seupeljewj

Answered

2022-06-24

Suppose $f$ is continuous on [0,2] and that $f(0)=f(2)$. Then $\mathrm{\exists}$$x,$$y\in [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

$g(0)g(1)=-(f(0)-f(1){)}^{2}<0$

since $f(0)=f(2)$. 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?

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

$g(0)g(1)=-(f(0)-f(1){)}^{2}<0$

since $f(0)=f(2)$. 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?

Answer & Explanation

alisonhleel3

Expert

2022-06-25Added 23 answers

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.

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

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