If $f^{''}$ is bounded then $f$ is uniformly continuous.

Question

Let $f:(0,1) \to \mathbb{R}$ be twice differentiable function such that $|f^{''}(x)| \leq M$ for some non negative real $M$ for all $x \in (0,1)$. Then how do I show that $f$ is uniformly continuous on $(0,1).$

If $M =0$ then it is easy to show. But I need some help. Thanks.

Answer 1

If you write the Taylor polynomial around $1/2$, then $$ f(x)=f(1/2)+f'(1/2)\,(x-1/2)+\frac{f''(c)}2\,(x-1/2)^2. $$ Then, with $c_1$ and $c_2$ depending on $y$ and $x$ respectively, \begin{align} |f(y)-f(x)|&=\left|f'(1/2) (y-x)+\frac{f''(c_1)-f''(c_2)}2\,[(y-1/2)^2-(x-1/2)^2 \right| \\ \ \\ &=\left|f'(1/2) (y-x)+\frac{f''(c_1)-f''(c_2)}2\,[(y-x)(x+y+1) \right| \\ \ \\ &\leq f'(1/2)\,|y-x|+3M\,|y-x|\\ \ \\ &=(f'(1/2)+3M)\,|y-x|. \end{align} In particular, $f$ is uniformly continuous.

Answer 2

By the MVT, for all $x\in (0,1)$ we have

$$|f'(x)|\le |f'(x)-f'(1/2)| + |f'(1/2)| = |f''(c)(x-1/2)| + |f'(1/2)|$$ $$ \le M|x-1/2| + |f'(1/2)| < M + |f'(1/2)|.$$

Another application of the MVT then shows

$$|f(y)-f(x)| = |f'(c)(y-x)| \le (M + |f'(1/2)|)|y-x|$$

for all $x,y\in (0,1).$ Thus $f$ is Lipschitz on $(0,1),$ which is more than enough for uniform continuity.