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Let $f$ be a continuous and integrable function on $[a,b]$ such that

$$\int_a^b f(x)\,\mathrm{d}x = 2$$

and for every $t_1,t_2$ such that $\displaystyle t_2 -t_1 = \frac{b-a}2$

$$\int_{t1}^{t2} f(x)\,\mathrm{d}x =1$$

Prove that $f$ is periodic and then determine its period.

The previous subquestion asked to prove that if $f$ is continuous then there exists an $F$ such that $F'(x)=f(x)$ for every $x$ and it's supposed to help.

Frankly I have no idea where to even start from.

I want to prove that there exists a $T$ such that $f (x+T) = f (x)$ for every $x$ that's all I can say, any insights please?

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Define $G(x)=\int_{x}^{x+\frac{b-a}{2}} f(t) dt$ for $a\le x\le a+(b-a)/2$. Then $G(x)=1$ by assumption. Hence,

$$0=G^\prime(x)=\frac{d}{dx} \int_x^{x+\frac{b-a}{2}} f(t) dt=f\left(x+\frac{b-a}{2}\right)-f(x)$$

so $f$ is periodic with period $\frac{b-a}{2}$.

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  • $\begingroup$ Looks correct!! $\endgroup$ – user3001408 Sep 21 '14 at 11:00
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Prove that f is periodic and then determine its period.

$f\left(x+\frac{b-a}{2}\right)-f(x)=0 $ $ $ for $ a\le x\le a+(b-a)/2$ $ $ proves that $f$ is periodic and that $T=(b-a)/2$ is a period.
What about its least positive period ?
See this page: What determines if a function has a least positive period?
Two cases:
1. $f$ is a constant function, then there is no least positive period. Note that in that case:
$f(x)=\frac{2}{(b-a)}$

2.$f$ is not a constant function. Then there exists a smallest period $T_m>0$ and a positive integer $n$ such that:
$f(x+T_m)=f(x)$ for $ a\le x\le a+(b-a)/2$ and $T_m=\frac{b-a}{2n}$

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