1
Homework #6
1. Consider
f
(
x
) =
e
x
for
x
∈
R
.
(a) Show that
f
is convex and compute the Legendre transform
f
*
(
ξ
).
Solution
It is clear that
f
(
x
) =
e
x
>
0, hence
f
is convex. Set
ξ
=
f
(
x
) =
e
x
and solve for
x
: ln
ξ
=
x
. Then
f
*
(
ξ
) =
xξ

f
(
x
) =
ξ
ln
ξ

f
(ln
ξ
) =
ξ
(ln
ξ

1).
(b) Prove that
e
n
x
≤
e
x
+(
n

1)
e
n
for all
x
∈
R
, and all positive integers
n
.
Solution
Consider the function
g
(
y
) =
e
y

y
. Then
g
(
y
) =
e
y

1
and has only one root when
y
= 0.
So
g
(
y
) =
e
y
and
g
(0)
>
0.
Hence,
y
= 0 is global minimum of
g
, by the second derivative test.
Therefore,
g
(
y
)
≥
g
(0) = 1 =
⇒
e
y

y
≥
1.
Let
y
=
x

n
.
Then
e
x

n

x
+
n
≥
1.
Since
e
n
>
0, multiply
expression by
e
n
:
e
x

xe
n
+
ne
n
≥
e
n
=
⇒
xe
n
≤
e
x
+ (
n

1)
e
n
,
as desired.
Also note that Young’s Inequality
xξ
≤
f
(
x
) +
f
*
(
ξ
) with
ξ
=
e
n
proves the desired result as well.
2. Consider
f
(
x, y
) =
e
(
x
2
+
y
2
)
/
2
for (
x, y
)
∈
Ω, where
Ω =
(
x, y
)
∈
R
2
 
x

<
1 and

y

<
1
.
(a) Show that the Hessian matrix
Hf
(
x, y
) =
f
xx
f
xy
f
yx
f
yy
is a positive definite matrix for all (
x, y
)
∈
Ω.
Solution
One way to show the matrix is positive definite is to show
all the principle minors are positive. In this case, show that
f
xx
>
0
and
f
xx
f
yy

f
2
xy
>
0. Then
f
xx
=
e
x
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 Spring '08
 staff
 Trigraph, 2m, 2k, Convex function

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