How prove this integral inequality $\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$?

Solution 1:

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 $f(x) = 1- \frac{260}{283}\phi(x)$.

Solution 2:

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 $\mathbb{R}^2$

$\hspace{5cm}$

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 $$ \int_0^1f(t)^2\,\mathrm{d}t\ge\frac{849}{4}\left(\int_0^1f(t)\,\mathrm{d}t\right)^2\tag{13} $$