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Let ,$f : [0,a] \to \mathbb{R}$ be a funtion satisfy $f(0)=f(a)$.Prove that there exist $x_1,x_2\in [0,a]$ with $x_2-x_1=1$ such that $f(x_1)=f(x_2)$.

I think to prove that there exist a $c\in [0,a] $ so that$ f'(c)=0$.

But nothing about continuity or differentiability of $f$ is given .I stuk here

QC_QAOA
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Trishan Mondal
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  • Well by the logic that nothing is given about continuity, not much is given about the function at all. I recommend you try first to solve it making some assumptions and then make it more general. – Certainly not a dog Dec 21 '19 at 04:29
  • The fact that we require $x_2 - x_1 = 1$ makes it unlikely, what if $a<1$? – fGDu94 Dec 21 '19 at 04:30
  • If $a=\frac{1}{2}$ then you won't even be able to find $x_1,x_2 \in [0,a]$ such that $x_2-x_1=1$. – Lee Mosher Dec 21 '19 at 04:30
  • It is not true if $f$ is not continuous. E.g. $f(x)=x$ if $x<a$and $f(a)=0$. – Alain Remillard Dec 21 '19 at 04:35
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    Please fix the question. You need $a \ge 1$. Also, if you are to conclude that $f'(c)= 0$ for some $c$ then $f$ must be assumed differentiable. – copper.hat Dec 21 '19 at 04:40
  • If $a\in\Bbb N$ and $f$ is continuous, then there are the required $x_1$ and $x_2$. This is a nice application of the Intermediate Value Theorem. – Angina Seng Dec 21 '19 at 04:45
  • Look up the universal chord theorem. You need $a$ to be an integer for the statement to be true. One question is here – Ross Millikan Dec 21 '19 at 14:56

2 Answers2

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What if $a < 1$. You can't have $x_2 - x_1 = 1$.

I think that your function has to be continuous, at least. Indeed, the function $f:[0,a] \to \mathbb{R}$ s.t $f(x) = x$ on $(0,a)$ and $f(0) = f(a) = -1$ follows your hypotheses, but not the conclusion. ($a \neq 1$)

Guillaume Garnier
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When $a$ is an integer, Ross's argument works.

If $a>1$ isn't an integer, then there is a continuous $f:[0,a]\to\Bbb R$ such that $f(0)=f(a)$ and $f(x+1)-f(x)\ne0$ for all $x\in[0,a-1]$. If we take $$f(x)=\lambda x+\cos(2\pi x)$$ then $f(x+1)-f(x)=\lambda$ for all $x$. As $a$ is not an integer, if we take $$\lambda=\frac{1-\cos(2\pi a)}a$$ then $\lambda>0$, and $f(0)=f(a)=1$.

Angina Seng
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