I think this problem is kind of a famous problem...
A function $f$ is real valued function from real line.
Suppose that both of the absolute value of $f$ and absolute value of the second derivative of $f$ are bounded by 1.
Then, show that the absolute value of the first derivative of $f$ is bounded by 2.
Are there anyone who can solve this?
Firstly refer to Inequality involving the sup of a function and its first and second derivatives for a description of the theorem I am about to use.
Using the notation of the previous question if $M_0 = \sup_{x \in \mathbb{R}}|f(x)|$ and $M_2 = \sup_{x \in \mathbb{R}} |f''(x)|$ then by the least upper bound property we have $$ M_0 \leq 1, \ M_2 \leq 1$$ this is because both $|f(x)|$ and $|f''(x)|$ are bounded above by 1. Furthermore if we denote $M_1 = \sup_{x \in \mathbb{R}} |f'(x)|$ then by virtue of being an upper bound for $|f'(x)| : x \in \mathbb{R}$ we have that $$ \forall x \in \mathbb{R} : |f'(x)| \leq M_1$$ Now combining the two we have $$ \forall x \in \mathbb{R} : M_1^2 \leq 4 M_0 M_2 = 4 \implies M_1 \leq 2 \implies |f'(x)| \leq 2$$ As desired.
