An inequality

**bigli** · June 7th 2010, 06:25 AM

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 $\int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|$ .

**Ackbeet** · June 7th 2010, 06:55 AM

What ideas have you already tried?

**Unknown008** · June 7th 2010, 06:56 AM

And why did you double post?

http://www.mathhelpforum.com/math-he...-equality.html

**bigli** · June 7th 2010, 08:44 AM

Because, anyone doesn't answered me and I guess the question is probably a question in analysis.

**Drexel28** · June 7th 2010, 08:44 AM

Originally Posted by bigli

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 $\int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|$ .

You're clearly going to need to use the MVT and the fact that an integral is less than the measure of the interval times the supremum of it on that interval.

**mohammadfawaz** · June 7th 2010, 09:31 AM

First, let $F(x)=\int_0^x f(t)dt$ . Apply MVT to this function on $[0,1]$ : $F(1)-F(0)=F'(c_1)$ where c between 0 and 1.
Clearly, $F(0)=0$ and $F(1)=\int_0^1 f(t)dt$ . Also, by the FTC, $F'(c_1)=f(c_1)$ . Hence, we have: $\int_0^1 f(t)dt = f(c_1)$ .
Let's now apply the MVT for the function $f$ on the interval $[c_1, \frac{1}{2}]$ (or the other way around). We get: $|f(c_1)-f(0.5)|=|f'(c_2)(c_1-0.5)|\leq \frac{1}{2}|f'(c_2)|$ .
Therefore, $|f(c_1)|\leq \frac{1}{2}|f'(c_2)|$ i.e. $|\int_0^1 f(t)dt|\leq \frac{1}{2}|f'(c_2)|$ .
That what I was able to reach so far...Inspire from this. The continuation must be in a similar manner.

Hope this helps

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