What is the maximum value of $C$ such that $\max_I |f'(x)|\geq C\int_0^1|f(x)|dx$? Suppose $f$ is continuously differentiable on $I=[0,1]$, and $f(0)=f(1)=0$. How could you determine the greatest $C>0$ such that
$$
\max_I |f'(x)|\geq C\int_0^1|f(x)|dx?
$$
I know that since $f$ is continuously differentiable, $f'$ is integrable, and $\int_0^1 f'(x)dx=f(1)-f(0)=0$. But I am only able to derive some useless inequalities, like
$$
\max_I|f'(x)|\geq\int_0^1|f'(x)|dx\geq\int_0^1 f'(x)dx=0.
$$
How can I relate it to the integral of $|f|$ to determine $C$?
 A: With $M= \max_I |f'(x)|$ we have $|f(x)| \le Mx$ and $|f(x)| \le M(1-x)$ on $[0, 1]$ from the mean value theorem. It follows that
$$
 |f(x)| \le g(x) = M \cdot \max(x, 1-x)
$$
on $[0, 1]$ and therefore
$$
\int_0^1 |f(x)| \, dx \le \int_0^1 g(x) \, dx = \frac M4 
$$
or
$$
 M \ge 4 \int_0^1 |f(x)| \, dx \, .
$$
To see that $C=4$ is the largest bound consider for a fixed $M > 0$ the functions
$$
f_\epsilon(x) = M \cdot \begin{cases}
x & \text {if } 0 \le x \le \frac 12 - \epsilon \\
\frac 12 - \frac \epsilon 2 - \frac{1}{2 \epsilon}(x-\frac 12)^2\, ,
& \text {if } \frac 12 - \epsilon \le x \le \frac 12 + \epsilon \, ,\\
1-x & \text{if } \frac 12 + \epsilon  \le x \le 1\, .
\end{cases}
$$
(The idea is to modify the function $g$ slightly to make it differentiable at $x=1/2$).
Show that $f_\epsilon$ is differentiable with $\max |f_\epsilon'(x)| = M$ and that $\int_0^1 |f_\epsilon(x)| \, dx$ is arbitrary close to $M/4$ for $\epsilon \to 0$.
(Of course one can assume that $M=1$ to simplify the notation.)
