An Equality ?

**bigli** · June 5th 2010, 08:52 PM

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 $\int_0^1\!f(x)\,dx\leqslant\frac{1}{4}|f'(c)|$ .

**choovuck** · June 5th 2010, 09:06 PM

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

**rebghb** · June 5th 2010, 09:16 PM

Try the mean value inequality

**bigli** · June 6th 2010, 02:54 AM

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

**bigli** · June 7th 2010, 05:43 AM

Please HINT me.

**choovuck** · June 8th 2010, 04:04 AM

Originally Posted by bigli

Please HINT me.

is there f(0)=f(1/2)=0 or f(0)=f(1)=0?

**choovuck** · June 8th 2010, 04:27 AM

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.

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