# 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 $$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$

## closed as off-topic by Carl Mummert, Namaste, Mohammad Riazi-Kermani, user296602, Cyclohexanol.Feb 19 '18 at 4:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Namaste, Mohammad Riazi-Kermani, Community, Cyclohexanol.
If this question can be reworded to fit the rules in the help center, please edit the question.

## 3 Answers

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]\\ &=12\int_{0}^{1}f'^2(x)dx \end{align}

• The final factor $12$ should be $\tfrac{1}{12}$. – WimC Mar 31 '13 at 17:49
• you used $(a+b)^2 \le 2a^2+2b^2$ on 3rd line of the derivation, which gives the final constant $1/12$, using the Cauchy-Schwarz inequality on the 3rd line itself with $$g(x) = \begin{cases} x & \textrm{ for } x \in[0,1/2] \\ 1-x & \textrm{ for } x \in[1/2,1] \end{cases}$$ (note g is continuous in [0,1]) gives the constant $1/24$ – r9m Nov 15 '14 at 1:57

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.

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

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