Introduction to Mathematical Analysis I - Second Edition

135 Thus, u ≤ lim x → ¯ x + f ( x ) − f ( ¯ x ) x − ¯ x = f 0 + ( ¯ x ) . Similarly, we have u · ( x − ¯ x ) ≤ f ( x ) − f ( ¯ x ) for all x < ¯ x . Thus, u ≥ f ( x ) − f ( ¯ x ) x − ¯ x for all x < ¯ x . This implies u ≥ f 0− ( ¯ x ) . So ∂ f ( ¯ x ) ⊂ [ f 0 − ( ¯ x ) , f 0 + ( ¯ x )] . To prove the opposite inclusion, take u ∈ [ f 0− ( ¯ x ) , f 0 + ( ¯ x )] . By Theorem 4.7.2 sup x < ¯ x φ ¯ x ( x ) = f 0 − ( ¯ x ) ≤ u ≤ f 0 + ( ¯ x ) = inf x > ¯ x φ ¯ x ( x ) . Using the upper estimate by f 0 + ( ¯ x ) for u , one has u ≤ φ ¯ x ( x ) = f ( x ) − f ( ¯ x ) x − ¯ x for all x > ¯ x . It follows that u · ( x − ¯ x ) ≤ f ( x ) − f ( ¯ x ) for all x ≥ ¯ x . Similarly, one also has u · ( x − ¯ x ) ≤ f ( x ) − f ( ¯ x ) for all x < ¯ x . Thus, ( 4.16 ) holds and, hence, u ∈ ∂ f ( ¯ x ) . Therefore, ( 4.18 ) holds. Corollary 4.7.4 Let f : R → R be a convex function and ¯ x ∈ R . Then f is differentiable at ¯ x if and only if ∂ f ( ¯ x ) is a singleton. In this case, ∂ f ( ¯ x ) = { f 0 ( ¯ x ) } . Proof: Suppose f is differentiable at ¯ x . Then f 0 − ( ¯ x ) = f 0 + ( ¯ x ) = f 0 ( ¯ x ) . By Theorem 4.7.3 , ∂ f ( ¯ x ) = [ f 0 − ( ¯ x ) , f 0 + ( ¯ x )] = { f 0 ( ¯ x ) } . Thus, ∂ f ( ¯ x ) is a singleton. Conversely, if ∂ f ( ¯ x ) is a singleton, we must have f 0− ( ¯ x ) = f 0 + ( ¯ x ) . Thus, f is differentiable at ¯ x .