Introduction to Mathematical Analysis I - Second Edition

164 Solutions and Hints for Selected Exercises (b) By Theorem 4.7.5 , ∂ 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 4.7.3 . 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 = 2 m − 1, then f has an absolute minimum at x = m . If n is even with n = 2 m , then f has an absolute minimum at any point x ∈ [ m , m + 1 ] . Let us prove the first conclusion. In this case, f ( x ) = 2 m − 1 ∑ i = 1 | x − i | = 2 m − 1 ∑ i = 1 f i ( x ) , where f i ( x ) = | x − i | . Consider ¯ x = m . Then ∂ f m ( ¯ x ) = [ − 1 , 1 ] , ∂ f i ( ¯ x ) = { 1 } if i < m , ∂ f i ( ¯ x ) = {− 1 } if i > m . The subdifferential sum rule yields ∂ f ( ¯ x ) = [ − 1 , 1 ] which contains 0. Thus, f has an absolute minimum at ¯ x . If ¯ x > m , we can see that ∂ f ( ¯ x ) ⊂ ( 0 , ∞ ) , which does not contain 0. Similarly, if ¯ x < m , then ∂ f ( ¯ x ) ⊂ ( − ∞ , 0 ) . Therefore, f has an absolute minimum at the only point ¯ x = m . The case where n is even can be treated similarly. Exercise 4.7.5 . Fix a , b ∈ R with a < b . By Theorem 4.7.9 , there exists c ∈ ( a , b ) such that f ( b ) − f ( a ) b − a ∈ ∂ f ( c ) ⊂ [ 0 , ∞ ) . This implies f ( b ) − f ( a ) ≥ 0 and, hence, f ( b ) ≥ f ( a ) . Therefore, f is monotone increasing on R ,