TITLE: real analysis derivatives and continuity
QUESTION [5 upvotes]: Suppose $g \colon[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and integrable with $g(a)=g(b)=0$ and that $g'' \colon (a,b)\rightarrow \mathbb{R}$ is continuous on $(a,b)$ and bounded by $K$. Prove that $|g(x)| \leq K|x-a||x-b|$ for all  $x \in [a,b]$. I tried using the definition of derivatives and I'm leaning toward the Mean Value Theorem. Help please.

REPLY [2 votes]: Using Taylor’s formula, we can get\begin{equation}g(a)=g(x)+g'(x)(a-x)+\frac{g''(x_1)(a-x)^2}{2}\end{equation}and\[g(b)=g(x)+g'(x)(b-x)+\frac{g''(x_2)(b-x)^2}{2}.\]Note that $g(a)=g(b)=0$, we have\[-(b-a)g(x)=\frac{g''(x_1)(a-x)^2(b-x)}{2}-\frac{g''(x_2)(b-x)^2(a-x)}{2},\]hence\begin{align}|(b-a)g(x)|=&\left|\frac{(a-x)(b-x)}{2}\right|(|g''(x_2)(a-x)|+|g''(x_1)(b-x)|)\\\leqslant&\left|\frac{(a-x)(b-x)}{2}\right|(K(x-a)+K(b-x))=K(b-a)\left|\frac{(a-x)(b-x)}{2}\right|\end{align}where $K=\sup|g''(x)|$.
Therefore\[|g(x)|\leqslant \left|\frac{K(a-x)(b-x)}{2}\right|.\]