If $f(1)=f(0)$, then show that $|f'(x)|\leq 1$ for all $x\in [0,1]$ Let the function $f$ be twice differentiable on $[0,1]$ such that $|f''(x)|\leq 1$ for all $x\in [0,1]$. If $f(1)=f(0)$, then show that $|f'((x)|\leq 1$ for all $x\in [0,1]$.   
My effort 
Applying Lagrange's mean value theorem we have
$f(1)-f(0)=f'(c)$ for some $c\in (0,1)\implies f'(c)=0$.
 Next, I don't know how to show the required result. 
 A: We have $f'(c) = 0$ for some $c \in (0,1)$. By the MVT, 
$$\left|\frac{f'(x)}{x-c}\right| \leq 1$$
Since $0<|x-c| \leq 1$, the result follows. 
Note that this does not assume the integrability of $f''$. 
A: By Taylor expansion with Lagrange remainder term, for any $x \in (0, 1)$:
\begin{align}
& f(0) = f(x) - f'(x)x + \frac{1}{2}f''(\xi)x^2, \tag{1} \\
& f(1) = f(x) + f'(x)(1 - x) + \frac{1}{2}f''(\eta)(1 - x)^2, \tag{2}
\end{align}
where $\xi \in (0, x)$, $\eta \in (x, 1)$.
Subtract $(2)$ from $(1)$ and use the condition $f(0) = f(1)$ yields
$$0 = -f'(x) + \frac{1}{2}f''(\xi)x^2 - \frac{1}{2}f''(\eta)(1 - x)^2,$$
which implies that
\begin{align}
|f'(x)| = \left|\frac{1}{2}f''(\xi)x^2 - \frac{1}{2}f''(\eta)(1 - x)^2\right| \leq \frac{1}{2} + \frac{1}{2} = 1.
\end{align}
To treat the end points case, directly expand $f(0)$ at $1$ yields
$$f(0) = f(1) - f'(1)(-1) + \frac{1}{2}f''(\zeta),$$
where $\zeta \in (0, 1)$, hence
$$|f'(1)| = |f''(\zeta)/2| \leq 1/2 \leq 1.$$
$|f'(0)|$ can be bounded similarly.
A: With $c$ so chosen, $f'(x)$ can only get so far away from $f'(c)$.  What I mean is that
\begin{equation}
f'(c) + m(x-c) \leq f'(x) \leq f'(c) + M(x-c),
\end{equation}
where $m$ and $M$ are the smallest and largest possible values of $f''(x)$, respectively.  Explicitly:
\begin{equation}
m = \inf_{x\in[0,1]} f''(x) \qquad\text{and}\qquad M=\sup_{x\in[0,1]} f''(x).
\end{equation}
Given what you know about $f''(x)$ and about $f'(c)$, can you make the desired conclusion about $f'(x)$?
A: This conjecture is wrong. Just take the counterexample
$$f:x\mapsto y=3\cdot\left(x-\frac12\right)^2.$$
Obviously holds $f(0)=\tfrac34=f(1)$, but with $f^\prime(x)=6\cdot\left(x-\tfrac12\right)$
you have $f^\prime(\tfrac34)=\tfrac32>1$.
