Bound on the integral of a function with multiple zeros

Question

This is a follow-up to this If $f(0)=f(1)=f(2)=0$, $\forall x, \exists c, f(x)=\frac{1}{6}x(x-1)(x-2)f'''(c)$

Let $f:[0,2]\to \mathbb R$ be a $C^3$ function such that $f(0)=f(1)=f(2)=0$

a. Prove that $\forall x\in[0,2], \exists c\in[0,2], f(x)=\frac{1}{6}x(x-1)(x-2)f'''(c)$

b. Prove that $\displaystyle \int_0^2|f|\leq \frac{1}{12} \sup_{t\in[0,2]} |f'''(t)|$

c. Let $M=\sup_{t\in[0,2]}f'''$ and $m=\inf_{t\in[0,2]}f'''$.

Prove that $\displaystyle \left| \int_0^2f\right|\leq \frac{M-m}{24}$

While b. follows directly from a., I haven't been able to prove c.

Since $x\to \frac{1}{6}x(x-1)(x-2)$ oscillates over $[0,2]$, I tried to get different bounds of $\int_0^1f$ and $\int_1^2 f$, but that doesn't yield the desired inequality.

Answer 1

As stated, part (c) is false (as Alex Ravsky shows). Suppose we change the definitions to: \begin{align} &M = \sup_{t\in [0,2]} f'''(t)\\ &m = \inf_{t\in[0,2]} f'''(t) \end{align} Then part (c) can be proven from part (b) by defining $h(x) = f(x) - cx(x-1)(x-2)$ for a suitable value of $c$ such that $h'''(x) \in [-(M-m)/2, (M-m)/2]$ for all $x \in [0,2]$.

Answer 2

It seems the following.

Without additional assumptions (for instance, that some derivative of the function $f$ is bounded at $[0,2]$ by a given constant), the bound (c) does not hold, because for each $\varepsilon>0$ there exists a $C^3$ function $f:[0,2]\to [0,1]$ such that $f(0)=f(1)=f(2)=0$, but $f(x)=1$ for each $x\in [0,2]\setminus ([0,\varepsilon)\cup (1-\varepsilon,1+\varepsilon)\cup (2-\varepsilon,2])$.