1. Quizz 10
Definition 1.1. Let S be a nonempty set and f : S → R be a function. Suppose that f is
bounded above1 We say that f attains its maximum on S if there exists s0 ∈ S such that
f (s0 ) = sup f (S). In this case, f (s0 ) is called a maximum of f. Similarly, if f is bounded
below, f attains its minimum on S if there exists s1 ∈ S such that f (s1 ) = inf f (S).
(1) Let S be a nonempty subset of R. Suppose that S is bounded above2. Show that
sup S is an adherent point of S. Furthermore, if sup S 6∈ S, then sup S is an accumulation point of S.
Proof. Let M denotes sup S. When M ∈ S, M ∈ S. Now, let us consider the case
when M 6∈ S. We will give two different proofs.
Proof I For > 0, M − is no longer an upper bound for S. There exists y ∈ S
such that s > M − . Since M 6∈ S, y 6= M. We see that 0 < M − y < . This implies
that y ∈ B0 (M, ) ∩ S. We find that B0 (M, ) ∩ S 6= ∅ for any > 0. This shows that
M is an accumulation point of S and thus M ∈ S
Proof II For any n ≥ 1, we can find yn ∈ S such that yn > M − 1/n. Since
M∈
6 S, we obtain a sequence (yn ) in S such that
1
M − < yn < M for any n ≥ 1.
n
1
Since lim M −
= lim M = M, by Sandwich principle, lim yn = M. This
n→∞
n→∞
n→∞
n
shows that M is a limit point of S and hence M ∈ S.
Remark. When S is nonempty and bounded below, the above statement holds
when we replace sup S by inf S.
(2) Let f : K → R be a continuous function on a sequentially compact metric space
(K, d). Show that f attains its maximum and minimum on K.
Proof. Since K is sequentially compact and f : K → R is continuous, f (K) is
sequentially compact in R. By Bolzano-Weierstrass Theorem, f (K) is closed and
bounded in R. Since f (K) is bounded, sup f (K) and inf f (K) exists. By the previous
exercise, sup f (K) and inf f (K) are both adherent to f (K). Since f (K) is closed,
it contains all of its adherent points. Thus sup f (K) and inf f (K) both belong to
f (K).
Remark. Choose sequences (xn ) and (yn ) in K such that limn→∞ f (xn ) = sup f (K)
and limn→∞ f (yn ) = inf f (K). We can do this because sup f (K) and inf f (K) are
adherent points of f (K).
1A function f : S → R is bounded above (below) if its range f (S) is bounded above (below).
2A nonempty subset S of R is bounded above if there exists U ∈ R such that x ≤ U for any x ∈ S. Such an U is
called an upper bound for S. If S is bounded above, the supremum sup S of S is the smallest U so that U is an upper
bound for S.
1
2
(3) Let f : K → R be a continuous function on a sequentially compact metric space.
Prove that there exist no sequences (xn ) and (yn ) in K satisfying the following
properties.
(a) d(xn , yn ) < 1/n for any n ≥ 1.
(b) There exists > 0 such that |f (xn ) − f (yn )| ≥ for any n ≥ 1.
Proof. Since K is sequentially compact, we can choose a subsequence (xnj ) of (xn )
such that (xnj ) is convergent to some x ∈ K. By sequentially compactness of K
again, (ynj ) has a subsequence (ynjk ) convergent to some y in K. By properties, we
have that for any k ≥ 1,
1
1
1
≤
d(xnjk , ynjk ) <
≤
njk
jk
k
and
|f (xnjk ) − f (ynjk )| ≥ .
By triangle inequality,
d(x, y) ≤ d(xnjk , x) + d(xnjk , ynjk ) + d(ynjk , y)
for any k ≥ 1. Since (xnj ) is convergent to x, its subsequence (xnjk ) is also convergent
to x. For any > 0, we can choose N0 ∈ N and N00 = [3/] + 1 and N000 ∈ N such
that
d(xnjk , x) < for k ≥ N0 ,
3
d(xnjk , ynjk ) < for k ≥ N00 ,
3
d(ynjk , y) < for k ≥ N000 .
3
000
00
0
Take N = max{N , N , N }. For k ≥ N , d(x, y) < . We see that d(x, y) < for
any > 0. Therefore d(x, y) = 0 and thus x = y.
(4) Let k : [0, 1] × [0, 1] → R be a continuous function.
(a) Prove that for any > 0 there exists δ > 0 such that
|k(x1 , y1 ) − k(x2 , y2 )| < whenever k(x1 , y1 ) − (x2 , y2 )k < δ with (xi , yi ) ∈ [0, 1] × [0, 1] for i = 1, 2.
Proof. Suppose not. We can find > 0 and choose a sequence (an ) and a
sequence (bn ) in [0, 1] × [0, 1] such that kan − bn k < 1/n for any n ≥ 1 while
|k(an ) − k(bn )| ≥ . Since [0, 1] × [0, 1] is closed and bounded, by BolzanoWeierstrass Theorem, [0, 1] × [0, 1] is sequentially compact. Using Exercise 3,
since k is continuous and [0, 1] × [0, 1] is compact, sequences (an ) and (bn ) do
not exist.
Remark. Suppose not. Use (3) to prove the result by contradiction.
(b) Let f : [0, 1] → R be continuous. Define g : [0, 1] → R by
Z 1
g(x) =
k(x, y)f (y)dy, x ∈ [0, 1].
0
3
Prove that g is also continuous on [0, 1]. Hint: consider
Z 1
g(x1 ) − g(x2 ) =
(k(x1 , y) − k(x2 , y))f (y)dy
0
and use (a).
Proof. When f = 0, the statement is obvious. Assume that f 6= 0. By (a), for
any > 0, choose δ > 0 such that
|k(x1 , y1 ) − k(x2 , y2 )| <
kf k∞
whenever k(x1 , y1 ) − (x2 , y2 )k < δ with (xi , yi ) ∈ [0, 1] × [0, 1] for i = 1, 2. For
this δ, if |x1 − x2 | < δ, then k(x1 , y) − (x2 , y)k < for any y ∈ [0, 1]. This implies
that
|k(x1 , y) − k(x2 , y)| <
,
kf k∞
whenever |x1 − x2 | < δ for any y ∈ [0, 1]. Thus if |x1 − x2 | < δ ,
Z 1
|g(x1 ) − g(x2 )| ≤
|(k(x1 , y) − k(x2 , y))|f (y)|dy ≤ kf k∞ ·
= .
kf
k∞
0
This implies that g is (uniformly) continuous.