Fall 2006
ARE211
Midterm Exam  Answer key
Problem 1
: Metrics and continuity
. (10 points)
Let
X
be an arbitrary universe, let
d
be an arbitrary metric. For arbitrary
y
∈
X
, consider the
function
f
y
:
X
⇒
R
n
defined by
f
y
(
·
) =
d
(
·
, y
).
Hint:
To provide a counterexample to either of the conjectures below, you will need to specify an
X
, a metric
d
, an element
y
∈
X
, a sequence
{
x
n
}
and a ¯
x
, and show that for these variables, the
selected conjecture is false.
a) Prove, or provide a counterexample, to the following conjecture.
Conjecture:
The function
f
y
is continuous when
d
is the metric on the domain and the metric
on
R
n
is the Pythagorian metric.
Ans:
Fix a sequence
{
x
n
}
that converges to
¯
x
∈
X
, w.r.t.
the metric
d
.
We need to show
that
{
f
y
(
x
n
)
}
converges to
f
y
(¯
x
)
.
Fix
epsilon1 >
0
arbitrarily and pick
N
∈
N
such that for all
n > N d
(
x
n
,
¯
x
)
< epsilon1
. From the triangle inequality, we have
d
(
x
n
, y
)
≤
d
(
x
n
,
¯
x
) +
d
(¯
x, y
)
, so that
d
(
x
n
, y
)

d
(¯
x, y
)
≤
d
(
x
n
,
¯
x
)
< epsilon1
. Hence

f
y
(
x
n
)

f
y
(¯
x
)

=
d
(
x
n
, y
)

d
(¯
x, y
)
< epsilon1
.
b) Prove, or provide a counterexample, to the following conjecture:
Conjecture:
(
{
f
y
(
x
n
)
}
converges to
f
y
(¯
x
) in the Pythagorian metric) implies (
{
x
n
}
converges
to ¯
x
in the metric
d
).
Ans:
Let
X
=
R
,
d
be the Pythagorian metric and
y
= 0
. Let
x
n
= (

1)
n
and let
¯
x
= 1
. For
each
n
,
f
y
(
x
n
) =
d
(
x
n
,
0) = 1
. Also
f
y
(¯
x
) =
d
(¯
x,
0) = 1
. Therefore, the sequence
{
f
y
(
x
n
)
}
trivially converges to
f
y
(¯
x
)
. However, the sequence
{
x
n
}
does not converge to 1.
Problem 2
: Extrema
. (20 points)
Let
F
denote the set of all continuous functions from
R
to
R
and consider the following five
properties:
P :
f
is strictly quasiconcave
Q :
f
is concave
R :
f
attains a global maximum
S :
f
attains a global minimum
T :
f
is bounded (that is, the image
f
(
R
) is bounded above and bounded below.)
There are 2
5
= 32 ways of combining these five properties (i.e., each can be satisfied or not). We
have selected ten of these thirtytwo.
For each one of these ten, say whether or not there is an
element of
F
that satisfies the combination.
If your answer is “yes,” provide an example; if it
is “no,” prove that the combination is internally inconsistent.
(Note that you may not need to
use
all five
properties in order to show internal inconsistency.) For example, for the combination
(
P
Q
¬
R
¬
S
¬
T)
there does exist
f
∈
F
satisfying these conditions:
f
(
x
) =
x
.
The ten
combinations are
a) (
¬
P
¬
Q
¬
R
¬
S
¬
T)
Ans:
Yes.
f
(
x
) =
x
(
x
+ 1)(
x

1)
.
b) (
¬
P
¬
Q
R
S
¬
T)
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Ans:
No. R and S implies T. To see this, let
f
be maximized at
¯
x
and minimized at
x
¯
and pick
N
∈
N
such that
N >
max(

f
(¯
x
)

,

f
(
x
¯
)

)
. But this means that
f
is bounded above by
N
and
below by

N
.
c) (
¬
P
¬
Q
R
S
T)
Ans:
Yes.
f
(
x
) =
braceleftBigg
0
if

x

>
1
x
(
x
+ 1)(
x

1)
if

x
 ≤
1
.
d) (
¬
P
Q
¬
R
¬
S
T)
Ans:
No.
¬
P and Q implies R. Specifically, the condition
¬
P implies that there exists
x
*
,
y
*
,
z
*
and
t
*
such that
z
*
=
t
*
x
*
+(1

t
*
)
y
*
and
f
(
z
*
)
≤
min
{
f
(
x
*
)
, f
(
y
*
)
}
. The condition Q implies
that
for
all
x
,
y
s.t.
x
negationslash
=
y
and
all
t
∈
(0
,
1)
,
f
(
tx
+ (1

t
)
y
)
≥
tf
(
x
) + (1

t
)
f
(
y
)
. First suppose
f
(
x
*
)
negationslash
=
f
(
y
*
)
and, without loss of generality, let
f
(
x
*
)
<
f
(
y
*
)
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 Fall '07
 Simon
 Convex function, blue line, Extrema., satisfy¯P

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