Let f be continuous on [ 0 , 1 ] and f ( 0 ) = f ( 1 ). Also, n ∈ N . Prove that there exists some x s.t. f ( x ) = f ( x + 1 n ).I think I need to assume towards a contradiction that if g ( x ) = f ( x ) − f ( x + 1 n ), then g ( x ) ≠ 0 ∀ x.So, g ( 0 ) = f ( 0 ) − f ( 1 n ), g ( 1 ) = f ( 1 ) − f ( 1 + 1 n ). Since f ( 0 ) = f ( 1 ), g ( 0 ) − g ( 1 ) = f ( 1 + 1 n ) − f ( 1 n ). For context, I am only up to Ch. 8 in Spivak, so I cannot use any derivative-based theorems which is why I am stuck here.

Question

Let   f be continuous on   [  0  ,  1  ] and   f  (  0  )  =  f  (  1  ). Also,   n  ∈      N  . Prove that there exists some   x s.t.   f  (  x  )  =  f  (  x  +      1    n    ).I think I need to assume towards a contradiction that if   g  (  x  )  =  f  (  x  )  −  f  (  x  +      1    n    ), then   g  (  x  )  ≠  0   ∀  x.So,   g  (  0  )  =  f  (  0  )  −  f  (      1    n    ),   g  (  1  )  =  f  (  1  )  −  f  (  1  +      1    n    ). Since   f  (  0  )  =  f  (  1  ),  g  (  0  )  −  g  (  1  )  =  f  (  1  +      1    n    )  −  f  (      1    n    ). For context, I am only up to Ch. 8 in Spivak, so I cannot use any derivative-based theorems which is why I am stuck here.

Emma Singleton · Accepted Answer

Assume the contrary that   g  (  x  ) is not 0 on   [  0  ,  1  −      1    n    ], which means either   g  (  x  )  &amp;gt;  0 or   g  (  x  )  &amp;lt;  0 on   [  0  ,  1  −      1    n    ] (since, g is continuous). If,   g  (  x  )  &amp;gt;  0    ⟹    f  (  0  )  &amp;gt;  f  (      1    n    )  &amp;gt;  f  (      2    n    )  &amp;gt;  ⋯  &amp;gt;  f  (  1  −      1    n    )  &amp;gt;  f  (  1  ), contradiction !! Similarly, for   g  &amp;lt;  0, we get a contradiction. Therefore,   g  (  x  ) has a zero in   [  0  ,  1  −      1    n    ].

Jairo Hodges · Accepted Answer

Hint: What is  g  (  0  )  +  g      (          1      n        )    +  .  .  .  g      (                  n        −        1            n        )    ?How does that help prove that   g  (  x  )  =  0 for some   x?

Let f be continuous on [0,1] and f(0)=f(1). Also, n in N. Prove that there exists some x s.t. f(x)=f(x+1/n).

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