# Thread: An Equality ?

1. ## An Inequality ?

Let $f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $f(0)=f(\frac{1}{2})=0$ . If $f$ be differentiable over $(0,1)$, prove that there exist $c\in (0,1)$ such that $\int_0^1\!f(x)\,dx\leqslant\frac{1}{4}|f'(c)|$.

2. Originally Posted by bigli
Let $f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $f(0)=f(\frac{1}{2})$ . If $f$ be differentiable over $(0,1)$ prove that there exist a $c\in (0,1)$ such that $\Big|\int_0^1\!f(x)\,dx\Big|\leqslant\frac{1}{4}|f '(c)|$.
I think you made some mistake writing the question. The way you wrote, it doesn't make sense. Take f to be a constant 1 function. Then f'(c)=0 for any c, but the integral is positive

3. Try the mean value inequality

4. Yes, rebghb was right and I fixed it.NOW, How does I solve it?

5. Please HINT me.

6. Originally Posted by bigli
Please HINT me.
is there f(0)=f(1/2)=0 or f(0)=f(1)=0?

7. ah, ok, i think f(0)=f(1/2)=0 is the right condition.

i am assuming that f is actually a positive on [0,1] function, since if not then we can essentially take |f| instead of f.

denote $M=\int_0^1 f(x)dx$. assume for any $c\in (0,1)$, $|f'(c)|\le 4M$

first consider interval $x\in[0,1/4]$. by the mean value theorem there: $f(x)=f'(\theta_x) (x-0)$ (since f(0)=0), where $\theta_x\in[0,x]$. therefore $f(x)=|f(x)|\le 4Mx$ for $x\in[0,1/4]$. therefore $\int_0^{1/4}f(x)dx\le 4M\int_0^{1/4}xdx=M/8$.

same arguments for $x\in[1/4,1/2]$ (using mean value theorem at the point 1/2, where f is equal to zero) show $\int_{1/4}^{1/2}f(x)dx\le M/8$.

finally consider $x\in[1/2,1]$. again, by the mean value theorem: $f(x)=f'(\theta_x) (x-1/2)$ (since f(1/2)=0), where $\theta_x\in[1/2,x]$. therefore $f(x)=|f(x)|\le 4M(x-1/2)$ for $x\in[1/2,1]$. therefore $\int_{1/2}^{1}f(x)dx\le 4M\int_{1/2}^{1}(x-1/2)dx=M/2$.

combining, we get $M=\int_{0}^{1}f(x)dx\le M/8+M/8+M/2=3M/4$, contradiction.