Ex4-8

Question:

(a) Let

f

{\displaystyle f}

 and

g

{\displaystyle g}

 be two functions that are usc at

x
0

{\displaystyle x\_0}

. Show that

f
g

{\displaystyle f+g}

 is usc at

x
0

{\displaystyle x\_0}

. Is

f
−
g

{\displaystyle f -g}

 usc at

x
0

{\displaystyle x\_0}

? When is

f
g

{\displaystyle f g}

 usc at

x
0

{\displaystyle x\_0}

?

(b) If

{

f
k

}

{\displaystyle \{ f\_k \}}

 is a sequence of functions that are usc at

x
0

{\displaystyle x\_0}

, show that

inf
k

f
k

(
x
)

{\displaystyle \inf\_k f\_k(x)}

 is usc at

x
0

{\displaystyle x\_0}

.

(c) If

{

f
k

}

{\displaystyle \{ f\_k \}}

 is a sequence of functions that are usc at

x
0

{\displaystyle x\_0}

 and converge uniformly near

x
0

{\displaystyle x\_0}

, show that

lim

f
k

{\displaystyle \lim f\_k}

 is usc at

x
0

{\displaystyle x\_0}

.

Solutions:

Recall: Let

f

{\displaystyle f}

 be defined on

E

{\displaystyle E}

,and let

x
0

{\displaystyle x\_0}

 be a limit point of

E

{\displaystyle E}

 that lies in

E

{\displaystyle E}

. Then is said to be upper semicontinuous (usc) at

x
0

{\displaystyle x\_0}

 if

lim sup

x
→

x
0

;
x
∈
E

f
(
x
)
≤
f
(

x
0

)
.

{\displaystyle \limsup\_{x\to x\_0;x\in E} f(x)\leq f(x\_0).}

(a) Given

f

{\displaystyle f}

 and

g

{\displaystyle g}

 be two functions that are usc at

x
0

{\displaystyle x\_0}

. This implies that

lim sup

x
→

x
0

f
(
x
)
≤
f
(

x
0

)

{\displaystyle \limsup\_{x\to x\_0} f(x)\leq f(x\_0)}

 and

lim sup

x
→

x
0

g
(
x
)
≤
g
(

x
0

)
.

{\displaystyle \limsup\_{x\to x\_0} g(x)\leq g(x\_0).}

 If

lim sup

x
→

x
0

f
(
x
)

lim sup

x
→

x
0

g
(
x
)

{\displaystyle \limsup\_{x\to x\_0} f(x)+\limsup\_{x\to x\_0} g(x)}

 is well-defined, that is not

(
∞
−
∞
)

{\displaystyle (\infty-\infty)}

 or

(
−
∞
∞
)

{\displaystyle (-\infty+\infty)}

, then

lim sup

x
→

x
0

[
f
(
x
)
g
(
x
)
]

≤

lim sup

x
→

x
0

f
(
x
)

lim sup

x
→

x
0

g
(
x
)

≤
f
(

x
0

)
g
(

x
0

)

{\displaystyle \begin{align}
\limsup\_{x\to x\_0} [f(x)+g(x)]&\leq \limsup\_{x\to x\_0}f(x)+\limsup\_{x\to x\_0}g(x)\\
&\leq f(x\_0)+g(x\_0)
\end{align}}

Thus,

f
g

{\displaystyle f+g}

 is usc at

x
0

{\displaystyle x\_0}

.
For the function

f
−
g

{\displaystyle f -g}

, it is not usc at

x
0

{\displaystyle x\_0}

. One counterexample is:

Let

f
(
x
# )
0

for all

x

{\displaystyle f(x)=0\quad \text{for all}~x}

 and

g
(
x
# )

{

0
,

if
x
<
0

1
,

if
x
≥
0

{\displaystyle g(x) = \begin{cases} 0, & \text{if }x<0 \\ 1, & \text{if }x\geq 0 \end{cases}}

, then

f

{\displaystyle f}

 and

g

{\displaystyle g}

 are usc at

x
0

=
0

{\displaystyle x\_0 = 0}

. However,

f
(
x
)
−
g
(
x
# )

{

0
,

if
x
<
0

−
1
,

if
x
≥
0

{\displaystyle f(x)-g(x) = \begin{cases} 0, & \text{if }x<0 \\ -1, & \text{if }x\geq 0 \end{cases}}

, which is not usc at

x
0

=
0

{\displaystyle x\_0 = 0}

.

