173 SECTION 5.4 Exercise 5.4.4. Since limx→∞ f (x) =limx→−∞ f (x) =∞, there exists a>0 such that f (x) ≥f (0) whenever |x| >a. Since f is lower semicontinuous, by Theorem 5.4.3, it has an absolute minimum on [−a,a] at some point x0 ∈[−a,a]. Obviously, f (x) ≥f (x0) for all x ∈[−a,a]. In particular, f (0) ≥f (x0). If |x| >a, then f (x) ≥f (0) ≥f (x0). Therefore, f has an absolute minimum at x0. Observe that in this solution, we can use any number γ in the range of f instead of f (0). Since any continuous function is also lower semicontinuous, the result from this problem is applicable for continuous functions. For example, we can use this theorem to prove that any polynomial with even degree has an absolute minimum onR. Since Ris a not a compact set, we cannot use the extreme value theorem directly. SECTION 5.5 Exercise 5.5.3. We apply the definition to solve this problem. Given anyu,v ∈I and λ ∈(0,1), we have f (λu+(1−λ)v) ≤λf (u)+(1−λ)f (v) by the convexity of f . Since f (u), f (v) ∈J and J is an interval, λf (u) + (1−λ)f (v) ∈J. By the nondecreasing property and the convexity of φ, φ(f (λu+(1−λ)v)) ≤φ(λf (u)+(1−λ)f (v)) ≤λφ(f (u))+(1−λ)φ(f (v)). Therefore, φ◦ f is convex onI. The result from this problem allows us to generate convex functions. For example, consider f (x) =|x| and φ(x) =xp, p>1. We have seen that f is convex on R. The function φ is convex and increasing on [0,∞) which contains the range of the function f . Therefore, the composition g(x) =|x|p, p>1, is convex onR. Similarly, h(x) =ex 2 is also a convex function on R. Observe that in this problem, we require the nondecreasing property of φ. A natural question is whether the composition of two convex functions is convex. The answer is negative. Observe that f (x) =x2 andφ(x) =|x−1| are convex, but (φ◦ f )(x) =|x2 −1| is nonconvex. Exercise 5.5.4. Use Theorem 5.5.6 or Corollary 5.5.7. Exercise 5.5.5. (a) Use the obvious inequality (√a− √b)2 ≥0.
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