1. ## An inequality

Let $\displaystyle f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $\displaystyle f(0)=f(\frac{1}{2})=0$ . If $\displaystyle f$ be differentiable over $\displaystyle (0,1)$, prove that there exist $\displaystyle c\in (0,1)$ such that $\displaystyle \int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|$.

2. What ideas have you already tried?

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

4. Originally Posted by bigli
Let $\displaystyle f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $\displaystyle f(0)=f(\frac{1}{2})=0$ . If $\displaystyle f$ be differentiable over $\displaystyle (0,1)$, prove that there exist $\displaystyle c\in (0,1)$ such that $\displaystyle \int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|$.
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.

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

Hope this helps