Suppose f : R → R is continuous and periodic with period 2a for some a &gt; 0; that is, f ( x ) = f ( x + 2 a ) for all x ∈ R. Show there is some c ∈ [ 0 , a ] such that f ( c ) = f ( c + a ).The only way I see to do this question is to apply the intermediate value theorem, but I just don't know how to apply it to this question. I know that f ( x ) = f ( x + 2 a ). Therefore, if i can show that f ( c + a ) = f ( c + 2 a ) or f ( c + a ) − f ( c + 2 a ) = 0, I'd be done. But I just don't know where to go from there. I tried substituting x = a, getting f ( 2 a ) − f ( 3 a ), but I can't show that's less than, equal to, or greater than 0. Any ideas?

Question

Suppose   f  :      R    →      R   is continuous and periodic with period 2a for some   a  &amp;gt;  0; that is,   f  (  x  )  =  f  (  x  +  2  a  ) for all   x  ∈  R. Show there is some   c  ∈  [  0  ,  a  ] such that   f  (  c  )  =  f  (  c  +  a  ).The only way I see to do this question is to apply the intermediate value theorem, but I just don&#039;t know how to apply it to this question. I know that   f  (  x  )  =  f  (  x  +  2  a  ). Therefore, if i can show that   f  (  c  +  a  )  =  f  (  c  +  2  a  ) or   f  (  c  +  a  )  −  f  (  c  +  2  a  )  =  0, I&#039;d be done. But I just don&#039;t know where to go from there. I tried substituting   x  =  a, getting   f  (  2  a  )  −  f  (  3  a  ), but I can&#039;t show that&#039;s less than, equal to, or greater than 0. Any ideas?

Kalmukujobvg · Accepted Answer

Hint: Define a function   g  (  x  )  =  f  (  x  +  a  )  −  f  (  x  ). If   g  (  0  )  =  0, then you&#039;re done; if not, note that   g  (  a  )  =  −  g  (  0  ).