Exercise 4.5.6

Let f : [ 0 , 1 ] → R be continuous with f ( 0 ) = f ( 1 )

(a)
Show that there must exist x , y ∈ [ 0 , 1 ] satisfying | x − y | = 1 ∕ 2 and f ( x ) = f ( y ) .
(b)
Show that for each n ∈ N there exist x n , y n ∈ [ 0 , 1 ] with | x n − y n | = 1 ∕ n and f ( x n ) = f ( y n ) .
(c)
If h ∈ ( 0 , 1 ∕ 2 ) is not of the form 1 ∕ n , there does not necessarily exist | x − y | = h satisfying f ( x ) = f ( y ) . Provide an example that illustrates this using h = 2 ∕ 5 .

Answers

(a)
Let g ( x ) = f ( x ) − f ( x + 1 ∕ 2 ) and note that g ( x ) is continuous over [ 0 , 1 ∕ 2 ] . Note also that g ( 0 ) = f ( 0 ) − f ( 1 ∕ 2 ) = − g ( 1 ∕ 2 ) = f ( 1 ∕ 2 ) − f ( 1 ) , and therefore we can apply IVT to g ( x ) over [ 0 , 1 ∕ 2 ] to conclude that there must be a root of g ( x ) , and therefore f ( x ) = f ( x + 1 ∕ 2 ) , for some x ∈ [ 0 , 1 ∕ 2 ] .
(b)
Let g ( x ) = f ( x ) − f ( x + 1 ∕ n ) and note that g ( x ) is continuous over [ 0 , n − 1 n ] . Note also that g ( 0 ) = − ∑ ⁡ i = 1 n − 1 g ( i ∕ n ) , and since g ( 0 ) and ∑ ⁡ i = 1 n − 1 g ( i ∕ n ) have opposite sign there must be some natural number 1 ≤ i ≤ n − 1 where g ( i ∕ n ) is opposite in sign from g ( 0 ) , at which point we can apply IVT in a similar fashion to part (a) and find a root of g ( x ) , completing the proof.
(c)
Consider
f ( x ) = { − 10 ( x − 0 ) + 0 0 < x ≤ 1 ∕ 5 15 ( x − 1 ∕ 5 ) − 2 1 ∕ 5 < x ≤ 2 ∕ 5 − 10 ( x − 2 ∕ 5 ) + 1 2 ∕ 5 < x ≤ 3 ∕ 5 15 ( x − 3 ∕ 5 ) − 1 3 ∕ 5 < x ≤ 4 ∕ 5 − 10 ( x − 4 ∕ 5 ) + 2 4 ∕ 5 < x ≤ 1

You could go through the grunt work of verifying that this meets the requirements, but it’s easier to just plot the function (see graphic) - the function takes a corner every 1/5 along x .

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2022-01-27 00:00
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