174 Solutions and Hints for Selected Exercises Alternatively, consider the function f (x)= −ln(x), x ∈ (0,∞). We can show that f is convex on (0,∞). For a,b ∈ (0,∞), one has a + b f (a)+ f (b) f ≤ . 2 2 This implies a+ b −ln(a) − ln(b) √ −ln( ) ≤ = −ln( ab). 2 2 Therefore, √ a+ b ≥ ab. 2 This inequality holds obviously when a = 0 or b = 0. (b) Use Theorem 5.5.3 for the function f (x)= −ln(x) on (0,∞). SECTION 5.6 Exercise 5.6.1. (a) By Theorem 5.6.5, if x < 0; {−a}, ∂ f (x)= [−a,a], if x = 0; {a}, if x > 0. (b) By Theorem 5.6.5, {−2}, if x < −1; [−2,0], if x = −1; ∂ f (x)= {0}, if x ∈ (−1,1); [0,2], if x = 1; {2}, if x > 1. Exercise 5.6.3. To better understand the problem, we consider some special cases. If n = 1, then f (x)= |x−1|. Obviously, f has an absolute minimum at x = 1. If n = 2, then f (x)= |x−1|+|x−2|. The graphing of the function suggests that f has an absolute minimum at any x ∈ [1,2]. In the case where n = 3, we can see that f has an absolute minimum at x = 2. We then conjecture that if n is odd with n = 2m − 1, then f has an absolute minimum at x = m. If n is even with n = 2m, then f has an absolute minimum at any point x ∈ [m,m + 1]. Let us prove the frst conclusion. In this case, 2m−1 2m−1 f (x)= ∑ |x − i| = ∑ fi(x), i=1 i=1 where fi(x)= |x − i|. Consider x0 = m. Then ∂ fm(x0)=[−1,1], ∂ fi(x0)= {1} if i < m, ∂ fi(x0)= {−1} if i > m. The subdifferential sum rule yields ∂ f (x0)=[−1,1] which contains 0. Thus, f has an absolute minimum at x0. If x0 > m, we can see that ∂ f (x0) ⊂ (0,∞), which does not contain 0. Similarly, if x0 < m, then ∂ f (x0) ⊂ (−∞,0). Therefore, f has an absolute minimum at the only point x0 = m.
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