Suppose that f is bounded above1 We say that f attains its maximum on S if there exists s0 ∈ S such that f (s0

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 accu- mulation 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 y_n ∈ S such that y_n > M − 1/n. Since M 6∈ S, we obtain a sequence (y_n) in S such that

M − 1

n < yn< M for any n ≥ 1.

Since lim

n→∞

 M − 1

n



= lim

n→∞M = M, by Sandwich principle, lim

n→∞y_n = M. This 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 (x_n) and (y_n) in K such that lim_n→∞f (x_n) = 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.

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(3) Let f : K → R be a continuous function on a sequentially compact metric space.

Prove that there exist no sequences (x_n) and (y_n) in K satisfying the following properties.

(a) d(x_n, y_n) < 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 (yn_jk) convergent to some y in K. By properties, we have that for any k ≥ 1,

d(xn_jk, yn_jk) < 1 nj_k

≤ 1 j_k ≤ 1

k and

|f (x_n

jk) − f (y_njk)| ≥ .

By triangle inequality,

d(x, y) ≤ d(x_njk, x) + d(x_njk, y_njk) + d(y_njk, y)

for any k ≥ 1. Since (x_nj) is convergent to x, its subsequence (x_njk) is also convergent to x. For any  > 0, we can choose N_⁰ ∈ N and N⁰⁰ = [3/] + 1 and N_⁰⁰⁰ ∈ N such that

d(x_njk, x) < 

3 for k ≥ N_⁰, d(xn_jk, yn_jk) < 

3 for k ≥ N_⁰⁰, d(yn_jk, y) < 

3 for k ≥ N_⁰⁰⁰.

Take N_ = max{N_⁰, N_⁰⁰, N_⁰⁰⁰}. 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 (x_i, y_i) ∈ [0, 1] × [0, 1] for i = 1, 2.

Proof. Suppose not. We can find  > 0 and choose a sequence (an) and a sequence (b_n) in [0, 1] × [0, 1] such that ka_n− b_nk < 1/n for any n ≥ 1 while

|k(a_n) − k(b_n)| ≥ . Since [0, 1] × [0, 1] is closed and bounded, by Bolzano- Weierstrass Theorem, [0, 1] × [0, 1] is sequentially compact. Using Exercise 3, since k is continuous and [0, 1] × [0, 1] is compact, sequences (a_n) and (b_n) 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 g(x) =

Z 1 0

k(x, y)f (y)dy, x ∈ [0, 1].

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Prove that g is also continuous on [0, 1]. Hint: consider g(x₁) − g(x₂) =

Z 1 0

(k(x₁, y) − k(x₂, y))f (y)dy 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 (x_i, y_i) ∈ [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_,

|g(x₁) − g(x2)| ≤ Z 1

0

|(k(x₁, y) − k(x2, y))|f (y)|dy ≤ kf k∞· 

kf k_∞ = .

This implies that g is (uniformly) continuous.