Introduction to Mathematical Analysis I - Second Edition

163 SECTION 4.6 Exercise 4.6.3 . We apply the definition to solve this problem. Given any u , v ∈ I and λ ∈ ( 0 , 1 ) , we have f ( λ u +( 1 − λ ) v ) ≤ λ f ( u )+( 1 − λ ) f ( v ) by the convexity of f . Since f ( u ) , f ( v ) ∈ J and J is an interval, λ f ( u ) + ( 1 − λ ) f ( v ) ∈ J . By the nondecreasing property and the convexity of φ , φ ( f ( λ u +( 1 − λ ) v )) ≤ φ ( λ f ( u )+( 1 − λ ) f ( v )) ≤ λφ ( f ( u ))+( 1 − λ ) φ ( f ( v )) . Therefore, φ ◦ f is convex on I . The result from this problem allows us to generate convex functions. For example, consider f ( x ) = | x | and φ ( x ) = x p , 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 on R . Similarly, h ( x ) = e x 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 ) = x 2 and φ ( x ) = | x − 1 | are convex, but ( φ ◦ f )( x ) = | x 2 − 1 | is nonconvex. Exercise 4.6.4 . Use Theorem 4.6.6 or Corollary 4.6.7 . Exercise 4.6.5 . (a) Use the obvious inequality ( √ a − √ b ) 2 ≥ 0 . 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 when a = 0 or b = 0. (b) Use Theorem 4.6.3 for the function f ( x ) = − ln ( x ) on ( 0 , ∞ ) . SECTION 4.7 Exercise 4.7.1 . (a) By Theorem 4.7.5 , ∂ f ( x ) =   {− a } , if x < 0; [ − a , a ] , if x = 0; { a } , if x > 0 .