12
$\begingroup$

let $f:[0,1]\longrightarrow R $ be a continuous function, if $$\int_{0}^{1}x^2f(x)dx=-2\int_{\frac{1}{2}}^{1}F(x)dx$$ where $F(x)=\displaystyle\int_{0}^{x}f(t)dt,x\in [0,1]$,then prove that $$\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$$

I think this inequality is very interesting, becasue not long ago,I have solve this a little same problem:Funny integral inequality

I believe this two problem have same methods.Thank you everyone

$\endgroup$
9
$\begingroup$

We start with, $\displaystyle \int_{\frac{1}{2}}^1 F(x)\,dx = \int_{\frac12}^1\int_0^x f(t)\,dt \,dx = \frac{1}{2}\int_0^{\frac12} f(x)\,dx + \int_{\frac12}^1 (1-x)f(x)\,dx$

Thus we have, $$\displaystyle \int_0^1 x^2f(x)\,dx = -\int_0^{\frac12}f(x)\,dx - 2\int_{\frac12}^1 (1-x)f(x)\,dx$$

and, $$\displaystyle \int_0^1 f(x)\,dx = -\int_{\frac12}^1 (x-1)^2f(x)\,dx - \int_0^{\frac12} x^2f(x)\,dx$$

Hence, $\displaystyle \int_{\frac12}^1 \{1+(x-1)^2\}f(x)\,dx + \int_0^{\frac12} \{1+x^2\}f(x)\,dx = 0$

Say, $\phi(x) = 1+x^2 \textrm{ for } x \in [0,\frac12) \textrm{ and } 1+(x-1)^2 \textrm{ for } x \in [\frac12,1]$

Then, $\displaystyle \int_0^1 \phi(x)f(x)\,dx = 0$

Now by Cauchy-Schwarz Inequality :

$\displaystyle \left(\int_0^1 f(x)\,dx \right)^2 = \left(\int_0^1 f(x) + m\phi(x)f(x)\,dx \right)^2 \le \left(\int_0^1 f^2(x)\,dx\right)\left(\int_0^1 (1+m\phi(x))^2\,dx\right)$

Since, $\displaystyle \int_0^1 (1+m\phi(x))^2\,dx = 1 + \frac{13}{6}m +\frac{283}{240}m^2 \ge \frac{4}{849}$

We have, $\displaystyle \left(\int_0^1 f(x)\,dx \right)^2 \le \frac{4}{849}\int_0^1 f^2(x)\,dx$

The constant $\frac{4}{849}$ is the best possible since equality is attained for $f(x) = 1- \frac{260}{283}\phi(x)$.

$\endgroup$
  • 1
    $\begingroup$ Again we work on the same questions :-) Our methods seem pretty disjoint on this one. $\endgroup$ – robjohn Jun 27 '14 at 19:26
4
$\begingroup$

We have $$ \begin{align} \int_{1/2}^1F(x)\,\mathrm{d}x &=\int_{1/2}^1\int_0^xf(t)\,\mathrm{d}t\,\mathrm{d}x\\ &=\color{#00A000}{\int_0^{1/2}\int_{1/2}^1f(t)\,\mathrm{d}x\,\mathrm{d}t}+\color{#C00000}{\int_{1/2}^1\int_t^1f(t)\,\mathrm{d}x\,\mathrm{d}t}\\ &=\int_0^{1/2}\frac12f(t)\,\mathrm{d}t+\int_{1/2}^1(1-t)f(t)\,\mathrm{d}t\\ &=-\frac12\int_0^1t^2f(t)\,\mathrm{d}t\tag{1} \end{align} $$ since we are integrating $f(t)$ over the following region in $\mathbb{R}^2$

$\hspace{5cm}$enter image description here

Rearranging $(1)$, we have $$ \int_0^{1/2}\frac{t^2+1}{2}f(t)\,\mathrm{d}t+\int_{1/2}^1\frac{(1-t)^2+1}{2}f(t)\,\mathrm{d}t=0\tag{2} $$ Aside from a scale factor, we want to minimize $$ \int_0^1f(t)^2\,\mathrm{d}t\tag{3} $$ given $$ \int_0^1f(t)\,\mathrm{d}t=1\tag{4} $$ That means for every variation $\delta f$ so that $(2)$ and $(4)$ hold, that is, $$ \int_0^{1/2}\frac{t^2+1}{2}\delta f(t)\,\mathrm{d}t+\int_{1/2}^1\frac{(1-t)^2+1}{2}\delta f(t)\,\mathrm{d}t=0\tag{5} $$ and $$ \int_0^1\delta f(t)\,\mathrm{d}t=0\tag{6} $$ we have $(3)$ is stationary, that is, $$ \int_0^1f(t)\delta f(t)\,\mathrm{d}t=0\tag{7} $$ Orthogonality says that there must be constants $\mu$ and $\lambda$ so that $$ f(t)=\mu\cdot1+\lambda\left(\frac{t^2+1}{2}[0\le t\le1/2]+\frac{(1-t)^2+1}{2}[1/2\le t\le1]\right)\tag{8} $$ Plugging $(8)$ into $(2)$ says $$ \frac{13}{24}\mu+\frac{283}{960}\lambda=0\tag{9} $$ Plugging $(8)$ into $(4)$ says $$ \mu+\frac{13}{24}\lambda=1\tag{10} $$ Solving $(9)$ and $(10)$ gives $$ \mu=\frac{849}{4}\qquad\text{and}\qquad\lambda=-390\tag{11} $$ Plugging $(8)$ and $(11)$ into $(3)$ yields $$ \int_0^1f(t)^2\,\mathrm{d}t=\frac{849}{4}\tag{12} $$ Squaring $(4)$, to make things homogeneous, allows us to say $$ \int_0^1f(t)^2\,\mathrm{d}t\ge\frac{849}{4}\left(\int_0^1f(t)\,\mathrm{d}t\right)^2\tag{13} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.