To show this, simply note that
lim
α
→
0
1
α
[
∥
x
+
αh
∥ − 
x
(
t
0
)
 −
αδf
(
x
;
h
)] = lim
α
→
0
1
α
[
∥
x
+
αh
∥ − 
x
(
t
0
) +
αh
(
t
0
)

] = 0
where the first equality follows because for suﬃciently small
α
(since clearly
x
(
t
0
)
̸
= 0), the
sign of
x
(
t
0
) +
αh
(
t
0
) is determined by sgn[
x
(
t
0
)], and the equality to
zero
follows from the
Lemma just proved.
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 Derivative, lim, dt, Necessary and sufficient condition, Gâteaux derivative

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