Funny integral inequality [closed]

Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that: $$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$

and how to find the smallest constant $C$ which satisfies $$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$

Solution 1:

let $$\displaystyle\int_{0}^{\frac{1}{2}}f(x)=0\Longrightarrow \int_{0}^{\frac{1}{2}}xf'(x)dx=\dfrac{1}{2}f(\dfrac{1}{2})$$

so \begin{align} &(\int_{0}^{1}f(x)dx)^2=\left[\int_{\frac{1}{2}}^{1}(f(x)-f(\dfrac{1}{2}))dx+\dfrac{1}{2}f(\dfrac{1}{2})\right]^2=\left[\int_{\frac{1}{2}}^{1}\int_{\frac{1}{2}}^{x}f'(t)dtdx+\int_{0}^{\frac{1}{2}}xf'(x)dx\right]^2\\ &=\left[\int_{\frac{1}{2}}^{1}(1-t)f'(t)dt+\int_{0}^{\frac{1}{2}}xf'(x)dx\right]^2\\ &\le2\left[\int_{\frac{1}{2}}^{1}(1-t)f'(t)dt\right]^2+2\left[\int_{0}^{\frac{1}{2}}xf'(x)dx\right]^2\\ &\le 2\left[\int_{\frac{1}{2}}^{1}(1-t)^2dt\int_{\frac{1}{2}}^{1}f'^2(t)dt+\int_{0}^{\frac{1}{2}}x^2dx\int_{0}^{\frac{1}{2}}f'^2(t)dt\right]\\ &=\frac{1}{12}\int_{0}^{1}f'^2(x)dx \end{align}

Solution 2:

write $g=f'$ and observe that $f(0)=-\int_0^{1/2}(1-2t)g(t)dt$ from $\int_0^{1/2}f(x)dx=0.$ Therefore $$(\int_0^{1}f(x)dx)^2=(\int_0^1g(t)\min(t,1-t)dt)^2\leq \int_0^{1}g(t)^2dt\times \int_0^1(\min(t,1-t)^2dt$$ from Schwarz.

Solution 3:

solutin 2:

by Schwarz,we have $$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\int_{0}^{\frac{1}{2}}x^2dx\ge\left(\int_{0}^{\frac{1}{2}}xf'(x)dx\right)^2=\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$ so $$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$ the same methods,we have $$\int_{\frac{1}{2}}^{1}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$

and use $2(a^2+b^2)\ge (a+b)^2$

then we have

$$\int_{0}^{1}[f'(x)]^2dx\ge 12\left(\int_{0}^{1}f(x)dx-2\int_{0}^{\frac{1}{2}}f(x)dx\right)^2$$