Exercise 5.4.7

(a)
First prove the following general lemma: Let f be defined on an open interval J and assume f is differentiable at a ∈ J . If ( a n ) and ( b n ) are sequences satisfying a n < a < b n and lim ⁡ a n = lim ⁡ b n = a , show f ′ ( a ) = lim ⁡ n → ∞ f ( b n ) − f ( a n ) b n − a n .

(b)
Now use this lemma to show that g ′ ( x ) does not exist.

Answers

(a)
Keeping in mind the Sequential Criterion for Functional Limits (Theorem 4.2.3),
lim ⁡ n → ∞ f ( b n ) − f ( a n ) b n − a n = lim ⁡ n → ∞ f ( b n ) − f ( a ) b n − a b n − a b n − a n + lim ⁡ n → ∞ f ( a ) − f ( a n ) a − a n a − a n b n − a n = f ′ ( a ) lim ⁡ n → ∞ b n − a b n − a n + f ′ ( a ) lim ⁡ n → ∞ a − a n b n − a n = f ′ ( a ) ( lim ⁡ n → ∞ b n − a + a − a n b n − a n ) = f ′ ( a )
(b)
I claim that
g ( y n + 1 ) − g ( x n + 1 ) y n + 1 − x n + 1 = g ( y n ) − g ( x n ) y n − x n ± 1 ∀ ⁡ n ≥ 0

To see this, note first that h n is a straight line with slope -1 or 1 between p n ∕ 2 n and ( p n + 1 ) ∕ 2 n , and therefore

h n ( b ) − h n ( a ) b − a = ± 1 ∀ ⁡ a , b ∈ [ p n 2 n , p n + 1 2 n ]

Note also that h n ( k ) = 0 when k is of the form p ∕ 2 n − 1 with p ∈ Z . This fact, combined with how we chose x n and y n so that [ x n + 1 , y n + 1 ] ⊂ [ x n , y n ] means we can use the above constant for each term in g as it appears.

g ( y n + 1 ) − g ( x n + 1 ) y n + 1 − x n + 1 = ∑ k = 0 n + 1 h k ( y n + 1 ) − h k ( x n + 1 ) y n + 1 − x n + 1 = h n + 1 ( y n + 1 ) − h n + 1 ( x n + 1 ) y n + 1 − x n + 1 + ∑ k = 0 n h k ( y n + 1 ) − h k ( x n + 1 ) y n + 1 − x n + 1 = ± 1 + ∑ k = 0 n h k ( y n ) − h k ( x n ) y n − x n = g ( y n ) − g ( x n ) y n − x n ± 1

This implies that

lim ⁡ n → ∞ g ( y n ) − g ( x n ) y n − x n

does not exist, since the difference between consecutive elements does not converge to zero, and therefore by our lemma in part (a), g is not differentiable at x .

User profile picture
2022-01-27 00:00
Comments