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 f a+b 2 ≤ f (a)+f (b) 2 . This implies −ln( a+b 2 ) ≤ − ln(a)−ln(b) 2 =−ln( √ab). Therefore, a+b 2 ≥ √ab. This inequality holds obviously whena=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, ∂ f (x) = {−a}, if x <0; [−a,a], if x =0; {a}, if x >0. (b) By Theorem 5.6.5, ∂ f (x) = {−2}, if x <−1; [−2,0], if x =−1; {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 anyx ∈[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 first conclusion. In this case, f (x) = 2m−1 ∑ i=1 | x−i| = 2m−1 ∑ i=1 fi(x), 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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