Suppose f is continuous on [0,2] and that f ( 0 ) = f ( 2 ). Then ∃ x , y ∈ [ 0 , 2 ] such that | y − x | = 1 and that f ( x ) = f ( y ).Let g ( x ) = f ( x + 1 ) − f ( x ) on [0,1]. Then g is continuous on [0,1], and hence g enjoys the intermediate value property! Now notice g ( 0 ) = f ( 1 ) − f ( 0 ) g ( 1 ) = f ( 2 ) − f ( 1 )Therefore g ( 0 ) g ( 1 ) = − ( f ( 0 ) − f ( 1 ) ) 2 &lt; 0since f ( 0 ) = f ( 2 ). Therefore, there exists a point x in [0,1] such that g ( x ) = 0 by the intermediate value theorem. Now, if we pick y = x + 1, i think the problem is solved.I would like to ask you guys for feedback. Is this solution correct? Is there a better way to solve this problem?

Question

Suppose   f is continuous on [0,2] and that   f  (  0  )  =  f  (  2  ). Then   ∃  x  ,  y  ∈  [  0  ,  2  ] such that       |    y  −  x      |    =  1 and that   f  (  x  )  =  f  (  y  ).Let   g  (  x  )  =  f  (  x  +  1  )  −  f  (  x  ) on [0,1]. Then   g is continuous on [0,1], and hence   g enjoys the intermediate value property! Now notice  g  (  0  )  =  f  (  1  )  −  f  (  0  )  g  (  1  )  =  f  (  2  )  −  f  (  1  )Therefore  g  (  0  )  g  (  1  )  =  −  (  f  (  0  )  −  f  (  1  )      )    2    &amp;lt;  0since   f  (  0  )  =  f  (  2  ). Therefore, there exists a point   x in [0,1] such that   g  (  x  )  =  0 by the intermediate value theorem. Now, if we pick   y  =  x  +  1, i think the problem is solved.I would like to ask you guys for feedback. Is this solution correct? Is there a better way to solve this problem?

alisonhleel3 · Accepted Answer

It is enough to write   g  (  1  )  =  −  g  (  0  )..And, as Hagen commented, you only forgot to mention the case when already   f  (  0  )  =  f  (  1  ), but then it is readily done.