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I have a Homework question which is:

Let $f(x)$ be an integrable function in $[0,1]$ and there exists a value $M>0$ such that for $x\in[0,1]$ then $f(x)>M$. Let $a=\frac{1}{2}\int_{0}^{1}{f(x)dx}.$

Prove that there is a value $c\in[0,1]$ such that $a=\int_{0}^{c}{f(x)dx}$, and that there is only one number $c$ which satisfies that condition.

I am not really sure how to solve this, I can prove that $\frac{1}{2}\int_{0}^{1}{f(x)dx}\ge M/2$. But I don't think that is the right direction.

Can someone please help me out?

Thanks :)

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  • $\begingroup$ Put $F(x)=\int_0^xf(t)dt$. Then $F$ is continuous and $F(0)=0$, $F(1)=2a>a$ since $a>0$. $\endgroup$ Commented Jan 24, 2012 at 10:46

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As Davide suggested, put $F(x)=\displaystyle\int_0^x f(t)dt$. Then $F(0)=\displaystyle\int_0^0 f(t)dt=0$ and $F(1)=\displaystyle\int_0^1 f(t)dt=2a$ since $a=\displaystyle\frac{1}{2}\int_0^1 f(t)dt$, which shows that $F(0)=0\leq a\leq 2a=F(1)$. Note that $F$ is continuous since $f$ is integrable, by Intermediate value theorem, there exists $c\in [0,1]$ such that $$\int_0^c f(t)dt=F(c)=a.$$

To prove that $c$ is unique, suppose by contradiction that there exists $c'\neq c$ such that $F(c')=a$. By definiton of $F$, we have $$\int_0^{c'} f(t)dt=F(c')=a=F(c)=\int_0^{c} f(t)dt.$$ Without loss of generality, we assume $c<c'$. Hence, we have $\displaystyle\int_c^{c'} f(t)dt=0$. By this is a contradiction, because by assumption $f(x)>M$ we have $\displaystyle\int_c^{c'} f(t)dt\geq M(c'-c)>0$.

Note: In my previous answer, I made a mistake by assuming that $f$ is continuous. Thank you for Didier pointing out the mistake.

Note added: Proof of $F$ being continuous: $$|F(x)-F(y)|=\left|\int_0^{x} f(t)dt-\int_0^{y} f(t)dt\right|\leq\int_{[x,y]}|f(t)|dt \rightarrow 0\mbox{ as }x\rightarrow y$$ since $f$ is integrable.

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  • $\begingroup$ Great answer :) Thank you very much $\endgroup$
    – Jason
    Commented Jan 24, 2012 at 11:54

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