Introduction to Mathematical Analysis I - Second Edition

129 Proof: For any x ∈ I and t ∈ ( 0 , 1 ) , we have f ( ¯ x + t ( x − ¯ x )) − f ( ¯ x ) t = f ( tx +( 1 − t ) ¯ x ) − f ( ¯ x ) t ≤ t f ( x )+( 1 − t ) f ( ¯ x ) − f ( ¯ x ) t = f ( x ) − f ( ¯ x ) . Since f is differentiable at ¯ x , f 0 ( ¯ x )( x − ¯ x ) = lim t → 0 + f ( ¯ x + t ( x − ¯ x )) − f ( ¯ x ) t ≤ f ( x ) − f ( ¯ x ) , which completes the proof. Corollary 4.6.4 Let I be an open interval and let f : I → R be a convex function. Suppose f is differentiable at ¯ x . Then f has an absolute minimum at ¯ x if and only if f 0 ( ¯ x ) = 0. Proof: Suppose f has an absolute minimum at ¯ x . By Theorem 4.2.1 , f 0 ( ¯ x ) = 0. Let us prove the converse. Suppose f 0 ( ¯ x ) = 0. It follows from Theorem 4.6.3 that 0 = f 0 ( ¯ x )( x − ¯ x ) ≤ f ( x ) − f ( ¯ x ) for all x ∈ I . This implies f ( x ) ≥ f ( ¯ x ) for all x ∈ I . Thus, f has an absolute minimum at ¯ x . Lemma 4.6.5 Let I be an open interval and suppose f : I → R is a convex function. Fix a , b , x ∈ I with a < x < b . Then f ( x ) − f ( a ) x − a ≤ f ( b ) − f ( a ) b − a ≤ f ( b ) − f ( x ) b − x . Proof: Let t = x − a b − a . Then t ∈ ( 0 , 1 ) and f ( x ) = f ( a +( x − a )) = f a + x − a b − a ( b − a ) = f ( a + t ( b − a )) = f ( tb +( 1 − t ) a ) . By convexity of f , we obtain f ( x ) ≤ t f ( b )+( 1 − t ) f ( a ) . Thus, f ( x ) − f ( a ) ≤ t f ( b )+( 1 − t ) f ( a ) − f ( a ) = t [ f ( b ) − f ( a )] = x − a b − a ( f ( b ) − f ( a )) .