Let $F$ defined on $[0,1]$ be twice differentiable such that $\mid f^{\prime \prime}(x) \leq 1$ for all $x \in[0,1] .$ If $f(0)=f(1)$, then show that $\left|f^{\prime}(x)\right|<1$ for all $x \in[0,1]$ ?
As $f(0)=f(1)$ and $f$ is differentiable, hence by Rolles theorem:
$f^{\prime}(c)=0$ for some $c \in[0,1]$
let us now apply LMVT (as function is twice differentiable) for point $c$ and $x \in[0,1]$,
hence,
$\frac{\left|f^{\prime}(x)-f(c)\right|}{x-c}=f^{\prime \prime}(d)$
$\Rightarrow \frac{|f(x)-0|}{x-c}=f^{\prime \prime}(d)$
$\Rightarrow \frac{|f(x)|}{x-c}=f^{\prime \prime}(d)$
A given that $\left|f^{\prime \prime}(d)\right|<=1$ for $x \in[0,1]$
$\Rightarrow \frac{|f(x)|}{x-c} \leq 1$
$\Rightarrow|f(x)| \leq x-c$
Now both $x$ and $c$ lie in $[0,1]$, hence $x-c \in[0,1]$