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If $f$ be a real valued continuous function defined on $[0,2]$ such that $f(0)=f(2),$ then prove that there exist a $ x \in [0,1]$ such that $f(x)=f(x+1).$

I tried in the following way,

Since $f$ is continuous and $f(0)=f(2)$, any line parallel to $x-$axis meets $f$ at least two points. Is this the correct way?

How to proceed further?

Thanks in advance...

joelr
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2 Answers2

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This is an example of the intermediate value theorem. Let $$g(x)=f(x+1)-f(x)$$

Then $$g(0)=f(1)-f(0)$$ and $$g(1)=f(2)-f(1)=-g(0)$$

Now if $g(0)=0$ we are done with $x=0$ otherwise $g(0)\neq 0$ and by intermediate value there is $x\in (0,1)$ such that $g(x)=0$. And we are done.

Rene Schipperus
  • 39,526
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Let $h$ the function defined by

$$h(t)=f(t+1)-f(t)$$ then $h$ is continuous and

$$h(1)h(0)=(f(2)-f(1))(f(1)-f(0))=-(f(1)-f(0))^2\le0$$ so by the intermediate value theorem there's $x\in[0,1]$ such that $h(x)=0$.