Theorem 1 (Theorem of the Maximum) Let X ⊆ Rl and Y ⊆ Rm , let f : X×Y →R be a
continuous function, and let Γ : X→Y be a compact-valued correspondence. Then the function
h : X→R defined in (1) is continuous, and the correspondence G : X→Y defined in (2) is nonempty,
compact-valued and u.h.c.
1. G(x)6=∅.
Proof 1
Fix x,
∵ Γ(x) is compact and f (x, ·) is continuous,
∴f (x, Γ(x)) is compact, then maximum can be attained
∴G(x)6=∅
2. G(x) is compact.
(a) G(x) is bonded.
Fix x,
∵G(x)⊆Γ(x) and Γ(x) is compact,
∴G(x) is bounded.
(b) G(x) is closed.
Fix x,
Pick any sequence {yn }s.t.yn ∈G(x)and{yn }→y
∵G(x)⊆Γ(x),
∴{yn }⊆Γ(x).
∵Γ(x) is closed,
∴y∈Γ(x).
∵f (x, yn ) = h(x),
∴f (x, y) = h(x).
∴y∈h(x)
Since {yn } is arbitrary, G(x) is closed.
Since G(x) is bounded and closed, G(x) is compact.
Since x is arbitrary, G is compact-valued.
1
3. G is u.h.c.
Fix x, pick any sequence {xn }s.t.xn →x, pick any ”point” yn s.t.yn ∈G(xn ) for all n, then we
have the sequence {yn }.
∵Γ is u.h.c,
∴∃ a sequence {ynk }s.t.ynk →y and y∈Γ(x).
Pick z∈Γ(x),
∵Γ is l.h.c,
∴∃{znk }s.t.znk →z.
∵f is continuous and f (xnk , ynk )≥f (xnk , znk ),
∴f (x, y)≥f (x, z)
∵z can be anywhere on Γ(x),
∴f (x, y) is the maximum on Γ(x) and then y∈G(x).
Since this holds for any {xn } and any {yn |yn ∈G(xn )}, G is u.h.c.
4. h is continuous.(De La Fuenee, P301)
Note
h(x) = maxy∈Γ(x) f (y, x) = f (G(x), x)
and composition of two u.h.c correspondence is u.h.c.
∵ h is u.h.c(continuous)and we just proofed G is u.h.c,
∴ h is u.h.c.
∵ h is single-valued,
∴ h is continuous.
Theorem 2 Let X⊆Rl , and let C(X) be the set of bounded continuous functions f : X→R with
the sup norm,kf k = supx∈X |f (x)|. Then C(X) is a complete normed vector space.
Proof 2 C(X) is complete
⇐⇒ if {fn } is a cauchy sequence, there exists a function f ∈C(X) s.t.
2
for any > 0, there exists N s.t.
kfn − f k≤, all n≥N
There are three steps in this proof:
1. find a candidate function f
2. show that {fn }→f in sup norm
3. show that f ∈C(X)
Fix x∈X,
then there exists the sequence of real number {fn (x)}∞
n=1 satisfies
|fn (x) − fm (x)|≤supy∈X |fn (y) − fm (y)| = kfn − fm k
∴{fn (x)}∞
n=1 is a cauchy sequence.
∵R is a complete space,
∴{fn (x)}∞
n=1 converges to a limit point, call it f (x).
Then limit values define a function
f : X→R
that we can take as the candidate.
∵{fn (x)}∞
n=1 is a cauchy sequence,
∴ for any given > 0, we can choose N s.t.
kfn − fm k≤
2
all n, m≥N
Assume m≥n without loss of generality,
3
for any fixed x∈X and all m≥n≥N ,
|fn (x) − f (x)| ≤ |fn (x) − fm (x)| + |fm (x) − f (x)|
≤ kfn − fm k + |fm (x) − f (x)|
≤
+ |fm (x) − f (x)|
2
∵{fm (x)}∞
m=1 converges to f (x) for each x,
∴ we can choose m separately for each fixed x, s.t. |fm (x) − f (x)|≤ 2
∴|fn (x) − f (x)|≤, for all n≥N and all x
=⇒ kfn − f k = supx∈X |fn (x) − f (x)|≤ for all n≥N
Since > 0 was arbitrary, fn converges to f .
f is continuous
⇐⇒ for any > 0 and every x∈X, there exist δ > 0 s.t.
|f (x) − f (y)| < if kx − ykE < δ
where k · k is the Euclidean norm on Rl .
∵fn →f ,
∴ we can choose k s.t. kfk − f k≤ 3 .
∵fk is continuous,
∴ we can choose δ s.t. kx − ykE implies |fk (x) − fk (y)| < 3 .
4
Then
|f (x) − f (y)| ≤ |f (x) − fk (x)| + |fk (x) − fk (y)| + |fk (y) − f (y)|
≤ 2kf − fk k + |fk (x) − fk (y)|
≤ 5