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.
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.
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).
(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.
(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.
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].
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).
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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