Integral condition implies derivative greater than $4$ Let $f:[0,1]\to \mathbb{R}$ be a differentiable function such that $\displaystyle \int \limits _0^1f(t)\,dt=1$, $f(0)=0$, $f(1)=0$. Prove that there exists an $x_0\in (0,1)$ such $|f'(x_0)|\geq 4$.
I'm trying to use mean value theorem on this but its leading to a result I already know that the there would exist a point where tangent is $0$.
 A: The fundamental theorem of calculus is usually a good thing to try: if to the contrary $|f'(t)| < 4$ for all $t$ you would have $$f(x) = f(x) - f(0) = \int_0^x f'(t) \, dt < 4x$$ for all $x \in [0,1]$.  In particular,
$$\int_0^{1/2} f(x) \, dx < \int_0^{1/2} 4x \, dx = \frac 12.$$
This is only a partial solution, but it completely neglected the hypothesis that $f(1) = 0$. What will that imply?
A: the function
$$f(x) = \begin{cases} 3x \hspace{1cm} &  x \leq \frac{11}{15} \\
3x - \frac{6075}{32}(x-\frac{11}{15})^2 + \frac{70875}{128}(x-\frac{11}{15})^3 \hspace{0.5cm} &x \geq \frac{11}{15} \end{cases}$$
is such that
(i) $f \in C^1(\mathbb{R})$
(ii) $ f'(x) < 4 \;\; \forall x \in [0,1]$
(iii) $f(0) = 0$, $f(1) = 0\;\;\;, \;\;\;\int_0^1{f(x)dx} = 1$
Therefore what you're trying to prove is false.
Maybe if you substitute $4$ with a smaller number the statement becomes true.
To prove that (i),(ii) and (iii) holds I recommend to use a graphic calculator like wolfram alpha or matlab and to not prove it analytically, although you can do it if you wish but it is very tedious
A: As has been shown in the other answer this statement seems to be false. However if you have $\left|f'\left(x_0\right)\right| \ge 4$ instead of $f'(x_0) \ge 4$ and $x_0 \in[0,1]$. To prove that let $$g(x) = \int_0^x\left(f(t) + f(1-t)\right)\mathrm d t$$
\begin{align}
\frac 14\sup\limits_{t\in [0,1]} \left|f'(t)\right| &\ge \frac 14\times \frac12\sup\limits_{t\in \left[0,\frac 12\right]} \left|f'(t) - f'(1-t)\right|\\
&= \frac 12 \sup_{t\in \left[0,\frac12\right]} \left|g''(t)\right|\left(\frac 12\right)^2\\
 & \ge \left|g\left(\frac12\right)-g(0)-\frac12 g'(0)\right| = 1
\end{align}
and you have what you are looking for using the continuity of $f'$.
