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Let $f:[0,1]\to \mathbb{R}$ be a positive, continuous and strictly increasing function such that $$\int_0^1 f(x)dx=1$$ Let $\xi\in (0,1)$ be such that $$\int_0^{\xi}f(x)dx=\int_{\xi}^1f(x)dx=\frac{1}{2}$$ Prove that $$\int_0^1 xf(x)dx\leq \xi$$

This problem comes from physics and has to do with the center of gravity of a shape. But I'd like to see a direct, computational proof. Here is my attempt. To begin with, note that $$ \begin{aligned} \int_0^1xf(x)dx &=\int_0^1[\xi+(x-\xi)]f(x)dx \\ &=\xi\int_0^1f(x)dx+\int_0^1(x-\xi)f(x)dx \\ &=\xi+\int_{-\xi}^{1-\xi}yf(y+\xi)dy \end{aligned} $$ For the second term, we know that $$\int_{-\xi}^{1-\xi}yf(y+\xi)dy=\int_{-\xi}^0yf(y+\xi)dy+\int_0^{1-\xi}yf(y+\xi)dy:=I_1+I_2$$ In $I_1$, let $z=-y$. Since $f$ is strictly increasing, the defining properties of $\xi$ imply that $\xi>\frac{1}{2}$. So $1-\xi<\xi$, and hence $$ \begin{aligned} I_1 &=\int_{\xi}^0(-z)(-1)f(-z+\xi)dz \\ &=-\int_0^{\xi}zf(-z+\xi)dz \\ &=-\left[\int_0^{1-\xi}zf(-z+\xi)dz+\int_{1-\xi}^{\xi}zf(-z+\xi)dz\right] \end{aligned} $$ We want to show that $I_1+I_2\leq 0$. Equivalently, we have to prove that $$\int_0^{1-\xi}z\left[f(z+\xi)-f(-z+\xi)\right]dz\leq \int_{1-\xi}^{\xi}zf(-z+\xi)dz$$ If we perform a change of variables on the RHS by setting $$w=\frac{1-\xi}{2\xi-1}z-\frac{(1-\xi)^2}{2\xi-1}$$ then the RHS becomes $$\int_0^{1-\xi}\left[(2\xi-1)+\left(\frac{2\xi-1}{1-\xi}\right)^2w\right]f\left((2\xi-1)-\frac{2\xi-1}{1-\xi}w\right)dw$$ However, this is getting too complicated and is clearly not the right way to go. Can anyone share some ideas on how to solve to problem? Thank you!

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    $\begingroup$ The upper bound you need to prove seems to be very sharp (this is because, if $f \equiv 1$ - which is not the case, since $f$ is strictly increasing -, then $\xi = 1/2$ and $\int_0^1 xf(x)dx = 1/2$; however, you can take a sequence of linear functions that converges uniformly to the constant function $1$ to conclude that the upper bound is very sharp). $\endgroup$
    – C_M
    Jun 26, 2020 at 22:26
  • $\begingroup$ I will also post one idea we may use to get some insight: considering $F(t) = \int_0^t (x-\xi)f(x)dx$ and computing its derivative, we see that $\int_0^\xi xf(x)dx \leq \xi/2$, and taking $G(t) = \int_t^1 (x-\xi)f(x)dx$ and computing its derivative, we get that $\int_\xi^1 xf(x)dx \geq \xi/2$, so it seems that these rather "easy to compute" upper/lower bounds for $\int_0^\xi xf(x)dx$ and $\int_\xi^1 xf(x)dx$ are not very sharp, so we need something better. $\endgroup$
    – C_M
    Jun 26, 2020 at 22:28

1 Answer 1

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Define $F(x)=\int_0^x f(x)$ and note that is is convex because $f$ is increasing.

Let $\mu=\int_0^1 xf(x)dx$ denote the average. By Jensen's inequality,

$F(\mu)\leq \int_0^1 F(x)f(x)dx$

On the other hand, $\int_0^1 F(x)f(x)dx=\int_0^1 (F(x)^2/2)'dx=1/2$ so we are done.

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