Introduction to Mathematical Analysis I - Second Edition

137 Example 4.7.3 Let a 1 < a 2 < · · · < a n and let µ i > 0 for i = 1 , . . . , n . Define f ( x ) = n ∑ i = 1 µ i | x − a i | . Then f is a convex function. By Theorem 4.7.5 , we get ∂ f ( ¯ x ) = ( ∑ a i < ¯ x µ i − ∑ a i > ¯ x µ i , if ¯ x / ∈ { a 1 , . . . , a n } ∑ a i < ¯ x µ i − ∑ a i > ¯ x µ i + [ − µ i 0 , µ i 0 ] , if ¯ x = a i 0 . Theorem 4.7.6 Let f i : R → R , i = 1 , . . . , n , be convex functions. Define f ( x ) = max { f i ( x ) : i = 1 , . . . , n } and I ( u ) = { i = 1 , . . . , n : f i ( u ) = f ( u ) } . Then f is a convex function. Moreover, ∂ f ( ¯ x ) = [ m , M ] , where m = min i ∈ I ( ¯ x ) f 0 i − ( ¯ x ) and M = max i ∈ I ( ¯ x ) f 0 i + ( ¯ x ) . Proof: Fix u , v ∈ R and λ ∈ ( 0 , 1 ) . For any i = 1 , . . . , n , we have f i ( λ u +( 1 − λ ) v ) ≤ λ f i ( u )+( 1 − λ ) f i ( v ) ≤ λ f ( u )+( 1 − λ ) f ( v ) . This implies f ( λ u +( 1 − λ ) v ) = max 1 ≤ i ≤ n f i ( λ u +( 1 − λ ) v ) ≤ λ f ( u )+( 1 − λ ) f ( v ) . Thus, f is a convex function. Similarly we verify that f 0 + ( ¯ x ) = M and f 0− ( ¯ x ) = m . By Theorem 4.7.3 , ∂ f ( ¯ x ) = [ m , M ] . The proof is now complete. Remark 4.7.7 The product of two convex functions is not a convex function in general. For instance, f ( x ) = x and g ( x ) = x 2 are convex functions, but h ( x ) = x 3 is not a convex function. The following result may be considered as a version of the first derivative test for extrema in the case of non differentiable functions. Theorem 4.7.8 Let f : R → R be a convex function. Then f has an absolute minimum at ¯ x if and only if 0 ∈ ∂ f ( ¯ x ) = [ f 0 − ( ¯ x ) , f 0 + ( ¯ x )] . Proof: Suppose f has an absolute minimum at ¯ x . Then f ( ¯ x ) ≤ f ( x ) for all x ∈ R .