Let $f:[0,1]→\mathbb{R}$ with $f'(x)$ continuous. It is known that $∫_0^1 f(x) dx=0$.
Prove that $∀α∈[0,1]$, $$|\int_{0}^{\alpha} f(x) dx |≤ \frac{1}{8} sup_{(0≤x≤1)}|f'(x) |$$
My answer so far :
For $\alpha = 1$ then of course the inequality applies, since it is known that $∫_0^1 f(x) dx=0$.
$$$$ Now for $\alpha \ne 1$,
Since $f:[0,1]→\mathbb{R}$ and $f'(x)$ continuous, then $f'(x)$ has extreme value for some $c \in [0,1]$ such that $|f'(c)| = sup_{(0≤x≤1)}|f'(x) |$. Let $M = |f'(c)|$.
Now since $∫_0^1 f(x) dx=0$, then there exist $k \in [0,1]$ such that $f(k)=0$.
Notice that for $x\ge k$, $f(x) = \int_{k}^{x} f'(t) dt \le (x-k) | M |$ so that $| \int_{0}^{\alpha} f(x) dx | \le(\frac{\alpha^2}{2} - \alpha k)|M| $,
Then i dont know what else to do...
Another way :
I have an idea to define a function $g(t) = \int_{0}^{t} f(x) dx$ then $g(1) = g(0) = 0$.
Any suggestion or hint for this problem??
Continuity of $f'$ is useless. Following your notation, let $$M:=\sup_{x\in[0,1]}|f'(x)|.$$ If $M=\infty$, there is nothing to prove, so let us suppose $M<\infty$. Following your thoughts, let $$g(t)=\int_0^t f(x)dx=-\int_t^1f(x)dx,\quad t\in[0,1].$$ By continuity, there exists $t_0\in [0,1]$, such that
$$|g(t_0)|=\max_{t\in[0,1]}|g(t)|,$$ and it suffices to show that $$|g(t_0)|\le \frac{M}{8}.$$ If $g(t_0)=0$, there is nothing to prove, so let us suppose $g(t_0)\ne 0$. Then $t_0\in(0,1)$ and $g$ attains either its maximum value or its minimum value at $t_0$, so $$f(t_0)=g'(t_0)=0. $$ If $t_0\le \frac{1}{2}$, $$|f(x)|\le M(t_0-x),\ \forall x\le t_0\Longrightarrow |g(t_0)|\le M\int_0^{t_0}(t_0-x)dx=\frac{M}{2}t_0^2\le \frac{M}{8}.$$ If $t_0\ge \frac{1}{2}$, $$|f(x)|\le M(x-t_0),\ \forall x\ge t_0\Longrightarrow |g(t_0)|\le M\int_{t_0}^1(x-t_0)dx=\frac{M}{2}(1-t_0)^2\le \frac{M}{8}.$$