An application of intermediate value theorem to prove the existence of solutions

Asked Jan 15 '16 at 01:08

Active Jan 15 '16 at 02:33

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Let $f$ be a continuous function on $[0,1]$ with $f(0)=f(1)=0$. Prove that, for each $\alpha \in (0,1)$, there exists $x_1, x_2\in (0,1)$ such that $$f(x_1)=f(x_2)$$ with $x_1-x_2=\alpha$ or $x_1-x_2=1-\alpha$.

The main difficulty in this problem is that I don't know how to set a new function in order to apply intermediate value theorem? Any one can help me?

edited Jan 15 '16 at 02:20

asked Jan 15 '16 at 01:08

Richkent

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1

Is there a differentiability assumption concerning $f$? – Oliver Jones Jan 15 '16 at 01:23
No assumption on differentiability of $f$. – Richkent Jan 15 '16 at 01:37
I guess the thing that you were going to need on this problem is intermediate value theorem instead of rolle . – Amir Naseri Jan 15 '16 at 01:44
You need differentiability on $(0,1)$ if you want to apply Rolle's theorem. – porridgemathematics Jan 15 '16 at 02:12
Sorry, I have edited the assumption. As @Amir Naseri, we just use mean value theorem. – Richkent Jan 15 '16 at 02:17
Again, mean value theorem requires some form of differentiability. You mean intermediate value theorem, or somehow taking advantage of uniform continuity? – porridgemathematics Jan 15 '16 at 02:18
Start with the Extreme Value Theorem. – Oliver Jones Jan 15 '16 at 03:16
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Hint: define $g\colon [0,2]\to\Bbb R$ by $g(x)=f(x)$ for $x\in[0,1]$, while $g(x) = f(x-1)$ for $x\in[1,2]$. (In other words, extend $f$ periodically to $[0,2]$.) Then consider $g(x+\alpha)-g(x)$. – Greg Martin Jan 15 '16 at 03:17

An application of intermediate value theorem to prove the existence of solutions

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