For the product function

f
g

{\displaystyle f g}

, it is usc at

x
0

{\displaystyle x\_0}

 if

lim sup

x
→

x
0

[
f
(
x
)
g
(
x
)
]
≤
(

lim sup

x
→

x
0

f
(
x
)
)
(

lim sup

x
→

x
0

g
(
x
)
)
.

{\displaystyle \limsup\_{x\to x\_0} [f(x)g(x)]\leq (\limsup\_{x\to x\_0}f(x))(\limsup\_{x\to x\_0}g(x)).}

This inequality can only be true when both functions

f

{\displaystyle f}

 and

g

{\displaystyle g}

 are nonnegative.

(b) Given

{

f
k

}

{\displaystyle \{ f\_k \}}

 is a sequence of functions that are usc at

x
0

{\displaystyle x\_0}

. So,

lim sup

x
→

x
0

f
k

(
x
)
≤

f
k

(

x
0

)

for all

k
∈

N

.

{\displaystyle \limsup\_{x\to x\_0} f\_k(x)\leq f\_k(x\_0)\quad\text{for all}~k\in\mathbb{N}.}

 Now, for all

k

{\displaystyle k}

,

lim sup

x
→

x
0

[

inf
k

f
k

(
x
)
]

≤

lim sup

x
→

x
0

f
k

(
x
)

≤

f
k

(

x
0

)
.

{\displaystyle \begin{align}
\limsup\_{x\to x\_0}[\inf\_k f\_k(x)]&\leq \limsup\_{x\to x\_0} f\_k(x)\\
&\leq f\_k(x\_0).
\end{align}}

Since the above inequality is true for all

k

{\displaystyle k}

, we have

lim sup

x
→

x
0

inf
k

f
k

(
x
)
≤

inf
k

f
k

(

x
0

)

{\displaystyle \limsup\_{x\to x\_0}\inf\_k f\_k(x)\leq \inf\_k f\_k(x\_0)}

and thus

inf
k

f
k

(
x
)

{\displaystyle \inf\_k f\_k(x)}

 is usc at

x
0

{\displaystyle x\_0}

.
(c) Given

{

f
k

}

{\displaystyle \{ f\_k \}}

 is a sequence of functions that are usc at

x
0

{\displaystyle x\_0}

. So,

lim sup

x
→

x
0

f
k

(
x
)
≤

f
k

(

x
0

)

for all

k
∈

N

.

{\displaystyle \limsup\_{x\to x\_0} f\_k(x)\leq f\_k(x\_0)\quad\text{for all}~k\in\mathbb{N}.}

Taking limits on both sides,

lim

k
→
∞

lim sup

x
→

x
0

f
k

(
x
)
≤

lim

k
→
∞

f
k

(

x
0

)
.

{\displaystyle \lim\_{k\to \infty}\limsup\_{x\to x\_0} f\_k(x)\leq \lim\_{k\to \infty}f\_k(x\_0).}

Since

{

f
k

}

{\displaystyle \{ f\_k \}}

converge uniformly near

x
0

{\displaystyle x\_0}

, we can interchange *lim* and *limsup*, then

lim sup

x
→

x
0

lim

k
→
∞

f
k

(
x
)
≤

lim

k
→
∞

f
k

(

x
0

)

{\displaystyle \limsup\_{x\to x\_0}\lim\_{k\to \infty} f\_k(x)\leq \lim\_{k\to \infty}f\_k(x\_0)}

and thus

lim

f
k

{\displaystyle \lim f\_k}

 is usc at

x
0

{\displaystyle x\_0}

.