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Let $f:[0,1]\to\mathbb{R}$ a continuous function and $\int_0^1xf(x)dx=0$. Show that there exists a point $c\in(0,1)$ so that $f(c)=(\int_c^1f(x)dx)^2$.

As a potential solution, I tried assuming that no such point exists, then the function $g(x)=f(x)-(\int_x^1f(t)dt)^2$ would have constant sign for all $x\in(0,1)$. $g(x)>0$ can't be true for all $x\in(0,1)$ since then $\int_0^1xf(x)dx>0$. But I can't figure out how to prove that $g(x)<0$ for all $x$ can't be true.

Thanks

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    $\begingroup$ A possibly helpful restatement of the problem: defining $F(c) = \int_c^1 f(x)\,dx$, we see that we want to find $c\in(0,1)$ such that $-F'(c) = F(c)^2$, given that $F$ is differentiable, $F(1)=0$, and $\int_0^1 F(x)\, dx=0$. (The last equation is obtained via integration by parts.) $\endgroup$ – Greg Martin Mar 4 '14 at 17:19
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As suggested in Greg Martin's comment, let us introduce $F(x):=\int_x^1 f(t)dt$, which satisfies that $F'(x)=-f(x)$ and $F(1)=0$. Moreover, $\int_0^1 F(x)dx=\int_0^1xf(x)dx=0$, so by continuity, there exists $x_0\in (0,1)$ such that $F(x_0)=0$. Let $G(x):=e^{\int_0^xF(t)dt}F(x)$. Firstly, by definition, $$G'(x)=e^{\int_0^xF(t)dt}\big((\int_x^1f(t)dt)^2-f(x)\big).$$ Secondly, $G(x_0)=G(1)=0$, so by Rolle's theorem, there exists $c\in (x_0,1)$, such that $G'(c)=0$. The conclusion follows.

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    $\begingroup$ I must say it was a feat of great ingenuity to use the integrating factor $\exp(\int y\,dx)$ to get the term $(y^{2} + y')$ which was needed here. +1 $\endgroup$ – Paramanand Singh Mar 10 '14 at 3:11
  • $\begingroup$ @ParamanandSingh: Thank you for your appreciation and for correcting the typo. $\endgroup$ – user104254 Mar 10 '14 at 3:43

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